Equilibrium Part 2

IONIC EQUILIBRIUM (ACIDS & BASES)

Like chemical equilibrium that exists between two or more chemical species of a reversible chemical reaction, an ionic equilibrium exists between an electrolyte and its ions:

Some useful terms regarding ionic equilibrium:

1) Electrolyte: The substances which produce ions when dissolved in suitable solvents or when in molten state; so called, as they disintegrate when electricity is passed through their solution. [electro = electric current, lysis = disintegration] electrolytes can be divided into

a) Strong electrolytes: The substances which dissociate almost completely into ions in liquid state (molten or solution) are strong.  Since the charged particles are responsible for the flow of electric current, the solutions of strong electrolytes are good conductor of electricity.  Degree of dissociation or ionization of these substances is high i.e., $\alpha \approx 1$

e.g., HCl, NaOH, NaCl, CH₃COONH₄ etc.

$\begin{array}{l}\mathrm{HCl}\rightleftharpoons H^{+}+{\mathrm{Cl}}^{-}\\ \mathrm{NaOH}\rightleftharpoons {\mathrm{Na}}^{+}+{\mathrm{OH}}^{-}\\ \mathrm{NaCl}\rightleftharpoons {\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}\end{array}$

b) Weak electrolytes : The substances which dissociate upto small extent in their liquid state are weak electrolytes. Since there are very few ion which can carry current, these are poor conductor of electricity. We say that their degree dissociation is less i.e.,$\alpha <<1.$

e.g., CH₃COOH, NH₄OH etc.

$\begin{array}{l}{\mathrm{CH}}_3\mathrm{COOH}\rightleftharpoons H^{+}+{\mathrm{CH}}_3{\mathrm{COO}}^{-}\\ {\mathrm{NH}}_4\mathrm{OH}\rightleftharpoons {{\mathrm{NH}}_4}^{+}+{\mathrm{OH}}^{-}\end{array}$

Since strong electrolytes are almost completely ionized, they do not form an equilibrium between unionized and ionized forms, but weak electrolytes do form an equilibrium between its ions and unionized form called ionic equilibrium.

Alternatively electrolytes can also be categorized as:

a) True electrolytes: In true electrolytes positively charged cation and negatively charged anion exists in pure state, e.g., crystals of CuSO₄, MgS, NaCl etc. When ionic crystal is dissolved in a solvent, ions break off from the crystal and get surrounded by a sheath of solvent molecules called solvated ions break and the process, called salvation. In cases when water is a solvent, hydrated ions are produced and the process is called hydration.

All the hydrated species are represented by the term (aq) following the name of the chemical species.

$\underset{\left(\mathrm{undissolved}\right)}{{\mathrm{NaCl}}_{\left(\mathrm{s}\right)}}\stackrel{\mathrm{aq}}{\rightleftharpoons }\underset{\left(\mathrm{dissolved}\right)}{{\mathrm{NaCl}}_{\left(\mathrm{aq}\right)}}\rightleftharpoons \underset{\mathrm{solvated} \mathrm{ions}}{\underbrace{\mathrm{N}{{\mathrm{a}}^{+}}_{\left(\mathrm{aq}\right)}+\mathrm{C}{{\mathrm{l}}^{–}}_{\left(\mathrm{aq}\right)}}}$

True electrolytes are good conductor of electricity in pure liquid state (i.e. when no solvent is used).

b) Potential electrolytes: In pure state these are uncharged molecules but when dissolved in a solvent, they react upto some extent with solvent to furnish ions. e.g., CH₃CO₂H, HCl etc.

$\begin{array}{l}{\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{CH}}_3{\mathrm{COO}}^{-}\\ {\mathrm{HCl}}_{\left(g\right)}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{Cl}}^{-}\end{array}$

These are poor conductor of electricity in pure liquid state.

2) Non-electrolytes: A chemical species whose melt or solution doesn’t furnish any ion, is called non-electrolyte.  e.g., urea, sugar etc.

3) Polyelectrolytes: The electrolytes which are polymeric in nature are called polyelectrolytes. e.g., RNAs and DNAs, which have ionizable acidic (PO₄^3–) groups and proteins which have ionizable acidic monomers

Depending on their chemical nature, electrolytes can also be divided into acids, bases and salts, which requires an account of the theories/definitions put forth by variations workers to explain the acids and bases.

Various concepts of acids and basis:

1. Classical concept: This concept is based on the experiences of various early workers & there is no any explanation of their such properties:

Acids	Bases
Don’t have soapy touch Have sour taste Turn blue litmus red Conduct electricity Lose their acidic nature when react with bases	Have soapy touch Have bitter taste Turn red litmus blue Conduct electricity Lose their basic nature when react with acids

Acidus (Lat.) = Sour

The above operational definitions can be explained on the basis of conceptual definitions of acids and bases given by various chemists like Arrhenius, Bronsted-Lowry and Lewis.

2. Arrhenius concept of Acids & Bases:

A substance that dissociates to give H⁺ in aqueous solution is called an acid, like HCl, CH₃CO₂H etc.

$$\begin{gathered} \mathrm{HCl}\stackrel{\mathrm{aq}}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{Cl}}^{-} \\ {\mathrm{CH}}_3\mathrm{COOH}\stackrel{\mathrm{aq}}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{CH}}_3{\mathrm{COO}}^{-} \end{gathered}$$

In general, $\underset{\text{ acid }}{\mathrm{HA}}\stackrel{aq}{\rightleftharpoons }\begin{array}{c}{\mathrm{H}}^{+}\\ \left[{\mathrm{H}}^{+}\text{ion or proton }\right]\end{array}+\underset{\text{ anion }}{{\mathrm{A}}^{-}}$

And, bases are the substances that furnish OH^– in its aqueous solution e.g., NaOH, NH₄OH etc

$$\begin{gathered} \mathrm{NaOH}\stackrel{\mathrm{aq}}{\rightleftharpoons }{\mathrm{Na}}^{+}+{\mathrm{OH}}^{-} \\ {\mathrm{NH}}_4\mathrm{OH}\stackrel{\mathrm{aq}}{\rightleftharpoons }{\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{–} \end{gathered}$$

In general, 

$\underset{\text{ base }}{\mathrm{BOH}}\to \underset{\text{ cation }}{{\mathrm{B}}^{+}}+\underset{\left[\text{ hydroxyl ion }\right]}{{\mathrm{OH}}^{-}}$

Existence of H⁺ and OH^–:

High dielectric constant of water ($\epsilon =86.6$) lowers the attractive forces of between positive and negative ions, which results into dissociation of the electrolyte.

Because of salvation of H⁺, It exists as H₃O⁺ (hydronium ion); some other species present in protonated aq. solutions are ${\mathrm{H}}_5{\mathrm{O}}_2^{+}, {\mathrm{H}}_7{\mathrm{O}}_3^{+} \text{and }{\mathrm{H}}_9{\mathrm{O}}_4^{+ }\text{etc }$

Similarly, the hydration of OH^– results into the formation of ${\mathrm{H}}_3{\mathrm{O}}_2^{-},{\mathrm{H}}_5{\mathrm{O}}_3^{-} \text{and }{\mathrm{H}}_7{\mathrm{O}}_4^{-}$ etc.

Such that we can write:

$\begin{array}{l}\mathrm{HCl}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{Cl}}^{-}\\ {\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{CH}}_3{\mathrm{COO}}^{-}\end{array}$

Actually a H⁺ is surrounded by four H₂O molecules tetrahedrally.

Oxides of many non-metals dissolve in H₂O to give acids, these are called acidic oxides or acid anhydrides.

$\begin{array}{l}{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{H}}_2{\mathrm{CO}}_3\rightleftharpoons 2{\mathrm{H}}^{+}+{\mathrm{CO}}_3^{2-}\\ {\mathrm{N}}_2{\mathrm{O}}_5+{\mathrm{H}}_2\mathrm{O}⟶2{\mathrm{HNO}}_3\rightleftharpoons 2{\mathrm{H}}^{+}+2{\mathrm{NO}}_3^{-}\end{array}$

Similarly, oxides of many metals dissolve in H₂O to form bases are called basic oxides.

$\begin{array}{l}{\mathrm{Na}}_2\mathrm{O}+{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{NaOH}\rightleftharpoons 2{\mathrm{Na}}^{+}+2{\mathrm{OH}}^{-}\\ \mathrm{CaO}+{\mathrm{H}}_2\mathrm{O}\to \mathrm{Ca}(\mathrm{OH}{)}_2\rightleftharpoons {\mathrm{Ca}}^{2+}+2{\mathrm{OH}}^{-}\end{array}$

NH₃ and N₂H₄ are also base as they produce OH^– when in aq. solution.

${\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{NH}}_4\mathrm{OH}\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{-}$

Strength of acid/base:  More the number of H⁺/OH^– (ions) furnished by an acid or a base, greater is the strength of that acid/base (electrolyte).

Arrhenius described neutralization as the process in which H⁺ ion & OH^– ions combine to form unionized molecule of water.

e.g.,v${\mathrm{Na}}^{+}+{\mathrm{OH}}^{-}+{\mathrm{H}}^{+}+{\mathrm{Cl}}^{-}⟶{\mathrm{Na}}^{+}+{\mathrm{Cl}}^{-}+{\mathrm{H}}_2\mathrm{O}$

As the result of this process, the characteristic properties of acids and bases are destroyed.

Limitations of Arrhenius Theory:

1) Presence of water as a solvent is a necessary condition to display acidic & basic property according to the theory. So, HCl would not be an acid neither it will display acidic properties in any other solvents like acetone.

2) A supposition that H₂O reacts first with a substance and produces H⁺ or OH^– thus behaves like an acid (in case of oxide of non-metals) or base (in case of NH₃ & metallic oxides).

3) Acidic nature of AlCl₃, BF₃, CO₂ couldn’t be explained.

4) Acidic & basic characters of substances in non-aqueous solvents can’t be explained.

5) Neutralization process of acids & bases that don’t occur in aq. or any other solution, can’t be explained.

3. Bronsted–Lowry concept: (The proton – Donor – Acceptor concept)

In 1923, Bronsted & Lowry independently proposed a theory depending on the behaviour of a substance towards the proton. So that,

An acid is the chemical species that can donate a proton (H⁺).

$\underset{{\mathrm{H}}^{+} \mathrm{donor} \mathrm{Acid}}{\mathrm{HCl}}+\underset{{\mathrm{H}}^{+}\mathrm{acceptor} \mathrm{Base}}{{\mathrm{H}}_2\mathrm{O}}\to {\mathrm{H}}_3^{+}\mathrm{O} + {\mathrm{Cl}}^{-}$

And, the base is the chemical entity that can accept a proton, e.g.,

$\underset{{\mathrm{H}}^{+} \mathrm{acceptor} \mathrm{Base}}{{\mathrm{NH}}_3}+ \underset{{\mathrm{H}}^{+} \mathrm{donor} \mathrm{Acid}}{{\mathrm{H}}_2\mathrm{O}}\to {\mathrm{NH}}_4^{+} + {\mathrm{OH}}^{-}$

When an acid loses a proton it has tendency to regain its H⁺ and behave like a base called conjugate base. The same is true about a base which after accepting a proton turns into an acid called conjugate acid.

Conjugate acids and bases:

A pair of Bronsted acid and a bse that differs by one proton is known as conjugate acid – base pair.  A strong acid has a weak conjugate base and a strong base has a weak conjugate acid.

$\text{ Acid ⇌}{\mathrm{H}}^{+}+\text{Base }$

Conjugate acids and bases are also called Bronsted-Lowry (or generally Bronsted) acids and bases.

The above reaction can be understood by breaking it into 2 parts, called conjugate pairs.

$\underset{\mathrm{Acid}}{{\mathrm{HNO}}_3} \underset{+{\mathrm{H}}^{+}}{\overset{–{\mathrm{H}}^{+}}{\rightleftharpoons }}\underset{\mathrm{Conj}.\mathrm{base}}{{\mathrm{NO}}_3^{–}}$

$\underset{\mathrm{Base}}{{\mathrm{H}}_2\mathrm{O}}\underset{{\mathrm{H}}^{+}}{\overset{+{\mathrm{H}}^{+}}{\rightleftharpoons }}\underset{\mathrm{conj}.\mathrm{acid}}{{\mathrm{H}}_3{\mathrm{O}}^{+}}$

So we can write acid₁ changes into base₁ and base₂ changes into acid₂ after transfer of H⁺:

Examples:

$\mathrm{HCN}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{CN}}^{–}$

${\mathrm{H}}_2\mathrm{O}+{\mathrm{NH}}_3\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{–}$

Evidently, according to this theory a substance can behave like an acid only when another substance, capable of accepting the H⁺ is present.  For example, HNO₃ or acetic acid can donate their H⁺ to H₂O molecules, so they are acidic in aqueous solutions.  But if they are present in benzene solution, they won’t work as acid since benzene molecules do not accept the proton.

$\underset{\text{ (Base) }}{{\mathrm{CH}}_3\mathrm{COOH}+}\underset{\text{ (Acid) }}{\mathrm{HF}}\rightleftharpoons {\mathrm{CH}}_3{\mathrm{COOH}}_2^{+}+{\mathrm{F}}^{–}$

Acetic acid is weak acid in presence of water but in presence of stronger acid HF it has to accept H⁺ from it & thereby behaves like a base.

$\underset{\left(\mathrm{Base}\right)}{\mathrm{HNO}}+\underset{\text{(Acid })}{{\mathrm{H}}_2{\mathrm{SO}}_4}\rightleftharpoons {\mathrm{H}}_2{\mathrm{NO}}_3^{+}+{\mathrm{HSO}}_4^{–}$

${\mathrm{H}}_2{\mathrm{NO}}_3⟶_{+}{\mathrm{H}}_2\mathrm{O}+{\mathrm{NO}}_2^{+}$

More interesting is the example of the strong acid, HNO₃ which in (absence of water and) presence of H₂SO₄ can’t donate its proton to any other species, rather it being weaker than H₂SO₄, accepts H⁺ from the latter and becomes a base.  [Refer to the nitration of benzene, where NO₂⁺ (nitronium ions – the electrophiles) are generated as above and therein the main role of anhydrous H₂SO₄ is to produce NO₂⁺ along with absorption of H₂O formed during the reaction]

$\underset{\left(\text{acid }\right)}{{\mathrm{HCO}}^{–}}+\underset{\text{ (base })}{{\mathrm{H}}_2\mathrm{O}}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{CO}}_3^{2–}$

$\underset{\text{ (acid) }}{\mathrm{HCl}}+\underset{\left(\mathrm{base}\right)}{{\mathrm{HCO}}_3^{–}}\rightleftharpoons {\mathrm{H}}_2{\mathrm{CO}}_3+{\mathrm{Cl}}^{–}$

Any chemical species can be an acid as well as a base depending upon its ability to accept and/or lose the proton.

$\underset{\left(\mathrm{Acid}\right)}{{\mathrm{H}}_2\mathrm{O}}+\underset{\left(\mathrm{Base}\right)}{{\mathrm{H}}_2\mathrm{O}}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{OH}}^{–}$

$\underset{\left(\mathrm{acid}\right)}{{\mathrm{NH}}_3}+\underset{\left(\mathrm{Base}\right)}{{\mathrm{NH}}_3} \rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{NH}}_2^{–}$

Above reactions called autoionization explain that within a solvent some molecules of same substance can behave like an acid & the other as a base.

${\mathrm{HCO}}_3^{–}, {\mathrm{H}}_2\mathrm{O}, {\mathrm{NH}}_3$ etc are amphiprotic or amphoteric due to their ability to accept or donate H⁺.

Basicity or Protocity of a Bronsted Acid:

The number of H⁺ furnished (ionizable protons) by a molecule of an acid is known as basicity of acid. e.g.,

Acidity or Hydroxicity of Bronsted Based

The number of OH^– furnished by one molecule of a base is its acidity.e.g.,

Base	NaOH	Ca(OH)2	Al(OH)3
Acidity	1	2	3

Since H⁺ are transferred from an acid to a base, there is no free existence of H⁺.  Either they become part of the conjugate acid else combine with H₂O molecule to give H₃O⁺ (called hydronium ion due to similarly with ammonium ion) which itself is a conjugate acid.

Relative Strength of Bronsted Acid-Base Pairs:

More the tendency of an acid to lose a H⁺ more easily it will dissociate, i.e., greater the ionization or dissociation stronger the acid.

HCl can give its proton more easily to water in comparison to CH₃COOH, so HCl is stronger than CH₃COOH.

$\mathrm{HCl}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3^{+}\mathrm{O}+{\mathrm{Cl}}^{–}$ ionization or dissociation constant ${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}_3^{+}\mathrm{O}\right]\left[{\mathrm{Cl}}^{–}\right]}{\left[\mathrm{HCl}\right]};{\mathrm{K}}_{\mathrm{a}}>>1$

HCl is a stronger acid while $H_3{O}^{+}$ is weaker and H₂O is a stronger base compared to Cl^–.  This shows that Cl has little tendency to take up protons.

As already seen it can be inferred that always a stronger acid has a weaker conjugate base and a stronger base has weaker conjugate acid.

Therefore greater the K_a, more the concentration of H⁺, more the tendency to give up H⁺, stronger will be the acid.

${\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{CH}}_3{\mathrm{COO}}^{–}$

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \left[{\mathrm{CH}}_3{\mathrm{COO}}^{–}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]} {\mathrm{K}}_{\mathrm{a}}\ll 1$

Morever, stronger the acid, the weaker will be its conjugate base and vice versa.  e.g.,

$\underset{\mathrm{strong} \mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4} + \underset{\mathrm{weak} \mathrm{base}}{{\mathrm{H}}_2\mathrm{O}} \rightleftharpoons \underset{\underset{\mathrm{acid} \left(\mathrm{strong}\right)}{\mathrm{conjugate}}}{{\mathrm{H}}_3{\mathrm{O}}^{+}} + \underset{\underset{\left(\mathrm{weak}\right)}{\mathrm{conjugate} \mathrm{base}}}{{\mathrm{HSO}}_4^{–}}$

Solvent effect or Leveling effect:

Nature of solvent play a key role in the dissociation and hence in the relative strengths.  HCl, HNO₃, H₂SO₄, HBr & HClO₄ all are strong acids and can donate protons to the solvents.

If their comparative strength is to be measured, water as a solvent can’t serve the purpose, because it is stronger enough base so as to accept or distract protons from the above acids in almost same intensity.  So, if their solutions of equal normalities are taken, each of them will have equal number of H⁺.  That’s why they are almost equally strong in presence of water as a solvent.  This effect of water to treat all the above acids at same level of strength is called leveling or solvent effect.

If CH₃COOH is taken as a solvent instead of water, the former being an acid will not easily accept proton from an acidic species.  It can accept easily only from them which are stronger enough, and this ease of acceptance of H⁺ will be in the proportion of their strength.

Such solvent which differentiate the acidic/basic nature of the acids/bases having approximately equal strengths are called differentiating solvents.

Using conductivity methods the strength of above mentioned acids in acetic acid solution can be studied, which give the order of strength, as:

HClO₄ > HBr > H₂SO₄ > HCl > HNO₃

Ammonia, stronger base than water can easily accept protons from strong as well as weak acids.  So that even weak acid like CH₃COOH behaves a strong acid approximately equal in strength to the HCl, HNO₃ H₂SO₄ & HClO₄.

Classification of solvents:

Protophilic	Protogenic	Amphiprotic	Aprotic
· Accept protons	· Produce protons	· Can accept as well as produce H+	· Do not involve H+
· Water, liq. ammonia, alcohol	· Water, liq. HF/HCl, glacial CH3COOH	· Water, ammonia, ethanol	· Benzene, CCl4, N2O4, liq. SO2

Advantage of Bronsted – Lowry concept:

i) This concept explains the behaviour of acids and bases in both aqueous and non-aqeuous solvents.

ii) It explains the behaviour of NH₃.CaO etc, as bases and CO₂.SO₂ etc as acids.

iii)  This is a more general theory encompassing larger number of acids and bases than Arrhenius theory.

Draw backs of Bronsted – Lowry concept:

i) Proton donation or acceptance happens only in the presence of other substances.

ii) It does not explain the behaviour of electron deficient molecule like AlCl₃, BCl₃ etc are acids.

4. Lewis Theory/Concept:

Limitation of Bronsted theory to transfer of only H⁺  (proton) was extended by Lewis incorporating the concept of transfer (or more accurately donation forming a dative bond) of lone pair.   Since proton & electron are of opposite nature; the proton donor was acid in Bronsted theory, the electron (pair) donor will be a base in Lewis theory.  Similarly the proton acceptor in Bronsted was base; the electron pair acceptor will be base in Lewis Concept.  So that,

i) “A substance that can accept an electron pair to form a co-ordinate covalent bond with the donor”. Is called Lewis acid. g.,

ii) “A substance that can donate a lone pair of electrons to form a coordinate covalent bond with the acceptor” is called Lewis base. g., , ligands in complex compounds.

An acid base neutralization reaction is simply a co-ordination reaction involving the sharing of electron pair between acid & base.  The product so formed is a co-ordination compound, co-ordination complex or adduct.

 Examples of Lewis acids: 

i) All cations : e.g., ${\mathrm{Ag}}^{+},{\mathrm{Co}}^{3+},{\mathrm{Cu}}^{3+},{\mathrm{Fe}}^{3+}$ etc

ii) Compounds whose central atom has an incomplete octet and possessing an empty orbital. e.g., ${\mathrm{BF}}_3,{\mathrm{BCl}}_3,{\mathrm{AlCl}}_3,{\mathrm{FeCl}}_3$ etc.

iii) Compounds in which the central atom has available d-orbitals and may expand its octet. e.g., ${\mathrm{SiF}}_4,{\mathrm{SnCl}}_4,{\mathrm{SF}}_4,{\mathrm{TeF}}_4,{\mathrm{FeCl}}_3$ etc.

iv) Molecules having multiple bonds between atoms of dissimilar eelectronegativities. e.g. ${\mathrm{CO}}_2,{\mathrm{SO}}_2,{\mathrm{SO}}_3,{\mathrm{NO}}_2,{\mathrm{Cl}}_2{\mathrm{O}}_7,{\mathrm{P}}_4{\mathrm{O}}_{10}$ etc.

v) Elements with an electron sextet. g., :  S, O

Examples of Lewis Bases:

a) Neutral molecules with lone pairs: Compounds with N, P, O & S as a central atom behaves generally as Lewis base if they have at least one of their electron pair non-bonded. e.g., $\ddot{\mathrm{N}}{\mathrm{H}}_3, {\mathrm{PH}}_3$, amines ($\mathrm{R}–\ddot{\mathrm{N}}{\mathrm{H}}_2$), ${\mathrm{H}}_2\ddot{\mathrm{O}}:$, alcohols ($\mathrm{R}–\ddot{\mathrm{O}}–\mathrm{H}$), ethers ($\mathrm{R}–\ddot{\mathrm{O}}–\mathrm{R}$), thiols ($\mathrm{R}–\ddot{\mathrm{S}}–\mathrm{H}$) etc.

Order of Basic Strength

${\mathrm{NI}}_3>{\mathrm{NBr}}_3>{\mathrm{NCl}}_3>{\mathrm{NF}}_3$

b) Simple anions: ${\mathrm{OH}}^{–},{\mathrm{Cl}}^{–},{\mathrm{CN}}^{–},{\mathrm{HS}}^{–}$ etc.,

Order of Basic Strength:

i) ${\mathrm{F}}^{–} {\mathrm{Cr}}^{–} {\mathrm{Br}}^{–} {\mathrm{I}}^{–}$

ii) ${\mathrm{CH}}_3^{–}>{\mathrm{NH}}_2^{–}>{\mathrm{OH}}^{–}>{\mathrm{F}}^{–}$    [discussed earlier]

c) Molecules with carbon-carbon multiple bond: In some compounds there is partial charge transfer from alkene, alkyne or even benzene molecules to a molecule having vacant p or (in most cases) d-orbitals.

${\mathrm{C}}_2{\mathrm{H}}_4$ molecule in ${\left[\mathrm{Ag}\left({\mathrm{C}}_2{\mathrm{H}}_4\right)\right]}^{+}$   $\left[\begin{array}{c}{\mathrm{CH}}_2\\ \left|\right|\\ {\mathrm{CH}}_2\end{array}.............> {\mathrm{Ag}}^{+}\right]$ Charge transfer complex

d) Ligend species in complexes: All ligands [in co-ordination chemistry] donate electron pairs.  So, they are Lewis bases. e.g., CO in [Ni(CO)₄]; CN^– in K₄[Fe(CN)₆] etc.

Example of Acid Base (neutralization) reactions:

(i) $\underset{\underset{\left(\mathrm{Lewis} \mathrm{base}\right)}{\mathrm{Donor}}}{\ddot{\mathrm{N}}{\mathrm{H}}_3}+\underset{\underset{\left(\mathrm{Lewis} \mathrm{acid}\right)}{\mathrm{Acceptor}}}{{\mathrm{BF}}_3} \to \underset{\underset{\mathrm{Ammonia} – \mathrm{borontrifluoride}}{\mathrm{Adduct}}}{\left[{\mathrm{NH}}_3\to {\mathrm{BF}}_3\right]}$

(ii) $\underset{\left(\mathrm{Base}\right)}{{\mathrm{H}}_2\mathrm{O}}+\underset{\left(\mathrm{Acid}\right)}{{\mathrm{H}}^{+}} \to \underset{\mathrm{Hydronium} \mathrm{ion}}{{\left[{\mathrm{H}}_2\mathrm{O}\to \mathrm{H}\right]}^{+}}$

(iii) $\underset{\underset{\mathrm{Ligand}}{\left(\mathrm{Base}\right)}}{2{\mathrm{NH}}_3}+\underset{\underset{\mathrm{Acceptor} \mathrm{mental}}{\left[\mathrm{Acid}\right]}}{{\mathrm{Ag}}^{+}} \to \underset{\mathrm{Diammine} \mathrm{silver}\left(\mathrm{I}\right) \mathrm{ion}}{{\left[{\mathrm{NH}}_3\to \mathrm{Ag}\leftarrow {\mathrm{NH}}_3\right]}^{+}}$

Note :

All Bronsted bases are Lewis bases, but all Bronsted acid are not Lewis acids.

Acids				Bases
	Bronsted	Lewis			Bronsted	Lewis
HCl	✓	✗		NH3	✓	✓
CH3COOH	✓	✗		H2O	✓	✓
H2CO3	✓	✗		Cl–	✓	✓

 Limitations of Lewis Theory:

i) One of the serious defects in the theory is that it cannot explain the strengths of acids and bases.

ii) Acids like HCl : H₂SO₄ react with bases such as NaOH or KOH but do not form coordinate covalent bond.

iii) Generally acid-base interactions i.e., neutralization reactions are instantaneous reactions but some Lewis acid – base reactions go on very slowly.

iv) All the acid – base reactions do not involve coordinate bond formation.

v) H⁺ ion, as catalyst, cannot be explained by this theory.

Illustration 1:

Using Lewis concept, determine the order & acidic strength of HClO₄, HClO₃ and HClO₂.

Solution.

In general, the greater the number of oxygen atom around a central atom in a oxyacid, greater the acidity.  Since O is more electronegative than Cl, the former tends to pull away the electron from Cl atom, which in turn pulls electron pair from O-H bond.  This leads into easier removal of H⁺ from the acid making it more acidic.

MORE THEORIES FOR ACID–BASE CONCEPT:

5) Lux Flood Theory (concept):  The theory has been proposed to explain the high temperature acid-base reactions:

Molecules or compounds which can accept an anion (like Cl^–, O^{2 –}, S^2– etc. are called acids.  e.g. CO₂, SiO₂, NO₂ etc.  Molecules/compounds that donate their anions during a reaction are bases.  e.g. Na₂O, CaO, BaO, FeO etc.

Acid–Base Neutralization reaction :

$\underset{\mathrm{Base}}{\mathrm{BaO}}+\underset{\mathrm{Acid}}{{\mathrm{CO}}_2} \stackrel{∆}{\to } \underset{\mathrm{Salt}}{{\mathrm{BaCO}}_3} \left[{\mathrm{Ba}}^{2+}{\mathrm{CO}}_3^{2–}\right]$

$\underset{\mathrm{Base}}{\mathrm{FeO}} + \underset{\mathrm{Acid}}{{\mathrm{SiO}}_2} \stackrel{∆}{\to } \underset{\mathrm{Salt}}{{\mathrm{FeSiO}}_3}$

Such reactions generally take place at high temperature in absence of any solvent.  These reactions are the part of ceramic industry and metallurgy of elements (formation of slag).

Relative Strengths of Some Acids and Bases

Acids:

i) ${\mathrm{HClO}}_4>{\mathrm{HClO}}_3>{\mathrm{HClO}}_2>\mathrm{HClO}$

ii) $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}>\mathrm{HF}$

iii) ${\mathrm{HClO}}_3>{\mathrm{HBrO}}_3>{\mathrm{HIO}}_3$

iv) ${\mathrm{HClO}}_4>\mathrm{HBr}>{\mathrm{H}}_2{\mathrm{SO}}_4>\mathrm{HCl}>{\mathrm{HNO}}_3>{\mathrm{H}}_3{\mathrm{O}}^{+}$   or   ${\mathrm{H}}^{+}>{\mathrm{H}}_2{\mathrm{SO}}_3>{\mathrm{HNO}}_2>{\mathrm{H}}_2{\mathrm{CO}}_3>{\mathrm{CH}}_3\mathrm{COOH}>\mathrm{HCN}$

v) $\mathrm{HCOOH}>{\mathrm{CH}}_3\mathrm{COOH}>{\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{COOH}$

vi) ${\mathrm{CH}}_3\mathrm{COOH}>{\mathrm{CHCl}}_2\mathrm{COOH}>{\mathrm{CH}}_2\mathrm{ClCOOH}>{\mathrm{CH}}_3\mathrm{COOH}$

Bases:

i) $\mathrm{KOH}>\mathrm{NaOH}>\mathrm{LiOH}$

ii) $\mathrm{Sr}{\left(\mathrm{OH}\right)}_2>\mathrm{Ba}{\left(\mathrm{OH}\right)}_2>\mathrm{Ca}{\left(\mathrm{OH}\right)}_2>\mathrm{Be}{\left(\mathrm{OH}\right)}_2$

iii) $\mathrm{NaOH}>\mathrm{Ca}{\left(\mathrm{OH}\right)}_2>{\mathrm{NH}}_4\mathrm{OH}$

iv) $\mathrm{NaOH}>\mathrm{Mg}{\left(\mathrm{OH}\right)}_2>\mathrm{Al}{\left(\mathrm{OH}\right)}_3$

v) $\mathrm{NaOH}>{\mathrm{NH}}_3>{\mathrm{H}}_2\mathrm{O}$

vi) ${\mathrm{NH}}_3>{\mathrm{NH}}_2.{\mathrm{NH}}_2>{\mathrm{NH}}_2\mathrm{OH}$

vii)  $\mathrm{R}–{\mathrm{NH}}_2>{\mathrm{NH}}_3>{\mathrm{ArNH}}_2$            R = alkyl; Ar = aryl

viii) ${\mathrm{NH}}_3>{\mathrm{C}}_5{\mathrm{H}}_3\mathrm{N}\left(\mathrm{pyridine}\right)>{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2$

ix) ${\left({\mathrm{CH}}_3\right)}_2\mathrm{NH}>{\mathrm{CH}}_3{\mathrm{NH}}_2>{\left({\mathrm{CH}}_3\right)}_3\mathrm{N}>{\mathrm{NH}}_3$     $\left[2°1°3°\mathrm{N}–\mathrm{Me}\right]$

x) $\left({\mathrm{C}}_2{\mathrm{H}}_5\right)\mathrm{NH}>{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2>{\mathrm{NH}}_3>{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_3\mathrm{N}$   $\left[2°1°\mathrm{N}3°–\mathrm{Et}\right]$

6) Usanovich (Solvent) concept:

Acid base reactions of aqueous & non-aqueous solvents are well explained by this theory.  Autoionization of a solvent gives two ions – a cation characteristic ion and the other anion characteristic ion.  If any substance after being dissolved in a particle solvent produces cation characteristic ions of the solvent, the substance will behave like an acid and if anion characteristic, it will be a base in the solvent.

In water (solvent)

$\underset{\mathrm{Solvent}}{{\mathrm{H}}_2\mathrm{O}}+{\mathrm{H}}_2\mathrm{O} \rightleftharpoons {\underset{\mathrm{Cation}}{{\mathrm{H}}_3\mathrm{O}}}^{+}+\underset{\mathrm{Anion}}{{\mathrm{OH}}^{–}}$

The substance producing  in water are acids and those producing OH^– are bases.

$\mathrm{HCl}+{\mathrm{H}}_2\mathrm{O} \rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{Cl}}^{–}$

Since HCl when present in H₂O, gives cation characteristic of solvent i.e., H₃O⁺, it will be an acid.

${\mathrm{NH}}_3+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{–}$

$\mathrm{NaOH}\rightleftharpoons {\mathrm{Na}}^{+}+{\mathrm{OH}}^{–}$

Ammonia and NaOH give anion characteristic of H₂O i.e., OH^–, so they are  bases in it.

Other solvents like liq. NH₃, liq. SO₂ also represent the same type of reactions.

${\mathrm{NH}}_3+{\mathrm{NH}}_3 \rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{NH}}_2^{–}$

In liq. NH₃ substance furnishing ${\mathrm{NH}}_4^{+}$ will behave like acid and those giving ${\mathrm{NH}}_2^{–}$ will work as a base e.g.,

$\underset{\underset{\left(\mathrm{with} {\mathrm{NH}}_4^{+}\right)}{\mathrm{Acid}}}{{\mathrm{NH}}_4\mathrm{Cl}}+\underset{\underset{\left(\mathrm{having} {\mathrm{NH}}_2^{–}\right)}{\mathrm{Base}}}{{\mathrm{NaNH}}_2}\to \underset{\underset{\left(\mathrm{having} \mathrm{neither} \mathrm{of} \mathrm{ion} \mathrm{char}.\mathrm{og} \mathrm{solvent}\right)}{\mathrm{Salt}}}{\mathrm{NaCl}}+\underset{\mathrm{solvent}}{2{\mathrm{NH}}_3}$

Liq. SO₂ [non-aqueous as well as non-protonic (aprotic) solvent]

${\mathrm{SO}}_2+{\mathrm{SO}}_2 \rightleftharpoons {\mathrm{SO}}^{2+}+{\mathrm{SO}}_3^{2–}$

In liq. SO₂, the compounds giving SO²⁺ will be acids and those giving $S{O}_3^{2–}$ will be bases e.g.,

$\underset{\underset{\left(\mathrm{Acid}\right)}{\mathrm{Thionyl} \mathrm{choride}}}{{\mathrm{SOCl}}_2}+\underset{\underset{\left(\mathrm{Base}\right)}{\mathrm{Sodium} \mathrm{sulphite}}}{{\mathrm{Na}}_2{\mathrm{SO}}_3}\rightleftharpoons \underset{\left(\mathrm{Salt}\right)}{2\mathrm{NaCl}}+\underset{\mathrm{Solvent}}{2{\mathrm{SO}}_2}$

Pearson Theory of Hard & Soft Acids & Bases:

Stability & ease of complex are salt formation between Lewis acids & Lewis bases was suggested by Pearson.

Hard Acids :  Acceptor atoms have smaller atoms, high nuclear charge, low polarizibility and noble gas configuration.  e.g., H⁺, Li⁺, Be²⁺, Al³⁺ Fe³⁺ etc.

Soft Acids :  Acceptor atoms have larger size, low charge density high polarizability and donot have noble gas configuration. e.g., Cu⁺, Ag⁺, Hg⁺, Br⁺, I₂ etc.

Hard Bases:  Donor atoms have high electronegativity, low polarisability and smaller size.  e.g. H₂O, Cl^–, F^–, OH^–, NH₃, NO₃^–, CO₃^2– etc.  

Soft Bases :  Donor atoms have low electronegativity and high polarizability e.g., R-SH, I^– CN^–, CO, H^– etc.

A complex formed between soft acid & soft base as well as between hard acid and hard base is more stable in comparison to hard-soft or soft-hard combination.

Modern Theory of Acid and Bases:  Its is nothing but integration of all previous theories.

Acids are the substances, which

i) furnish H⁺ (protons) in the solution

ii) can accept electron are electron pair

iii) accept anions (like O^2–, S^2–, etc.)

iv) give cation characteristic of the solvent, when dissolved in it

Bases are the substances which

i) furnish OH^– in the solution

ii) can accept protons

iii) donate done pairs

iv) give anions (like O^2–, S^2– etc)

v) produce anion characteristic of the solvent

Salient Features of Acids & Bases:

Acids :  i)  Mineral acids are strong acids (HCl, HNO₃, H₂SO₄ etc).

ii) In oxyacids, the ionizable (replaceable) H atoms are those which are attached directly to O atom e.g.,

An acidic salt has some replaceable H atoms e.g., NaHCO₃, NaH₂PO₄, NaH₂PO₃ etc.

Note : Na₂HPO₃ is not acidic since H is attached to P atom.

Bases :

Hydroxides and oxides of IA & IIA group (alkali & alkaline earth metals) are strong base.

Rest of the hydroxides & oxides are weak base.

Oxides of metals are basic (few of them being amphoteric) while those of non-metals are acidic (some are neutral).

Basic	Amphoteric	Acidic	Neutral
${\mathrm{K}}_2\mathrm{O}, \mathrm{CaO}, \mathrm{BaO}$	BeO, Al2O3	CO2	CO
CuO, Fe2O3	ZnO, SnO2	N2O3, NO2, N2O5 SO2, SO3, P2O5, SiO2	N2O, NO

Acid anhydrides:

If some acids are dehydrated they will leave behind an oxide, with same oxidation number of central atom; alternatively, if this oxide is dissolved in water it gives the corresponding acid.  This oxide is called acid anhydride.

This list follows:

Anhydride	CO2	N2O5	N2O3	SO3		SO2
Acid	H2CO3	HNO3	HNO2	H2SO4		H2SO3
Anhydride	P2O5	P2O3	CrO3		CrO3		Mn2O7
Acid	H3PO4	H3PO3	H2CrO4		H2CrO4		HMnO4

Basic anhydrides:

Oxide	Na2O	CaO	Al2O3
Base	NaOH	Ca(OH)2	Al(OH)3

Degree of Dissociation (α):  It is the fraction of the molecules of a weak electrolyte which dissociates into ions in its solution.

$\mathrm{\alpha }=\frac{\mathrm{No}.\mathrm{of} \mathrm{molecules} \mathrm{dissociated}}{\mathrm{Total} \mathrm{no}.\mathrm{of} \mathrm{molecules}}$

Factors governing α :

1) Nature of solute :  For strong electrolytes α $\approx$ 1

For weak electrolytes α <<1

2) Nature of solvents:  Solvents with high dielectric constant (like H₂O; $\epsilon$ = 86.6) make the strong electrolytes highly ionizable (i.e., α $\approx$ 1) in comparison to solvents with low dielectric constant (like CH₃OH)

3) Temperature: α increases with increase in temperature.

4) Dilution: α increases with dilution of solution.

5) Presence of other species :  If some ions present in solution are common to the electrolyte’s and the electrolyte is weaker, its α decreases [ common ion effect].  If such an ion is present in the solution that reacts with either of ions of electrolyte to give some product the α of electrolyte increases.

Ostwald Dilution Law (Ionisation of weak electrolytes):

Let an electrolyte AB dissociate into ions A⁺ & B^–.  If its degree of dissociation is a, for the condition of equilibrium,

                                           AB          $\rightleftharpoons$         ${\mathrm{A}}^{+}$    +       ${\mathrm{B}}^{–}$

Initial concentration       C                          O                O

Conc. at equilibrium       C(1-α)                Cα                 Cα

Equilibrium constant,        $\mathrm{K}=\frac{\left[{\mathrm{A}}^{+}\right]\left[{\mathrm{B}}^{–}\right]}{\left[\mathrm{AB}\right]}=\frac{\mathrm{C\alpha } \mathrm{C\alpha }}{\mathrm{C}\left(1–\mathrm{\alpha }\right)}$        . . . . (i)

$\mathrm{K}=\frac{{\mathrm{\alpha }}^2.\mathrm{C}}{\left(1–\mathrm{\alpha }\right)}$    . . . . (ii)

For weak electrolytes   α << 1

So,     (1-α) $\approx$ 1

$\mathrm{K}={\mathrm{\alpha }}^2.\mathrm{C}$

$\mathrm{\alpha }=\sqrt{\frac{\mathrm{K}}{\mathrm{C}}}$    . . . .  (iii)

If V is the volume of solution having 1 mole of electrolyte, then $\mathrm{V}=\frac{1}{\mathrm{C}}$

Thence,  $\mathrm{\alpha }=\sqrt{\mathrm{KV}}$      . . . .  (iv)

or  $\mathrm{\alpha }\propto \sqrt{\mathrm{V}}$

“Degree of dissociation (α) of weak electrolytes is directly proportional to the square root of volume or dilution”.

Equations (iii & iv) describe that for the given volume or concentration the degree of dissociation (a) is proportional to K i.e., if K is greater, a will also be higher.

If deg. of disso. (α) for a weak electrolyte is greater than 0.05, (1-α) can’t be treated equal to unity.  In that very case, eq. (ii) must be followed.  For calculation of α, the quadric equation should then be used

i.e.,   ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0, \mathrm{x}=\frac{–\mathrm{b}\pm \sqrt{{\mathrm{b}}^2–4\mathrm{ac}}}{2\mathrm{a}}$

Illustration 2 :

A monobasic acid has Ka = 1.8 ⨯ 10^–5 at RT.  Calculate the degree of dissociation and concentration of H⁺ furnished by the dissociation of acid having concentration 0.2 M.

Solution.

Assuming a to the very small

$\mathrm{\alpha }=\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{C}}}=\sqrt{\frac{1.8\times {10}^{–5}}{0.2}}=0.009486$

$\left[{\mathrm{H}}^{+}\right]=\sqrt{{\mathrm{K}}_{\mathrm{a}}.\mathrm{C}}=\sqrt{1.8\times {10}^{–5}\times 0.2}$       {or, [H⁺] = a ⨯ C}

= 0.001897

Common Ion effect:  When a weak electrolyte is present in a solution, already having any of its ion common to that of another strong electrolyte, the degree of dissociation of weak electrolyte is further decreased.  This is called common ion effect.  For example if CH₃COOH (weak) is present in a solution along with HCl (strong), then

Strong electrolyte   HCl             $\rightleftharpoons$    ƒ         H⁺     +       Cl^–                (greater dissociation)

Weak electrolyte    CH₃COOH     ƒ$\rightleftharpoons$          H⁺     +       CH₃COO^–       (less dissociation)

                                                                    Common ions

Dissociation of acetic acid will further decrease.

Explanation from Le Chatelier :  HCl being strong will almost completely ionize to give H⁺ and Cl^–.  An equilibrium between undissociated HCl and dissociated (H⁺ & Cl^–) will be established.  Now, if we add CH₃COOH, a weaker acid to the solution it will produce H⁺ along with CH₃COO^–.  These new H⁺ from CH₃COOH can disturb the equilibrium towards left i.e., to minimize the a of HCl which is not possible as HCl is strong acid.  So, CH₃COOH acid will become much weaker acid in presence of a strong acid like HCl.

However the dissociation of CH₃COOH can be increased by an electrolyte producing OH^–.  Consider the addition of NH₄OH to the CH₃COOH:

H⁺ are removed from the reaction mixture, so as the equilibrium gets disturbed and shifts toward right hand side to produce more and more H⁺, thereby increasing a of CH₃COOH.  Now, since H₂O formed by union of H⁺ (from CH₃COOH) and OH^– (from NH₄OH) is a weak electrolyte, it will not ionize back to produce H⁺ & OH^– in the solution.

Relative strength of Acids & Bases:

For acids :

Relative strength = $\frac{\mathrm{Strength} \mathrm{of} {\mathrm{Acid}}_1}{\mathrm{Strength} \mathrm{of} {\mathrm{Acid}}_2}$

$\frac{{\mathrm{\alpha }}_1}{{\mathrm{\alpha }}_2}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}1}}{{\mathrm{K}}_{\mathrm{a}2}}}$     K_a = Dissociation constant for an acid

Similarly,

For bases                                 K_b = Dissociation constant. for a base

$\frac{{\mathrm{\alpha }}_1}{{\mathrm{\alpha }}_2}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{b}1}}{{\mathrm{K}}_{\mathrm{b}2}}}$

Ionic product of water (K_w):

Water is a weak electrolyte.  It dissociates feebly as follows:

$\mathrm{K}=\frac{\left[{\mathrm{H}}^{+}\right] \left[{\mathrm{OH}}^{–}\right]}{\left[{\mathrm{H}}_2\mathrm{O}\right]}$             K = Dissociation constant

Since, the concentration of water remains practically unchanged, [H₂O] will be constant.

K[H₂O] = [H⁺] [OH^–]

K_w = [H⁺] [OH^–]              K_w = Ionic product of water = K [H₂O]

Since K_w is the product of two constants, it will also be a constant.

“The product of concentration of H⁺ (or H₃O⁺) and OH^– in pure water, remains constant for the given temperature, called ionic product of water”.

At RT (25⁰C)  its value is found to be 10^–14 mole²/lit²

K_w increase with increase in temperature.

At 25⁰C, in 1 lit of H₂O having 55.5 moles only 10^–7 moles are dissociated.

Temp	Kw
00 C	n10–15
250C	1 ⨯ 10–14
600C	N10–13

Concept of pH:

S.P.L. Sorenson devised a more sophisticated method to express the [H⁺] of dilute solutions.  The letter ‘small p’ stands for power raised to (Fr. puissance; Ger. Potenz) and ‘capital H’ for the hydrogen ion.

pH is the negative power raised to the base 10 to express [H⁺] or

pH is the inverse to the log [H⁺]

[H⁺] = 10^–pH

or         pH = – log[H⁺]     $\left(\log \frac{1}{\left({\mathrm{H}}^{+}\right)}\right)=–\log \left[{\mathrm{H}}^{+}\right]$

So, pH is the measurement of [H⁺] or acidic nature of the solution.

In normal water, $\left[{\mathrm{H}}^{+}\right]={10}^{–7}$

Since   $\left[{\mathrm{H}}^{+}\right]={10}^{–\mathrm{pH}}; \mathrm{pH}=7$

So, we conclude that neutral solutions have pH = 7.

Suppose, a drop of a HCl is added to pure water, considering the equilibrium,

Due to H⁺ from HCl dissociation of H₂O is decreased, so that the concentration of H⁺ & OH^– from water falls appreciably.  But due to HCl itself, extra H⁺ are introduced to the solution.  The result is [H⁺] increases while [OH^–] decreases because the product of [H⁺] and [OH^–] is constant.  One can say that since K_w is constant; if conc of OH^– is decreased (due to common ion effect on H₂O], the concentration of H⁺ must increased so as to make its K_w fixed at 1 ⨯ 10^–14.

Consider an aq. solution of HCl.  Let the concentration of H⁺ from HCl in the solution be 10^–4 mole/lit.

The overall concentration of H⁺ in the solution will be due to the H⁺ from water as well as from HCl i.e., .

= ${10}^{–7}$(from water) + ${10}^{–4}$ (from HCl)

= 10^-4 (10^-4+1)

= 1.001 ⨯ 10^–4          or   $\approx$ 1 ⨯ 10^–4

Since,   K_w = [H⁺] [OH^–]

10^–14 = 10^–4 ⨯ [OH^–]

[OH] = 10^–10

i.e., concentration of [H⁺] is increased & those of OH^– is decreased.

Considering its pH,

pH = – log [H⁺]      = – log 10^–4     = + 4 log 10

= 4

Acidic solutions have their pH < 7.

Similarly, if 10^–4M NaOH is added to water and it contributes 10^–4 OH^–.  The concentration of OH^– in solution,

[OH^–] = 10^–4

[H⁺] = 10^–10                   [K_w = [H⁺] [OH^–]

pH = – log 10^–10 = 10

We conclude that basic solutions have their pH > 7

Acidic solution e.g., HCl, CH3COOH, AlCl3 (aq)		Neutral solution e.g., pure water, NaCl (aq), urea (aq)		Basic solution e.g., NH4OH, NaOH, NaHCO3 etc
[H+] > [OH–] (> 10–7)  (<10–7) pH < 7		[H+]  = [OH–] = 10–7 pH = 7		[H+] < [OH–] (< 10–7)  (> 10–7) pH > 7

Evidently, [H⁺] $\propto \frac{1}{\mathrm{pH}}$

Neutral solutions at 60⁰C have pH 6.5.  Human body, at an average temperature of 37⁰C has the neutral point at pH 6.8 (K_w < 10^–14)

pH range of some common solutions.

Chemical                                                                              Biological

System	pH		System	pH
Soft drinks	2.0 – 4.0		Gastric juice	1.0 – 3.0
Vinegar	2.4 – 3.4		Hunan urine	4.8 – 8.4
Sea water	8.5		Saliva	6.5 – 7.5
Milk of magnesia (Antacid)	10.5		Blood Tears	7.36 – 7.42 7.4

 pH of strong acids & bases

pH = – log [H⁺]

For acids [H⁺] = Normality

For bases [H⁺] = $\frac{{10}^{–14}}{\mathrm{Normality}}$

pH of weak acids & bases

For acids  [H⁺] = Normality ´ a

For bases [H⁺]  = $\frac{{10}^{–14}}{\mathrm{Normality}\times \mathrm{\alpha }}$

pK_a & pK_b

Like pH & pOH for strength of solutions of acids & bases, the relative strength of acid & bases can be compared using their pK values.

pK_a = – log K_a               (for acids)

pK_b = – log K_b               (for bases)

Higher the K value, lower the pK value, stronger the acid/base.

pH of very low concentration of an acid/base (e.g. 10^–8 MHCl)

Since water also give H⁺ in the solution i.e., [H⁺] = 10^–7.  While for normally dilute solutions we ignore this very low concentration of H⁺ from water since it has practically no effect on pH.  But for very dilute solution where there are only 10^–8 ions given by the electrolyte, the ions from H₂O must be considered to calculate the total concentration.

So that Net [H⁺] = 10^–7 (from H₂O) + 10^–8 (from HCl)

= 10^–7 (1 + 10^–1) = 1.1 ⨯ 10^–)

pH = + 7 log 10 – log 1.1

= 6.958

K_w doesn’t change by adding a salt, acid or base to a solution or neutral water.  It changes with temperature.

Calculation of pH or pOH of a solution:

Case i):  If H^– ion or OH^– ion concentration is given then

$\mathrm{pH}=–\log \left[{\mathrm{H}}^{+}\right]={10}^{–\mathrm{pH}}$   similarly

$\mathrm{pOH}=–\log \left[{\mathrm{OH}}^{–}\right] \mathrm{or} \left[{\mathrm{OH}}^{–}\right]={10}^{–\mathrm{pOH}}$

Case ii):  When the acid or the base is completely ionized in aqueous solution (i.e., for strong acid or strong base).

[H⁺] = normality of the acid

[OH^–] = normality of the base.

Case iii):  When a mixture of strong acids is present in the aqueous solution.  Then the normality of the mixture, N $=\left[\frac{\left({\mathrm{V}}_1{\mathrm{N}}_1+{\mathrm{V}}_2{\mathrm{N}}_2\right)}{\left({\mathrm{V}}_1+{\mathrm{V}}_2\right)}\right]$ where V₁ and N₁ are the volume and normality of the first acid and V₂ and N₂ are those of the second acid.

$\left[{\mathrm{H}}^{+}\right]$ in the mixed solution = normality of the mixed solution = N

$\therefore \mathrm{pH}=–\log \left(\mathrm{N}\right)=–\log \left[\frac{\left({\mathrm{V}}_1{\mathrm{N}}_1+{\mathrm{V}}_2{\mathrm{N}}_2\right)}{\left({\mathrm{V}}_1+{\mathrm{V}}_2\right)}\right]$

Case iv):  When strong alkalies are mixed o calculate the pH or pOH of the mixed solution.  The normality of the mixture of the strong bases is given by

$\mathrm{N}=\left[\frac{\left({\mathrm{V}}_1{\mathrm{N}}_1+{\mathrm{V}}_2{\mathrm{N}}_2\right)}{\left({\mathrm{V}}_1+{\mathrm{V}}_2\right)}\right]$ where V₁, N₁ are volume and normality of the first solution and V₂, N₂ are the values for the second solution.

Then [OH^–] in the mixture – Normality of the mixed solution = N.

$\therefore \mathrm{pOH}=–\log \left(\mathrm{N}\right)=–\log \left[\frac{\left({\mathrm{V}}_1{\mathrm{N}}_1+{\mathrm{V}}_2{\mathrm{N}}_2\right)}{\left({\mathrm{V}}_1+{\mathrm{V}}_2\right)}\right]$

at room temperature, pH + pOH = 14.     $\therefore$ pH = (14 – pOH)

Case v):  When a strong acid and a strong base are mixed, to calculate the pH of a mixed solution then normality of the mixture of the strong acid (A) and a strong base (B) is given by

$\left(\mathrm{N}\right)=\left[\frac{\left({\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}}–{\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}}\right)}{\left({\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}\right)}\right]$         (If the ${\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}}>{\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}}$)

or  $\mathrm{N}=\left[\frac{\left({\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}}–{\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}}\right)}{\left({\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}\right)}\right]$        (if V_BN_B > V_AN_A)

$\therefore$ [H⁺] = (N)  = OH

$\therefore \mathrm{pH}=–\log \left[\mathrm{H}\right]=\log \left(\mathrm{N}\right)=\log \left[\frac{\left({\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}}–{\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}}\right)}{\left({\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}\right)}\right]$

Similarly $\mathrm{pOH}=–\log \left[{\mathrm{OH}}^{–}\right]=–\log \left(\mathrm{N}\right)=–\log \left[\frac{\left({\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}}–{\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}}\right)}{\left({\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}\right)}\right]$

but, pH + pOH = 14

$\therefore$ pH = (14 – pOH)

Illustration 3:

What is the pH of a solution having a [H⁺] = 2.512 ⨯ 10^–6 M?

Solution:

$\log \left[{\mathrm{H}}^{+}\right]=\log {10}^{–6}+\log 2.512$

$\log \left[{\mathrm{H}}^{+}\right]=–5.60$

$\therefore –\log \left[{\mathrm{H}}^{+}\right]=5.60=\mathrm{pH}$

Illustration 4:

Find the hydrogen ion concentration in a solution of pH = 3.75?

Solution:

$\mathrm{pH}=–\log \left[{\mathrm{H}}^{+}\right]=3.75$

$\therefore \log \left[{\mathrm{H}}^{+}\right]=–3.75 \mathrm{or} 4.2500$

$\therefore \left[{\mathrm{H}}^{+}\right]=\mathrm{antilog}\left[4.2500\right]=1.8\times {10}^{–4}$

Illustration 5:

What is the K_w value in an aqueous solution of pK_w = 13.725?

Solution:

${\mathrm{pK}}_{\mathrm{w}}=–{\mathrm{logK}}_{\mathrm{w}}=13.725$

$\therefore {\mathrm{logK}}_{\mathrm{w}}=–13.725=–14.275$

$\therefore {\mathrm{K}}_{\mathrm{w}}=1.884\times {10}^{–14}$

Illustration 6:

What is the pH of a solution containing 0.63g of HNO₃ in 100 ml of solution?

Solution:

Concentation of  ${\mathrm{HNO}}_3\left(\mathrm{N}\right)=\left[\frac{0.63}{63}\times \frac{1000}{100}\right]=\frac{1}{10}=0.1\mathrm{N}$

$\therefore \left[{\mathrm{H}}^{+}\right]=0.1\mathrm{N}; \log \left[{\mathrm{H}}^{+}\right]=1.0000$

$–\log \left[{\mathrm{H}}^{+}\right]=1.0000=\mathrm{pH}$

Illustration 7:

Ba(OH)₂ is a strong base.  Then

Solution:

$\left[{\mathrm{OH}}^{–}\right]=\left[\mathrm{base}\right]$ = normality of the base.

$=0.2\mathrm{N}=2\times {10}^{–1}; \log \left[{\mathrm{OH}}^{–}\right]=\log 2+\log {10}^{–1}$

$–\log \left[{\mathrm{OH}}^{–}\right]=0.3010–1=0.70=\mathrm{pOH}$

$\mathrm{pH}=14–0.7=13.3$

Illustration 8:

Calculate the pH of an aqeusous solution having 2g of NaOH per 500 ml it.

Solution:

Molarity (M) of the solution= $\left[\frac{2}{40}\times \frac{1000}{500}\right]=\frac{1}{10}\mathrm{M}={10}^{–1}\mathrm{M}$

NaOH is a strong base.  Hence

$\left[{\mathrm{OH}}^{–}\right]=\left[\mathrm{Base}\right]$= Normality of the base = 10^–1 N

$\log \left[{\mathrm{OH}}^{–}\right]=–1.0$(here Molarity = Normality)

$–\log \left(10\right)\Rightarrow \mathrm{pOH}=1.0$

But, $\mathrm{pH}+\mathrm{pOH}=14.0; \mathrm{pH}+1=14.0; \mathrm{pH}=13.0$

Illustration 9:

150ml of 0.5N HCl and 100 ml of 0.2N HCl are mixed.  Find the pH of the resulting solution.

Solution:

Normality of the mixed solution is given by

$\left[\frac{\left({\mathrm{V}}_1{\mathrm{N}}_1+{\mathrm{V}}_2{\mathrm{N}}_2\right)}{\left({\mathrm{V}}_1+{\mathrm{V}}_2\right)}\right]=\left[\frac{\left(150\times 0.5\right)+\left(100\times 0.2\right)}{\left(150+100\right)}\right]=\frac{75+20}{250}$ = 0.38N

$\therefore$ pH = – log 0.38 = 0.42

Illustration 10:

Equal volumes of 0.5N NaOH and 0.3N KOH are mixed in a experiment.  Find the pOH and pH of the resulting solution.

Solution:

Normality of the mixed solution

$\mathrm{N}=\left[\frac{\left({\mathrm{V}}_1{\mathrm{N}}_1+{\mathrm{V}}_2{\mathrm{N}}_2\right)}{\left({\mathrm{V}}_1+{\mathrm{V}}_2\right)}\right]=\frac{V_1\left(0.5\right)+V_1\left(0.3\right)}{2V_1}=\frac{0.5+0.3}{2}=0.4\mathrm{N}$

$\therefore \mathrm{pOH}=–\log \left(\mathrm{N}\right)=–\log 0.4=0.398$

$\therefore$ pH = 14 – 0.398 = 13.6

Illustration 11:

50ml of 0.2M HCl is added to 30ml of 0.1M KOH solution.  Find the pH of the solution.

Solution:

In the mixed solution ${\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}}>{\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}}$

$\therefore \mathrm{N}=\left[\frac{\left({\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}}–{\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}}\right)}{\left({\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}\right)}\right]=\left[\frac{\left(50\times 0.2\right)–\left(30\times 0.1\right)}{\left(50+30\right)}\right]=\frac{10–3.0}{80}=0.0875\mathrm{N}$

$$\begin{gathered} \therefore \left[{\mathrm{H}}^{+}\right]=0.875\mathrm{N}=0.875\mathrm{M} \\ \mathrm{pH}=–\log 0.0875=1.06 \end{gathered}$$

Illustration 12:

40ml of 0.2N HNO₃ when reacted with 60ml of 0.3M NaOH, gave a mixed solution.  What is the pH of the solution?

Solution:

$\therefore$ N = Normality of the mixed solution = $\left[\frac{\left({\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}}–{\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}}\right)}{\left({\mathrm{V}}_{\mathrm{A}}+{\mathrm{V}}_{\mathrm{B}}\right)}\right]$

= $\left[\frac{\left(60\times 0.3\right)–\left(40\times 0.2\right)}{\left(60+40\right)}\right]=\frac{10}{100}=0.1\mathrm{N}$

$\therefore \left[\mathrm{OH}\right]=0.1\mathrm{N}$

pOH = 1

$\therefore$ pH = 14-1 = 13

Exercise 1:

(i) The K_a of a weak acid is 10^–5.  The value of pK_b of its conjugate base will be

(A) 5          (B) 6          (C) 9          (D) 10^–9

(ii) K_a of a weak monobasic acid in aq. soln. is 8.0 ⨯ 10^–5.  The degree of dissociation of a 0.05 M solution of the acid is

(A) 3.52%          (B)  4.0%          (C)  4.52%          (D)  8.5%

(iii) The pH of an aq. solution is 4.  It [OH^–] will be

(A) 10          (B) 14          (C) 10^–4          (D)  10^–10

(iv) The pH of an aqueous solution having [H⁺] = 3.0 ⨯ 10^–3M is

(A) 2.471          (B) 2.523          (C) 3.0          (D) 6.433

 (v) The pH of a 5 ⨯ 10^–3 M aq. solution of a weak monobasic acid with dissociation constant 1.8 ⨯ 10^–5 will be

(A) 4          (B) 2.52          (C) 3.523          (D) 5.53

BUFFER SOLUTIONS:

If one drop of concentration acid is added to 1 lit H₂O (pH = 7), its pH falls rapidly; had a drop of base been added it would have increased abruptly.

But there are some mixtures or solutions whose pH doesn’t vary so easily (i.e, by a single drop of an acid or a base); they actually resist any change in pH.  Such solutions are called the buffer solutions.

Types of Buffer Solutions:

I) Neutral Buffers:

a) Strong acids and bases are buffers at higher concentrations

b) Proteins and amino acid solutions

c) Salts of weak acids and weak bases (CH₃COONH₄)

d) Mixture of a normal and an acidic salt of polybasic acids. (Na₃PO₄ + Na₂HPO₄)

II) Acidic Buffers:

Weak acids with their conjugate bases

III) Basic Buffers:

Weak bases with their conjugate acids (NH₄OH + NH₄⁺; glycine + glycine hydrochloride)

Buffer Action:

Consider a simple buffer – an aq. solution of CH₃COONH₄:

$\underset{\left(\mathrm{strong} \mathrm{electrolyte}\right)}{{\mathrm{CH}}_3{\mathrm{COONH}}_4} \rightleftharpoons {\mathrm{CH}}_3{\mathrm{COO}}^{–}+{\mathrm{NH}}_4^{–}$

Excess of CH₃COO^– & NH₄⁺ are present in the solution.  If a drop or so of an acid is added, the following changes will take place.

Dissociation of HCl adds H⁺ to the solution to increase (H⁺).  But these H⁺ ions react promptly with acetate ions to form CH₃COOH.  Acetic acid is weak acid and doesn’t dissociate back to H⁺ & CH₃COO^–.  So there is no effect on pH of the solution.

If a base is added,

This time, extra OH^–, which could have disturbed the pH of solution, react with NH₄⁺ of ammonium acetate to give NH₄OH.  The pH of the solution remains unaffected.

pH of a Buffer solution (Handerson Equation):

For an Acidic Buffer

Let HA be the weak acid and the salt of its conjugate base be NaA,

$\mathrm{HA}\rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{A}}^{–}$. . . . (i)

$NaA\rightleftharpoons N{a}^{+}+A^{–}$. . . . (ii)

For eq. (i)

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{–}\right]}{\left[\mathrm{HA}\right]}$   or    $\left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{a}}\left[\mathrm{HA}\right]}{\left[{\mathrm{A}}^{–}\right]}$               . . . . (iii)

$\log \left[{\mathrm{H}}^{+}\right]={\mathrm{logK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{HA}\right]}{\left[{\mathrm{A}}^{–}\right]}$

$–\log \left[{\mathrm{H}}^{+}\right]=–{\mathrm{logK}}_{\mathrm{a}}–\log \frac{\left[\mathrm{HA}\right]}{\left[{\mathrm{A}}^{–}\right]}$

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{A}}^{–}\right]}{\left[\mathrm{HA}\right]} \left[{\mathrm{pK}}_{\mathrm{a}}=–{\mathrm{logK}}_{\mathrm{a}}\right]$

Since HA is weak acid, [A^–] in solution is mainly due to the salt NaA. i.e., [A^–] = [salt, NaA]

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$. . . . (iv)

Similarly, for a basic Buffer  $\mathrm{pOH}={\mathrm{pH}}_{\mathrm{b}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]}$      . . . . (v)

Eq. (iv) & (v) are Handerson equations.

Illustration 13:

What will be the pH of a solution obtained by the mixing of 5g of acetic acid and 7.5 g of sodium acetate and making the volume to 500mL (K_a = 1.75 ⨯ 10^–5)

Solution.

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$

${\mathrm{pK}}_{\mathrm{a}}=–{\mathrm{logK}}_{\mathrm{a}}=–\log \left(1.75\times {10}^{–5}=4.76\right)$

$$\begin{gathered} \left[\mathrm{salt}\right]=\frac{7.5\times 1000}{82\times 500}=0.1829 {\mathrm{moIL}}^{–1} \\ \left[\mathrm{acid}\right]=\frac{5\times 1000}{60\times 500}=0.1667 {\mathrm{molL}}^{–1} \\ \mathrm{pH}=4.76+\log \frac{0.1829}{0.1667}=4.80 \end{gathered}$$

Illustration 14:

How much volume of 0.1M HAc should be added to 50 mL of 0.2M NaAc solution if we want to prepare a buffer solution of pH 4.91.  Given pK_a for acetic acid is 4.76.

Solution.

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$

i.e., 4.91 = 4.76 + $\log \frac{\left[\mathrm{Salt}\right]}{\mathrm{Acid}}$

of  $\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}=0.15$

or $\frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}=\mathrm{Antilog} 0.15=1.41$

$\frac{\mathrm{Moles} \mathrm{of} \mathrm{Salt}}{\mathrm{Moles} \mathrm{of} \mathrm{Acid}}=1.41$

$\frac{{\displaystyle \frac{0.2}{1000}}\times 50}{{\displaystyle \frac{0.1}{1000}}\times \mathrm{V}}=1.44 \mathrm{or} \frac{0.01}{0.0001 \mathrm{V}}=1.41$

$\therefore$ V = 100/1.41 = 70.92 mL

Illustration 15:

50ml of 0.1M sodium acetate, 25 ml of 0.2 M acetic acid were added together to form the buffer solution pK_a of CH₃COOH is 4.8.  Find the pH of the solution.

Solution.

The buffer solution is an acid buffer.  Hendersen’s equation for an acid buffer is

${\mathrm{p}}_{\mathrm{H}}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]}=4.8+\left\{\log \frac{{\displaystyle \frac{\left(0.1\times 50\right)}{75}}}{{\displaystyle \frac{\left(0.2\times 25\right)}{75}}}\right\}=4.8+\log \left(1\right)$

$\therefore \mathrm{pH}=4.8$

llustration 16:

50ml of 0.1M NH₄Ohm 25 ml of 2M NH₄Cl were used to make a buffer.  What is the pH if pK_b is 4.8?

Solution.

The given buffer is a basic buffer

For a basic buffer pOH = ${\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]}$

$\therefore \mathrm{pOH}=4.8+\log \left\{\frac{{\displaystyle \frac{\left(25\times 2\right)}{75}}}{{\displaystyle \frac{\left(50\times 0.1\right)}{75}}}\right\}=4.8+\log \left(10\right)=4.8+1=5.8$

$\therefore \mathrm{pH}=\left(14–\mathrm{pOH}\right)=\left(14–5.8\right)=8.2$

Illustration 17:

The pH of buffer prepared by mixing 50 ml of 0.2 M CH₃COOH and 25 ml of CH₃COONa is 4.8.  If the pK_a = 4.8, what is the strength of CH₃COONa?

Solution.

The given solution is an acid buffer.

The pH of the buffer is given by, $\mathrm{pH}=\mathrm{pKa}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{Acid}\right]}$

i.e., $4.8=4.8+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]} \mathrm{or} \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}=0.0$

$\therefore \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}=1.0 \mathrm{or} \left[\mathrm{Salt}\right]=\left[\mathrm{Acid}\right]$

or   $\frac{25\times \mathrm{M}}{75}=\frac{50\times 0.2}{75} \mathrm{or} \mathrm{M}=\frac{50\times 0.2}{25}=0.4\mathrm{M}$

$\therefore$ [CH₃COONa] = 0.4M

 Illustration 18:

When 20ml of 0.1M NH₄OH are added to 20ml of 1M NH₄Cl solution, the pH of the buffer formed is 8.2, what is the pK_b of the NH₄OH?

Solution.

For basic buffer pH is given by

$\mathrm{pH}=14–\mathrm{pOH} \mathrm{and} \mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]}$

given the pH of the buffer = 8.2;

$\therefore$ 14 – 8.2 = ${\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]}$

or  5.8 = ${\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]}$

$={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left({\displaystyle \frac{20\times 1}{40}}\right)}{\left({\displaystyle \frac{20\times 0.1}{40}}\right)}={\mathrm{pK}}_{\mathrm{b}}+\log 10$

or 5.8 = pK_b + 1

$\therefore$ pK_b = 4.8

Exercise  2:

(i) What will be pH of a solution of 0.1M HA (acid) and 0.1M NaA?  (K_a = 10^–6)

(A) 3          (B) 5          (C) 6          (D) 7

(ii) A 1 lit solution of 0.25 mol NH₄Cl and 0.15 mol NH₄OH (dissociation constant, 1.98 ´ 10^–5) will have a pH of

(A) 9          (B) 9.2          (C) 10          (D) 10.2

Buffer Capacity or Buffer Index:

No. of moles of an acid or a base required to change the pH of one litre of a buffer solution by unity is called as the buffer capacity.

Buffer capacity, $\mathrm{\varphi }=\frac{\mathrm{Number} \mathrm{of} \mathrm{mole} \mathrm{of} \mathrm{acid} \mathrm{or} \mathrm{base} \mathrm{added} \mathrm{to} \mathrm{one} \mathrm{litre} \mathrm{of} \mathrm{the} \mathrm{solution}}{\mathrm{change} \mathrm{in} \mathrm{pH}}, \mathrm{\varphi }=\frac{\mathrm{n}}{∆\mathrm{pH}.\mathrm{V}}$

Applications of Buffer Solutions:  Buffer solutions are widely used in those solutions where reactions proceeds only in a particular pH range, for example in

– Qualitative analysis – removal of PO₄^3–  after IInd group (CH₃COOH + CH₃ COONa buffer)

– Precipitation of IIIrd group (NH₄OH + NH₄Cl buffer)

– Blood (H₂CO₃ + HCO₃^– buffer)

– Medical formulations

– Preservation of foods, dairy products & fruits

– Industries like dyes, inks, paints, paper

SALT

A salt is a compound formed by the neutralization reactions between an acid and a base.  Its +ve part comes from the base while negative part from the acid.

Types of salts: 

(a) Simple Salts: Normal :  The salts without any replaceable H⁺ or OH^– (complete neutralization) e.g., NaCl, KNO₃, CaSO₄ Na₃BO₃ etc.

Acidic :  They have replaceable H⁺, ( incomplete neutralization) e.g., NaHCO₃, KHSO₄, NaH₂PO₄ etc.

Basic :  They have replaceable OH^– (incomplete neutralization) e.g., Zn(OH)Cl, Mg (OH) Br etc.

b) Double Salts: These are formed by reunion of two salts and have properties of each ion of both the constituting salts. g., Potash alum, K₂SO₄. Al₂(SO₄)₃.24H₂O; Mohr salt, (NH₄)₂SO₄.FeSO₄. 7H₂O etc.

c) Complex Salts: These are formed by coordination of ions to a central metal atom.  The species with square bracket is an independent entity and it doesn’t further dissociate into its constituents. e.g.${\mathrm{K}}_4\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right], \left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]$Cl etc.

d) Mixed Salts: These salts furnish more than one cation or anion in solution e.g., Bleaching power, CaOCl₂; Rochelle’s salt (CHOHCOONa – CHOHCOOK).

While Mohr salt gives the tests of all of its constituent ions i.e., of NH₄⁺, Fe²⁺ & SO₄^2– ; K₄ [Fe(CN)₆] gives tests of only K⁺ & [Fe(CN)₆]^–4 i.e., ferrocyanide, but not that of Fe²⁺ & CN^– ions separately.

Salt Hydrolysis: 

Reverse to the neutralization reaction and involves the treatment of the salt with the solvent.  The cations and anions of a salt interact with the anions and cations of solvent respectively furnishing bases and acids.  The process is called salt hydrolysis .

$\underset{\left(\mathrm{salt}\right)}{{\mathrm{B}}^{+}{\mathrm{A}}^{–}} + \underset{\left(\mathrm{solvent}\right)}{{\mathrm{H}}^{+}{\mathrm{OH}}^{–}} \rightleftharpoons \underset{\left(\mathrm{Base}\right)}{\mathrm{BOH}} + \underset{\left(\mathrm{Acid}\right)}{\mathrm{HA}}$

The resulting mixture can be acidic, basic or neutral depending upon the relative strengths of the acid and base formed during the reaction.

Being reverse to neutralization, it’s always endothermic.

Normal salts can further be divided into 4 groups:

Salts of strong acid and strong base e.g., NaCl

Salts of strong acid and weak base e.g., NH₄Cl

Salts of weak acid and strong base e.g. CH₃COONa

Salts of weak acid and weak base e.g., CH₃COONH₄

Salts of Strong Acids and Strong Bases: (NaCl, KNO₃ CaSO₄, Ba(NO₃)₂ etc.)

These salts don’t hydrolyse. The cations and anions that these salts furnish in solution don’t interact upto appreciable extent with H⁺ & OH^– of H₂O, so that the concentration of H⁺ and OH^– remains unchanged, leaving behind a neutral solution (pH = 7).  e.g.,

Since NaOH and HCl both are strong, they completely ionize back to produce Na⁺, OH^–, H⁺ and Cl^–.  Of these , H⁺ combines with OH^– to give H₂O, a weak electrolyte.  Hence Na⁺ and Cl^– practically don’t interact with H₂O (H⁺ & OH^–).

Salts of strong acidic and weak bases: (NH₄Cl, CISO₄, FeCl₃, AlCl₃ etc.)

These salts hydrolyse to give acidic solutions (pH < 7).  Cations undergo hydrolysis called cationic hydrolysis. e.g.,

Although equal amounts of ions from H₂O (i.e., H⁺ & OH) are offering the reaction, there should be number change used up in H⁺ or OH^– concentration as per Le Chatelier principle, more H₂O will dissociate now to make up the equilibrium condition.  However, HCl is strong acid, it dissociates completely into H⁺ & Cl^– while NH₄OH being weak base will give only few OH^–.  Further due to common ion effect, its dissociation is further decreased, making the solution scare of OH^– & plenty of H⁺, hence it will become acidic:

The result:

${\mathrm{NH}}_4\mathrm{OH}\rightleftharpoons {\mathrm{NH}}_4^{+}+{\mathrm{OH}}^{–}$

i.e., equilibrium shifts to left; decreasing OH^– concentration

Salts of Weak Acids and Strong Bases:  (KCN, Na₂S, Na₂CO₃, CH₃COONa etc.)

These salts hydrolyse to give basic solution (pH > 7).  Anions are hydrolysed, called anionic hydrolysis.

The concentration of H⁺ is decreased in the solution due to poor dissociation of HCN coupled with common ion effect of KCN.  The [OH^–] becomes greater due to presence of strong base KOH somewhat like cationic hydrolysis.

Salts of Weak Acids and Weak Bases: (CH₃COONH₄, Ag₂CO₃, NH₄CN, (CH₃COO)₂Cu etc.)

These salts hydrolyse to produce almost neutral solution (pH $\approx$ 7).  In fact, the nature of solution depends upon the relative strengths of the acids & bases so formed i.e., the resulting solution could be feebly acidic, basic or neutral.

Both the cations and anions undergo hydrolysis, called cationic – anionic hydrolysis.

Since strengths of acetic acid & ammonium hydroxide are approximately equal, the resulting solution is almost neutral (actually very little basic)

${\mathrm{Ag}}_2{\mathrm{CO}}_3+2{\mathrm{H}}_2\mathrm{O} \rightleftharpoons \underset{{\mathrm{K}}_{\mathrm{b}}=1.1\times {10}^{–4}}{2\mathrm{AgOH}}+\underset{{\mathrm{K}}_{\mathrm{a}}=4.52\times {10}^{–7}}{{\mathrm{H}}_2{\mathrm{CO}}_3}$

The aq. solution of Ag₂CO₃, thus will be basic.

${\left({\mathrm{NH}}_4\right)}_2{\mathrm{C}}_2{\mathrm{O}}_4+{\mathrm{H}}_2\mathrm{O} \rightleftharpoons \underset{{\mathrm{K}}_{\mathrm{b}}=1.8\times {10}^{–5}}{2{\mathrm{NH}}_4\mathrm{OH}}+\underset{{\mathrm{K}}_{\mathrm{a}}=5.0\times {10}^{–2}}{{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4}$

The ammonium carbonate solution will become acidic due to stronger nature of acid (H₂C₂O₄) in comparison to NH₄OH.

Degree of hydrolysis (h) and hydrolysis constant (K_h) :

The fraction of a salt that undergoes hydrolysis in comparison to the total number of moles taken, is called degree of hydrolysis (h).

$\mathrm{h}=\frac{\mathrm{no}.\mathrm{of} \mathrm{moles} \mathrm{hydrolysed}}{\mathrm{total} \mathrm{no}.\mathrm{of} \mathrm{moles} \mathrm{dissolved} \mathrm{in} \mathrm{solution}}$

                                                    BA     +     ${\mathrm{H}}_2\mathrm{O}$     $\rightleftharpoons$   HA     +     BOH

Moles before hydrolysis           1                …                 0                0

Moles after hydrolysis           (1-h)                                  h                 h

Conc. at equilibrium            C(1-h)                                Ch              Ch

Applying law of mass action,

$\mathrm{K}=\frac{\left[\mathrm{HA}\right]\left[\mathrm{BOH}\right]}{\left[\mathrm{BA}\right]\left[{\mathrm{H}}_2\mathrm{O}\right]}$

${\mathrm{K}}_{\mathrm{h}}=\frac{\left[\mathrm{HA}\right]\left[\mathrm{BOH}\right]}{\left[\mathrm{BA}\right]}=\frac{{\mathrm{Ch}}^2}{\left(1–\mathrm{h}\right)}$      $\left({\mathrm{K}}_{\mathrm{h}}, \mathrm{hydrolysis} \mathrm{constant}=\mathrm{K}\left[{\mathrm{H}}_2\mathrm{O}\right]\right)$

K_h for salt of strong acid and weak base (cationic hydrolysis)

${\mathrm{B}}^{+}+{\mathrm{H}}_2\mathrm{O} \rightleftharpoons {\mathrm{H}}^{+}+\mathrm{BOH}$

${\mathrm{K}}_{\mathrm{h}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[\mathrm{BOH}\right]}{\left[{\mathrm{B}}^{+}\right]}$

${\mathrm{K}}_{\mathrm{h}}=\frac{{\mathrm{K}}_{\mathrm{w}}\left[\mathrm{BOH}\right]}{\left[{\mathrm{OH}}^{–}\right]\left[{\mathrm{B}}^{+}\right]} \left(\left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{–}\right]}\right)$

${\mathrm{K}}_{\mathrm{h}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}} \left({\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{B}}^{+}\right]\left[{\mathrm{OH}}^{–}\right]}{\left[\mathrm{BOH}\right]}\right)$

Since,  ${\mathrm{K}}_{\mathrm{h}}=\frac{{\mathrm{Ch}}^2}{\left(1–\mathrm{h}\right)}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}}$

For weak base h << 1; so (1 – h) $\approx$ 1

$\therefore {\mathrm{Ch}}^2=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}}$

$\mathrm{h}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}.\mathrm{C}}}$

$\because \left[{\mathrm{H}}^{+}\right]=\mathrm{C}.\mathrm{h}=\mathrm{C}\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}.\mathrm{C}}}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}\mathrm{C}}{{\mathrm{K}}_{\mathrm{b}}}}$

$–\log \left[{\mathrm{H}}^{+}\right]=\frac{1}{2}\left[–{\mathrm{logK}}_{\mathrm{w}}–\mathrm{logC}+{\mathrm{logK}}_{\mathrm{b}}\right]$

$\mathrm{pH}=\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{w}}–{\mathrm{pK}}_{\mathrm{b}}–\mathrm{logC}\right]$

K_h for salt of weak acid and strong base

${\mathrm{A}}^{–}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{HA}+{\mathrm{OH}}^{–}$

${\mathrm{K}}_{\mathrm{h}}=\frac{\left[\mathrm{HA}\right]\left[{\mathrm{OH}}^{–}\right]}{\left[{\mathrm{A}}^{–}\right]}=\frac{\left[\mathrm{HA}\right]{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{A}}^{–}\right] \left[{\mathrm{H}}^{+}\right]}$

${\mathrm{K}}_{\mathrm{h}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}=\frac{{\mathrm{Ch}}^2}{\left(1–\mathrm{h}\right)}$

$\mathrm{h}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}.\mathrm{C}}}$            $\left(\because \mathrm{h}<<1\right)$

$\left[{\mathrm{OH}}^{–}\right]=\mathrm{C}.\mathrm{h}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}.\mathrm{C}}{{\mathrm{K}}_{\mathrm{a}}}}$

$\left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{–}\right]}={\mathrm{K}}_{\mathrm{w}}\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{{\mathrm{K}}_{\mathrm{w}}\mathrm{C}}}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}.{\mathrm{K}}_{\mathrm{w}}}{\mathrm{C}}}$

$\mathrm{pH}=\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{a}}+{\mathrm{pK}}_{\mathrm{w}}+\mathrm{logC}\right]$

K_h for salt of weak acid and weak base

${\mathrm{B}}^{+}+{\mathrm{A}}^{–}+{\mathrm{H}}_2\mathrm{O} \rightleftharpoons \mathrm{HA}+\mathrm{BOH}$

${\mathrm{K}}_{\mathrm{h}}=\frac{\left[\mathrm{HA}\right] \left[\mathrm{BOH}\right]}{\left[{\mathrm{B}}^{+}\right] \left[{\mathrm{A}}^{–}\right]}$

= $\frac{\left[\mathrm{HA}\right]}{\left[{\mathrm{H}}^{+}\right] \left[{\mathrm{A}}^{–}\right]}.\frac{\left[\mathrm{BOH}\right]}{\left[{\mathrm{B}}^{+}\right] \left[{\mathrm{OH}}^{–}\right]}.\left[{\mathrm{H}}^{+}\right] \left[{\mathrm{OH}}^{–}\right]$     (multipying by the factor $\frac{\left[{\mathrm{H}}^{+}\right] \left[{\mathrm{OH}}^{–}\right]}{\left[{\mathrm{H}}^{+}\right] \left[{\mathrm{OH}}^{–}\right]}$)

$=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}.{\mathrm{K}}_{\mathrm{b}}}$

${\mathrm{K}}_{\mathrm{h}}=\frac{{\mathrm{C}}^2{\mathrm{h}}^2}{{\mathrm{C}}^2{\left(1–\mathrm{h}\right)}^2}=\frac{{\mathrm{h}}^2}{{\left(1–\mathrm{h}\right)}^2}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}.{\mathrm{K}}_{\mathrm{b}}}$

$\mathrm{h}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}.{\mathrm{K}}_{\mathrm{b}}}} \left(\because \mathrm{h}\ll 1\right)$

${\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right] \left[{\mathrm{A}}^{–}\right]}{\left[\mathrm{HA}\right]}$

$\left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{a}}.\left[\mathrm{HA}\right]}{\left[{\mathrm{A}}^{–}\right]}=\frac{{\mathrm{K}}_{\mathrm{a}}.\mathrm{Ch}}{\mathrm{C}\left(1–\mathrm{h}\right)}$

$\left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{a}}.\mathrm{h}}{\left(1–\mathrm{h}\right)}$

= K_a. h                $\left(\because \mathrm{h}<<1, \left(1–\mathrm{h}\right)\approx 1\right)$

= ${\mathrm{K}}_{\mathrm{a}}\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}.{\mathrm{K}}_{\mathrm{b}}}}$

= $\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}.{\mathrm{K}}_{\mathrm{a}}}{{\mathrm{K}}_{\mathrm{b}}}}$

$\mathrm{pH}=\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{w}}+{\mathrm{pK}}_{\mathrm{a}}–{\mathrm{pK}}_{\mathrm{b}}\right]$

If

 $$\begin{gathered} {\mathrm{pK}}_{\mathrm{a}}={\mathrm{pK}}_{\mathrm{b}} \mathrm{pH}=\frac{1}{2}{\mathrm{pK}}_{\mathrm{w}}=7 \mathrm{i}.\mathrm{e}., \mathrm{Neutral} \\ {\mathrm{pK}}_{\mathrm{a}}>{\mathrm{pK}}_{\mathrm{b}} \mathrm{pH}>7 \mathrm{i}.\mathrm{e}., \mathrm{Basic} \\ {\mathrm{pK}}_{\mathrm{a}}<{\mathrm{pK}}_{\mathrm{b}} \mathrm{pH}<7 \mathrm{i}.\mathrm{e}., \mathrm{Acid} \end{gathered}$$

Note :  The pK_a value and strength of acid are inversely related i.e., higher the pK_a value, weaker the acid.

Illustration 19:

A weak acid, HX forms a salt with NaOH.  Find the hydrolysis constant?  Given K_a = 2 $\times$ 10^–6.

Solution.

${\mathrm{K}}_{\mathrm{h}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}$ since the acid is weak

$=\frac{1\times {10}^{–14}}{2\times {10}^{–6}}$ (because ${\mathrm{K}}_{\mathrm{w}}=1\times {10}^{–14}$ at room temperature)

= $0.5\times {10}^{–8} \mathrm{or} 5\times {10}^{–9}$

 Illustration 20:

A weak acid (K_a = 2 $\times$ 10^–6) and a weak base (K_b = 5 $\times$ 10^–7) form a salt.  Calculate the hydrolysis constant.

Solution.

${\mathrm{K}}_{\mathrm{h}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}+{\mathrm{K}}_{\mathrm{b}}}$ for weak acid – weak base salts.

${\mathrm{K}}_{\mathrm{h}}=\frac{1\times {10}^{–14}}{\left(2\times {10}^{–6}\right)\left(5\times {10}^{–7}\right)}=\frac{1\times {10}^{–14}}{10\times {10}^{–13}}=1\times {10}^{–2}=0.01$

 Illustration 21:

Calculate the degree of hydrolysis of a 0.01M solution of NH₄Cl.  [K_w = 1.008 $\times$ 10^–14, K_b (NH₄OH) = 1.81 $\times$ 10^–15] _25°C

Solution.

${\mathrm{K}}_{\mathrm{h}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{b}}}=\frac{1.008\times {10}^{–14}}{1.81\times {10}^{–5}}=5.569\times {10}^{–10}$

Assuming the degree of hydrolysis to be very small,

$\mathrm{h}=\sqrt{\raisebox{1ex}{${\mathrm{K}}_{\mathrm{h}}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{C}$}\right.}$

$=\sqrt{\frac{5.569\times {10}^{–10}}{{10}^{–2}}}$

= 2.359 $\times$ 10^–4

Indicators:

The substances that inform us about the completion of a reaction by (generally change in their colour are called indicators.  Generally this completion point of a reaction is called the end point.

Types of indicators:

1) Acid base indicators:  Used in neutralization reactions

Change their colour depending on pH of solution. e.g. Phenolphthalein, Methyl red, litmus etc.

2) Redox indicators:  Used in oxidation – reduction reactions.

Colour changes with the change of emf e.g., KMnO₄, N-phenyl anthranilic acid.

3)  Adsorption indicators:  Change of colour due adsorption of a substance on its surface.

4)  Radioactive indicators:  Used in the study of variety of reactions as tracer.

Their presence is detected by suitable devices.  e.g., D², O⁸, I¹²⁷etc.

1 g equivalent acid neutralizes 1g equivalent base completely.  But in practice, there is no any visible indication that the reaction is complete. Indicators do the job pretty well.  They show a variation in their colour between certain pH limits, called indicator range or pH range.

Features of a Good Indicator:

• It should change the colour at a pH which falls in proximity to the end point for a given reaction.
• Colour change should be very sharp.
• Colour should be stable and brilliant.
• The two colours of an indicator should be contrasting so that they can easily be distinguished.

Indicators	pH range	Colour in
		Acidic	Basic
Methyl orange	3.1 – 4.5	Pink	Yellow
Methyl red	4.2 – 6.3	Red	Yellow
Litmus	5.4 – 7.5	Red	Blue
Phenolphthalein	8.0 – 9.8	Colourless	Colourless
Thymolphthalein	9.3 – 10.5	Colourless	Blue

Internal indicator: Added into the titration mixture before titration.  e.g., Phenolphthalein, N-Phenyl anthranilic acid, Starch + KI etc.

External indicator:  Only a small fraction of the reaction mixture is taken out and checked for end point by adding the indicator e.g., ${\mathrm{K}}_3\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$.

Self indicator :  When any of the titrant itself changes colour during a titration, there is no need to use other external substances. e.g., KMnO₄ in redox reactions decolourises at end point.

Titration	Indicator used
Strong acid   Vs   strong base Strong acid    Vs  weak base Weak acid    Vs   strong base Weak acid    Vs   weak base	Phenolphthalein, methyl red Methyl orange, methyl red Phenolphthalein, thymolphthalein No suitable indicator (carried out using conductivity methods)

Universal Indicator (Multiple Range Indicator)

It is a mixture of certain indicators in a definite proportion, which shows different colour changes over a wide range of pH.

Common universal indicator is prepared by dissolving

1 g phenolphthalein	In 500 mL absolute ethanol and then adding NaOH solution till it turns yellow.
2 g Methyl orange
3 g Methyl yellow
4 g Bromothymol blue
5 g Thymol blue

It’s colour change ranges are:

pH	2	4	6	8	10
Colour	Red	Orange	Yellow	Green	Blue

Titration:  An experiment carried out in order to know the strength/concentration of a substance is called titration.

Theories of indicators:  How, it is possible for a same compound to have two colours is explained using following theories:

1) Ostwald Theory

2) Benzenoid–Quinonoid theory

1. Ostwald Theory:

All indicators are weak acids or weak bases.  In unionized (moleculer) form and ionized (ionic) form they have different colours.  Being weak electrolytes their dissociation into corresponding positive & negative ions is largely affected by other ions present in the solution.

An indicator changes its colour when it is 50% dissociated.

a) Phenolphthalein : It is a weak acid considering it to be HPh, it dissociates into H⁺ (colourless) and Ph^– (pink) ions, but only upto very small extent (as $\alpha$ << 1) i.e.,

$\underset{\left(\mathrm{colourless}\right)}{\mathrm{HPh}} \rightleftharpoons {\mathrm{H}}^{+}+\underset{\left(\mathrm{pink}\right)}{{\mathrm{Ph}}^{–}}$

$\mathrm{K}=\frac{\left[{\mathrm{H}}^{+}\right] \left[{\mathrm{PH}}^{–}\right]}{\left[\mathrm{HPh}\right]}$

i) In presence of an acid:

Equilibrium of HPh shifts to the backward direction i.e., dissociation is decreased.  Very few Ph^– ions are left in the solution hence it remains colorless.

ii) In presence of a base:

Since H⁺ & OH^– are removed, equilibrium shifts to the forward direction.  So that more Ph^– are produced.  Although they may combine with Na⁺ but NaPh is highly ionized being strong electrolyte, it dissociates almost completely to produce excess of Ph^–.  These Ph^– provide pink colour to the solution.

b) Methyl Orange: The weak basic nature of methyl orange (MeOH) gives very small amount of Me⁺ (red) and OH^– (colourless) ions. Since concentration of Me⁺ is very less, it remains yellow in neutral solution.

$\underset{\underset{\left(\mathrm{Yellow}\right)}{\mathrm{weak} \mathrm{base}}}{\mathrm{MeOH}} \rightleftharpoons \underset{\left(\mathrm{Red}\right)}{{\mathrm{Me}}^{+}\mathrm{O}} + {\mathrm{OH}}^{–}$

In acidic medium, (H⁺) increases the dissociation of MeOH and thus produces more Me⁺ which make the solution red. (Le Chatelier principle)

The basic medium, (OH^–), because of common ion effect, decreases the dissociation of MeOH by which only few Me⁺ are left in the solution.  Major portion is the molecular form i.e., MeOH, which appears yellow.

2. Quiuonoid Theory:

According to this theory indicators are aromatic organic compounds.  Each indicator has two tautomeric forms i.e., benzenoid & quinonoid.  Quinonoid form has deeper colour than benzenoid.  Colour transition is observed due to these tautomeric inter-conversions.  The tautomeric conversion is pH dependent.

a) Phenolphthalein :

b)

Solubility Product:

For a saturated solution of an electrolyte A_xB_y:

${\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}} \rightleftharpoons {\mathrm{xA}}^{\mathrm{y}+}+{\mathrm{yB}}^{\mathrm{x}–}$

$\mathrm{K}=\frac{{\left[{\mathrm{A}}^{\mathrm{y}+}\right]}^{\mathrm{x}}{\left[{\mathrm{B}}^{\mathrm{x}–}\right]}^{\mathrm{y}}}{\left[{\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}\right]}$

For saturated solution, $\left[{\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}\right]$ = constant

$\therefore {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{A}}^{\mathrm{y}+}\right]}^{\mathrm{x}}{\left[{\mathrm{B}}^{\mathrm{x}–}\right]}^{\mathrm{y}}$     ${\mathrm{K}}_{\mathrm{sp}}$ = Solubility product = $\mathrm{K}\left[{\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}\right]$

Since [A_xB_y] i.e., solubility of an electrolyte is temperature dependent, K_sp is constant only for given temperature

Ex. $\mathrm{AgCl} \rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{–} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right] \left[{\mathrm{Cl}}^{–}\right]$

${\mathrm{Ag}}_2{\mathrm{CrO}}_4 \rightleftharpoons 2{\mathrm{Ag}}^{+}{\mathrm{CrO}}_4^{–}$       ${\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Ag}}^{+}\right]}^2\left[{\mathrm{CrO}}_4^{2–}\right]$

For a solution product of concentration of its cations and anions raised to their stoichiometric coefficients is called ionic product (IP).

If, K_sp < IP $\Rightarrow$ solution starts to precipitate.

If, K_sp > IP $\Rightarrow$ solution remains as such

and      if K_sp = IP $\Rightarrow$ solution exists in equilibrium with its precipitate.

Application of solubility product:

a) Predicting the formation of ppt.: A solution of  is added to a solution of 10^–7 M CaCl₂.  If K_sp for AgCl is 1.7 ´ 10^–10, predict whether precipitate will form.

K_sp = [Ag⁺] [Cl^–]

= 10^–3 $\times$ 2 $\times$ 10^–7                                          (For CaCl₂ [Cl^–] = 2 $\times$ [CaCl₂])

= 2 $\times$ 10^–10

Precipitation will take place.

b) Predicting solubility of sparingly soluble salts:

$$\begin{gathered} {\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}} \rightleftharpoons {\mathrm{xA}}^{\mathrm{y}+} + {\mathrm{yB}}^{\mathrm{x}–} \\ 1 0 0 \\ \left(1–\mathrm{s}\right) \mathrm{x}.\mathrm{s} \mathrm{y}.\mathrm{s} \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{A}}^{\mathrm{y}+}\right]{\left[{\mathrm{B}}^{\mathrm{x}–}\right]}^{\mathrm{y}} \\ ={\left(\mathrm{xs}\right)}^{\mathrm{x}} {\left(\mathrm{ys}\right)}^{\mathrm{y}} \end{gathered}$$

Ex.

$$\begin{gathered} \mathrm{AgCl} \rightleftharpoons {\mathrm{Ag}}^{+} + {\mathrm{Cl}}^{–} \\ 1 0 0 \\ 1–\mathrm{s} \mathrm{s} \mathrm{s} \end{gathered}$$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Cl}}^{–}\right]={\mathrm{s}}^2 \\ \mathrm{s}=\sqrt{{\mathrm{K}}_{\mathrm{sp}}} \end{gathered}$$

$$\begin{gathered} {\mathrm{Ag}}_2{\mathrm{CrO}}_4 \rightleftharpoons 2{\mathrm{Ag}}^{+} + {\mathrm{CrO}}_4^{2–} \\ 1 0 0 \\ 1–\mathrm{s} \mathrm{s} \mathrm{s} \\ {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Ag}}^{+}\right]}^2\left[{\mathrm{CrO}}_4^{2–}\right]={\left(2\mathrm{s}\right)}^2.\mathrm{s}=4{\mathrm{s}}^3 \\ \mathrm{s}=\sqrt[3]{\raisebox{1ex}{${\mathrm{K}}_{\mathrm{sp}}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \end{gathered}$$

$$\begin{gathered} {\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2 \rightleftharpoons 3{\mathrm{Ca}}^{2+} + 2{\mathrm{PO}}_4^{3–} \\ 1 0 0 \\ 1–\mathrm{s} 3\mathrm{s} 2\mathrm{s} \\ {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Ca}}^{2+}\right]}^3 {\left[{\mathrm{PO}}_4^{3–}\right]}^2={\left(3\mathrm{s}\right)}^3.{\left(2\mathrm{s}\right)}^2=108{\mathrm{s}}^5 \\ \mathrm{s}=\sqrt[5]{\frac{{\mathrm{K}}_{\mathrm{sp}}}{108}} \end{gathered}$$

c) Purification of common salt : This is achieved by passing dry HCl gas through brine (conc. NaCl) solution.  Since excess Cl^– are now available, [H⁺]. [Cl^–] i.e., ionic product becomes greater than K_sp; NaCl precipitates out and can be filtered off from the solution.

$\mathrm{NaCl} \rightleftharpoons {\mathrm{Na}}^{+} + {\mathrm{Cl}}^{–}$

$\mathrm{HCl} \rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{Cl}}^{–}$             sufficient Cl^– ions to ppt NaCl

d) Salting out of soap: Soaps are Na or K salts of higher fatty acids.  They can be made to ppt by adding NaCl or KCl.

$\mathrm{RCOONa} \rightleftharpoons {\mathrm{RCOO}}^{–}+{\mathrm{Na}}^{+}$

$\mathrm{NaCl} \rightleftharpoons {\mathrm{Cl}}^{–}+{\mathrm{Na}}^{+}$  effective concentration of Na⁺ increases

IP of soap i.e., $\left[{\mathrm{RCOO}}^{–}\right]\left[{\mathrm{Na}}^{+}\right]$ becomes greater than K_sp and hence the soap precipitates out.

e) Inorganic analysis:

The complete analysis of basic radicals excluding zero group (NH₄⁺) is based upon solubility product.

i) 1^st group: Pb²⁺, Ag⁺ and Hg₂²⁺ are precipitated in this group as chlorides.  K_sp of the chlorides of above ions are comparatively lower than the other radicals.  So when dil HCl is added, sufficient Cl^– are produced to precipitate these radicals, while those of all the other radicals/ions being highly soluble (with much higher K_sp values) remain in the solution in dissolved state.

ii) IInd group: IInd group and IV group both are precipitated as sulphides. But to limit the precipitation of only IInd group and not those of Vth or any higher group (III, V, VI etc) we make use of their solubility products. For IInd group radicals HCl is used along with H₂S, so as to produce only limited number of S^2–.

K_sp of IInd group            –    10^–72 – 10^–30

K_sp of IVth group           –    10^–28 – 10^–18

Comparison of the K_sp values of II & IV groups reveals that the K_sp of IInd group members are less than IV group.  So, they get precipitated even in the presence of these small number of ions present in the solution.

HNO₃ and H₂SO₄ can’t be used in the place of HCl here to reduce concentration of S^{2 –}; since they are oxidizing agents and oxidize H₂S to S.

iii) IIIrd group: Al³⁺, Fe³⁺ & Cr³⁺ are precipitated as hydroxides in presence of NH₄OH & NH₄Cl.  They are selectively precipitated as hydroxides using NH₄OH + NH₄Cl (called group reagent) ; since more OH^– may be produced in the solution otherwise which can precipitate even IV or higher group radicals/ions.  Their concentration is controlled (decreased) by using common ion effect, permitting to precipitate only IIIrd group hydroxides.

Illustration 22:

The solubility of AgCl in water at 25⁰C is 0.00179g L^–1. Calculate its solubility product at the temperature.

Solution.

Solubility = $\frac{0.00179\left({\mathrm{gL}}^{–1}\right)}{143.5\left({\mathrm{gmol}}^{–1}\right)}=0.0000125{\mathrm{molL}}^{–1}$

$\left[{\mathrm{Ag}}^{+}\right]=\left[{\mathrm{Cl}}^{–}\right]=\mathrm{solubility}=1.25\times {10}^{–5} {\mathrm{molL}}^{–1}$

K_sp [Ag⁺] [Cl^–] = (1.25 $\times$ 10^–5)² mol² L^–2

= 1.56 $\times$ 10^–10 mol^–2 L^–2

Illustration 23:

Calculate the solubility in gL^–1 of Al(OH)₃, if its solubility product is 8.5 $\times$ 10^–32.

Solution.

$\mathrm{Al}{\left(\mathrm{OH}\right)}_3\left(\mathrm{saturated} \mathrm{soln}\right) \rightleftharpoons {\mathrm{Al}}^{3+}\left(\mathrm{aq}\right)+3{\mathrm{OH}}^{–}\left(\mathrm{aq}\right)$

Let ‘s’ be the solubility,

$$\begin{gathered} {\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Al}}^{3+}\right]{\left[{\mathrm{OH}}^{–}\right]}^3=\left(\mathrm{s}\right){\left(3\mathrm{s}\right)}^3 \\ = 27{\mathrm{s}}^4 \end{gathered}$$

$\mathrm{s}=\frac{\left(8.5\times {10}^{–32}\right)}{27}=0.749\times {10}^{–8} {\mathrm{molL}}^{–1}$

$=0.749\times {10}^{–8}{\mathrm{molL}}^{–1}\times 78\mathrm{g} {\mathrm{mol}}^{–1}$

$=5.842\times {10}^{–7}{\mathrm{gL}}^{–1}$

Illustration 24 :

The solubility product of BaSO₄ is 1.5 $\times$ 10^-9.  Find its solubility in pure water and 0.1M BaCl₂ solution.

Solution.

${\mathrm{BaSO}}_4 \rightleftharpoons {\mathrm{Ba}}^{2+}+{\mathrm{SO}}_4^{2–}$

Let solubility be ‘s’ mol L^–1

$\mathrm{Ksp}=\left[{\mathrm{Ba}}^{2+}\right]\left[{\mathrm{SO}}_4^2\right]={\mathrm{s}}^2$

$1.5\times {10}^{–19}=s^2$

$\mathrm{s}=3.87\times {10}^{–5}{\mathrm{molL}}^{–1}$   (inpure water)

Let solubility in 0.1 M BaCl₂ be s’

Total Ba²⁺ concentration (S’ + C) mol L^–1

$S{O}_4^{2–}$ concentration (S’ + C) mol L^–1

Ksp = (s’ + C)s’

= s’ ² + s’ C

$\mathrm{s}{‘}^2+0.1\mathrm{s}‘=1.5\times {10}^{–9}$

BaSO₄ is sparingly soluble, hence neglecting the s’²

$0.1s‘=1.5\times {10}^{–9}$

$\mathrm{s}‘=1.5\times {10}^{–8} {\mathrm{molL}}^{–1} \left(\mathrm{in} {\mathrm{BaCl}}_2 \mathrm{soln}\right)$

llustration 25 :

The solubility of AgCl is $1.6\times {10}^{–10}$ moles²/lit².  What is its solubility?

Solution.

$K_{sp}=s^2 or s=\sqrt{1.6\times {10}^{–10}}=1.3\times {10}^{–5}$  moles/litre

Illustration 26 :

The solubility of Ag₂CrO₄ is $1.3\times {10}^{–4}$ mol/lit.  What is the solubility product?

Solution.

${\mathrm{Ag}}_2{\mathrm{CrO}}_4 \rightleftharpoons \underset{2\mathrm{S}}{2{\mathrm{Ag}}^{+}}+\underset{\mathrm{S}}{{\mathrm{CrO}}_4^{–2}}$

${\mathrm{K}}_{\mathrm{sp}}=4{\mathrm{S}}^3=4\times {\left(1.3\times {10}^{–4}\right)}^3=9\times {10}^{–12}$

Exercise 3:

(i) Solubility product of a sparingly soluble salt AB at a temperature is 1.21 ⨯ 10^–6 mol² L^–2.  Its molar solubility will be

(A) $1.21\times {10}^{–6} {\mathrm{molL}}^{–2}$          (B) $1.1\times {10}^{–4} \mathrm{M}$          (C) $1.1\times {10}^{–3} \mathrm{M}$          (D) $1.21\times {10}^{–3} \mathrm{M}$

(ii) The solubility product of a sparingly soluble salt is 1.08 ⨯ 10^–13 mol³ L^–3.  It its molecular mass is 125, the g L^–1 solubility will be

(A) 3 ⨯ 10^–5          (B) 3.75 ⨯ 10^–7          (C)  3.75 ⨯ 10^–5          (D) 3.75 ⨯ 10^–5

(iii) The solubility of KAl (SO₄)₂ is x mol L^–1, its K_sp will be

(A) x³          (B) 4x³          (C) x⁴          (D) 4x⁴

Answer to Exercises

Exercise – 1

i) C

ii) B

iii) C

iv) B

v) C

Exercise – 2

i) C

ii) A

Exercise – 3

i) C

ii) B

iii) D

FORMULAE AND CONCEPTS AT A GLANCE

1. $\mathrm{K}=\frac{\left[{\mathrm{A}}^{+}\right]\left[{\mathrm{B}}^{–}\right]}{\left[\mathrm{AB}\right]}$

2. Degree of Dissociation, $\mathrm{\alpha }=\frac{\mathrm{No}.\mathrm{of} \mathrm{molecules} \mathrm{dissociated}}{\mathrm{Total} \mathrm{no}.\mathrm{of} \mathrm{molecules}}$

3. $\mathrm{K}=\frac{{\mathrm{\alpha }}^2.\mathrm{C}}{\left(1–\mathrm{\alpha }\right)} \mathrm{\alpha }=\sqrt{\mathrm{KV}}$

4. Relative strength = $\frac{\mathrm{strength} \mathrm{of} {\mathrm{Acid}}_1}{\mathrm{strength} \mathrm{of} {\mathrm{Acid}}_2}$

$\frac{{\mathrm{\alpha }}_1}{{\mathrm{\alpha }}_2}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}1}}{{\mathrm{K}}_{\mathrm{a}2}}}$     K_a = Dissociation constant for an acid

$\frac{{\mathrm{\alpha }}_1}{{\mathrm{\alpha }}_2}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{b}1}}{{\mathrm{K}}_{\mathrm{b}2}}}$     K_b = Dissociation constant for a base

5. K_w = [H⁺] [OH^–] K_w = Ionic product of water = K [H₂O]

6. pH = – log[H⁺]              [H⁺] = 10^-pH; pH=7

7. pK_a = – log K_a   (for acids)

     pK_b = – log K_b  (for bases)

8. pH of a Buffer solution (Handerson Equation)

For an Acidic Buffer pH = ${\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$

For a basic Buffer pOH = ${\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]}$

9. $\mathrm{\varphi }=\frac{\mathrm{n}}{∆\mathrm{pH}.\mathrm{V}}$

10. $\mathrm{h}=\frac{\mathrm{no}.\mathrm{of} \mathrm{moles} \mathrm{hydrolysed}}{\mathrm{total} \mathrm{no}.\mathrm{of} \mathrm{moles} \mathrm{dissolved} \mathrm{in} \mathrm{solution}}$

11. ${\mathrm{K}}_{\mathrm{h}}=\frac{\left[\mathrm{HA}\right] \left[\mathrm{BOH}\right]}{\left[\mathrm{BA}\right]}=\frac{{\mathrm{Ch}}^2}{\left(1–\mathrm{h}\right)}$     (${\mathrm{K}}_{\mathrm{h}}, \mathrm{hydrolysis} \mathrm{constant} = \mathrm{K}\left[{\mathrm{H}}_2\mathrm{O}\right]$)

12. K_h for salt of strong acid and weak base $\mathrm{pH}=\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{w}}–{\mathrm{pK}}_{\mathrm{b}}–\log \mathrm{C}\right]$

13. K_h for salt of weak acid and strong base $\mathrm{pH}=\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{a}}+{\mathrm{pK}}_{\mathrm{w}}+\log \mathrm{C}\right]$

14. K_h for salt of weak acid and weak base $\mathrm{pH}=\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{w}}+{\mathrm{pK}}_{\mathrm{a}}–{\mathrm{pK}}_{\mathrm{b}}\right]$

15. $\therefore {\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{A}}^{\mathrm{y}+}\right]}^{\mathrm{x}} {\left[{\mathrm{B}}^{\mathrm{x}–}\right]}^{\mathrm{y}}$        ${\mathrm{K}}_{\mathrm{sp}}$ = Solubility product = $\mathrm{K}.\left[{\mathrm{A}}_{\mathrm{x}}{\mathrm{B}}_{\mathrm{y}}\right]$

16. In general for A_mB_n type salt Ksp = mⁿn^m (s)^m+n

SOLVED PROBLEMS-1

Prob 1.

0.1M aq. solution of a weak acid is ionized upto 2%.  What will be the concentration of H⁺ and OH^–? (K_w = 1 ⨯ 10^–14)

Sol.

For weak acids, $\left[{\mathrm{H}}^{+}\right]=\mathrm{C}.\mathrm{\alpha }=0.1\times 2\times {10}^{–2}$

= 2 ⨯ 10^–3

$\left[{\mathrm{OH}}^{–}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{H}}^{+}\right]}=\frac{1\times {10}^{–14}}{2\times {10}^{–3}}$

= 5.0 ⨯ 10^–12

Prob 2.

What is the initial concentration of a solution of CH₃COOH, if its [H⁺] is 3.4 ⨯ 10^–4? (K_a = 1.7⨯10^–5)

Sol.

$\left[{\mathrm{H}}^{+}\right]={\left(\raisebox{1ex}{${\mathrm{K}}_{\mathrm{a}}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{C}$}\right.\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\mathrm{C}=\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{{\mathrm{K}}_{\mathrm{a}}}=\frac{{\left(3.4\times {10}^{–4}\right)}^2}{1.7\times {10}^{–5}}$

= 6.8 ⨯ 10^–3 M

Prob 3.

In a mixture of weak acid and its salt, the ratio of the concentration of acid to salt is increased tenfolds.  What will be the change in pH?

Sol.

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]} \because \log \frac{1}{10}=–1$

pH will decrease by 1

Prob 4.

Calculate % hydrolysis of  NaCN in M/80 solution when dissociation constant for HCN is  1.3⨯10^–9.

Sol.

${\mathrm{CN}}^{–}+{\mathrm{H}}_2\mathrm{O} \rightleftharpoons \mathrm{HCN}+{\mathrm{OH}}^{–}$

$\mathrm{h}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}.\mathrm{C}}}$

$=\sqrt{\frac{1.0\times {10}^{–14}}{1.3\times {10}^9\times {\displaystyle \frac{1}{80}}}}=0.0248$

% hydrolysis = 2.48

Prob 5.

How K_sp of a sparingly soluble salt XY₄ is related with its (molar) solubility?

Sol.

                               ${\mathrm{XY}}_4 \rightleftharpoons {\mathrm{X}}^{\mathrm{m}+}+4{\mathrm{Y}}^{\mathrm{n}–}$

At equilibrium      1 – s            s         4s

${\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{X}}^{\mathrm{m}+}\right] \left[{\mathrm{Y}}^{\mathrm{n}–}\right]$

= $s\times {\left(4s\right)}^4$

= 256s⁵

or      s = (K_sp/256)^1/5

Prob 6.

At 25⁰C, K_b for a weak base BOH is 1.0⨯10^–12.  What will be the [OH^–] in 0.01M aqueous soln. of the base?

Sol.

$\underset{\mathrm{C}\left(1–\mathrm{\alpha }\right)}{\mathrm{BOH}} \rightleftharpoons \underset{\mathrm{C\alpha }}{{\mathrm{B}}^{+}} + \underset{\mathrm{C\alpha }}{{\mathrm{OH}}^{–}}$          C = concentration, $\alpha$ = degree of dissociation

${\mathrm{K}}_{\mathrm{b}}=\frac{\left[{\mathrm{B}}^{+}\right] \left[{\mathrm{OH}}^{–}\right]}{\left[\mathrm{BOH}\right]} \left[{\mathrm{B}}^{+}\right]=\left[{\mathrm{OH}}^{–}\right]={\left[{\mathrm{OH}}^{–}\right]}^2$     Assuming [OH^–] to be practically unchanged

$\left[{\mathrm{OH}}^{–}\right]={\left(1.0\times {10}^{12}\times {10}^{–2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$= 1.0⨯10^–7

Prob 7.

40 mL of 0.1M NH₄OH is mixed with 20 mL of 0.1 M HCl.  What is the pH of the mixture?  [pK_b (NH₄OH)) = 4.74]

Sol.

Half of the NH₄OH will be neutralized by the acid.

Hence,

[Salt] = [ Base]       $\mathrm{pOH}={\mathrm{pK}}_{\mathrm{b}}+\log \frac{\left[\mathrm{salt}\right]}{\left[\mathrm{base}\right]}$    = 4.74 + log 1  = 4.74

pH = 14 – 4.74 = 9.26

Prob 8.

K_a for ascorbic acid (HAc) is 5⨯10^–5.  Calculate the [H⁺] and percent hydrolysis in an aqueous solution in which concentration of ascorbate ions is 0.02 M

Sol.

$\mathrm{h}=\sqrt{\frac{{\mathrm{K}}_{\mathrm{w}}}{\mathrm{C}.{\mathrm{K}}_{\mathrm{a}}}}=\sqrt{\frac{{10}^{–14}}{0.02\times 5\times {10}^{–5}}}={10}^{–4}$      % hydrolysis = 0.01

$\mathrm{pH}=\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{w}}+{\mathrm{pK}}_{\mathrm{a}}+\mathrm{logC}\right]$= 8.3          $=\frac{1}{2}\left[14+\left(–\log 5\times {10}^{–5}\right)+\log 0.02\right]=8.3$

[H⁺] = – antilog 8.3.                                  $\left[{\mathrm{H}}^{+}\right]=5\times {10}^{–9} \mathrm{M}$

Prob 9.

Solubility of calcium phosphate (Mol. mass, M) in water is ‘w’ g per 100 mL at 25⁰C.  Calculate its solubility product at 25⁰C.  Calculate

Sol.

Solubility,  $\mathrm{s}=\frac{10\mathrm{w}}{\mathrm{M}}{\mathrm{molL}}^{–1}$          ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2 \rightleftharpoons 3{\mathrm{Ca}}^{2+}+2{\mathrm{PO}}_4^{3–}$

K_sp = (3s)³ ⨯ (2s)²                                           = 108 s⁵ = $108{\left(\frac{10\mathrm{w}}{\mathrm{M}}\right)}^5$

Prob 10.

The solubility product of lead iodide is 1.4⨯0^–8.  Calculate its molar solubility in 0.1 M KI solution.

Sol.

Let the solubility of PbI₂ be s

PbI₂ ƒ$\rightleftharpoons$  Pb²⁺ + 2I^–

s                  s         s

KI is strong electrolyte and hence is completely ionized. Concentration of I^– provided by it = 0.1 M.

$\left[{\mathrm{Pb}}^{2+}\right]=\mathrm{s}$                                   $\left[{\mathrm{I}}^{–}\right]=\left(2\mathrm{s}+01\right)\mathrm{M}$

K_sp = [Pb²⁺] [I^–]²                                = s ⨯ (2s + 0.1)²

= s ⨯ (4s² + 0.4s + 0.01)           = 4s³ + 0.4s² + 0.01s

Since s itself is smaller (PbI is sparingly soluble in water), s² & s³ can be neglected.

$1.4\times {10}^{–8}=0.01s$

$\mathrm{s}=\frac{1.4\times {10}^{–8}}{{10}^{–2}}=1.4\times {10}^{–6} {\mathrm{molL}}^{–1}$

SOLVED PROBLEMS-2

Prob 1.

Ostwald’s dilution law is valid for

(A) strong electrolyte

(B) weak electrolyte

(C) non-electrolyte

(D) strong as well as well electrolytes

Sol.

(B) The degree of dissociation of weak electrolytes increases considerably with dilution (decrease in concentration).  While non-electrolytes don’t dissociate at all, strong electrolytes dissociate completely at almost all ranges of concentrations.  Hence almost no variation is observed for change in concentration with strong electrolytes.

Prob 2.

Formic acid is 4.5% dissociated in a 0.1N solution at 20⁰C.  The ionization constant for the acid is

(A) 10^–10          (B) 10^–8          (C) 10^–6          (D) 10^–4

Sol.

(D) $\mathrm{K}=\frac{{\mathrm{\alpha }}^2\mathrm{C}}{1–\mathrm{\alpha }}=\frac{4.5\times 4.5\times 0.1}{100\times 100}=2.12\times {10}^{–4}$

Prob 3.

One litre of water contains 10^–7 moles of H⁺.  The degree of dissociation (ionization) of water is

(A) 1.8 ⨯ 10^–7%          (B) 0.8 ⨯ 10^–9%          (C) 3.6 ⨯ 10^–9%          (D) 3.6 ⨯ 10^–7%

Sol.

(D) The number of moles in one litre of water = $\frac{1000}{18}$

Out of 1000/18 moles of water, the ionized ones are, 10^-7

Out of 1000/18 moles of water, the ionized once will be = $\frac{{10}^{–7}}{{\displaystyle \raisebox{1ex}{$1000$}\!\left/ \!\raisebox{-1ex}{$18$}\right.}}$

= $\frac{18}{1000}\times {10}^{–7}$ = 1.8 ⨯ 10^–9

% dissociation = 1.8 ⨯ 10^–7

Prob 4.

The degree of dissociation of weak electrolytes increases on

(A)  increasing dilution

(B)  decreasing dilution

(C)  increasing concentration

(D)  decreasing pressure

Sol.

(A) ${\mathrm{\alpha }}^2=\sqrt{\frac{\mathrm{K}}{\mathrm{C}}}$ i.e., $\alpha \propto$ 1/C, i.e., $\alpha$ is inversely proportional to C that infers the degree of dissociation is directly proportional to dilution (as, concentration $\propto$ 1/dilution)

Prob 5.

When NH₄Cl is added to NH₄OH solution, the dissociation of NH₄OH is reduced due to

(A) oxidation

(B) reduction

(C) common ion effect

(D) hydrolysis

Sol.

(C) NH₄Cl as well as NH₄OH on dissociation produce NH₄⁺.  Since NH₄Cl is strong electrolyte, the almost of the NH₄⁺ ions produced in the solution would come from NH₄Cl only, thereby reducing the dissociation of NH₄OH, a weak electrolyte.

Prob 6.

Cl^– ion is the conjugate base of

(A) HCl          (B) HOCl          (C) HClO₃          (D) HClO₄

Sol.

(A) Acids after dissociation produce H⁺ and a conjugate base of the acid.

$\underset{\left(\mathrm{acid}\right)}{\mathrm{HCl}} \rightleftharpoons {\mathrm{H}}^{+}+\underset{\left(\mathrm{conjugate} \mathrm{base}\right)}{{\mathrm{Cl}}^{–}}$

Prob 7.

Phenolphthalein gives a pink colour in alkaline solution due to the fact that

(A) it is a coloured compound

(B) it ionizes to give coloured ions

(C) it is decomposed by alkali

(D) it forms a complex compound with alkali

Sol.

(B)  Phenolphthalein is weak acid.  In acidic medium its ionization is suppressed due to common ion effect, but in alkaline medium, OH^– produced by the base react with the H⁺ of the phenolphthalein and $\underset{\left(\mathrm{Colouless}\right)}{\mathrm{Hph}} \rightleftharpoons {\mathrm{H}}^{+}+\underset{\left(\mathrm{Pink}\right)}{{\mathrm{Ph}}^{–}}$

shifts the equilibrium towards right.  The Ph^– ions thus produced are pink which make the solution coloured.

Prob 8.

What will happen if 1.0M solution of a weak acid is diluted to 0.1M at constant temperature?

(A) [H⁺] will decrease to 0.01 M

(B) pH will decrease by 2 units

(C) Ka will increase

(D) percentage ionization will increase

Sol.

(D) For strong electrolytes which are 100% ionized, [H⁺] will decrease to 0.01M and pH will increase.  For weak electrolytes, the degree of ionization increases with dilution, K_a being constant for a given temperature.

Prob 9.

The aqueous solution of AlCl₃ is acidic due to

(A) cationic hydrolysis

(B) anionic hydrolysis

(C) hydrolysis of both the ions

(D) dissociation

Sol.

(A) AlCl₃ dissociates in water in Al³⁺ ions and Cl^–.  Since Cl^– are very stable being conjugate base of strong acid, they don’t get hydrolysed.  Al³⁺ ion are not very stable in aq. medium, they undergo hydrolysis to form strong Al – OH bonds, thereby releasing H⁺ which make the solution acidic.

Prob 10.

Which among the following species can act both as an acid as well as a base?

(A) PO₄^3–          (B) SO₄^2–          (C) HSO₄^–          (D) NH₃

Sol.

(C) ${\mathrm{HSO}}_4^{–}$ has an ionizable H atom, i.e.,

${\mathrm{HSO}}_4^{–} \rightleftharpoons {\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2–}$

Hence it behaves as an acid.  At the same time, it is produced by the deprotonation of H₂SO₄ (an acid), to, it is the conjugate base of the acid.