The degree of hydrolysis for a salt of strong acid and weak base

Increases with decrease in K b of the bases

Decreases with decrease in temperature

Table of Contents

What Is Hydrolysis?

Salts are strong electrolytes and ionise fully to form ions. These ions may react with water and form their conjugate pairs. Depending on the strength of the pairs, the cation or anion may undergo hydrolysis resulting in an increase in the hydrogen or hydroxide ion concentrations, making the solution acidic or basic or neutral. So, it is the reverse of the neutralisation reaction.

Consider hydrolysis of the ester.

CH₃COOCH₃ + H₂O → CH₃COOH + CH₃OH

Methyl acetate + Water → Acetic acid + Methanol

Consider a salt BA ionises as

\begin{array}{l} B A ⇌ B^{+} + A^{-} \end{array}

Cation hydrolysis

\begin{array}{l} B^{+} + 2 H_{2} O ⇌ B O H + H_{3} O^{+} \end{array}

Anion hydrolysis

\begin{array}{l} A^{-} + H_{2} O ⇌ H A + O H^{-} \end{array}

Hydrolysis of salt

\begin{array}{l} S a l t + W a t e r ⇌ A c i d + B a s e \end{array}

Factors Determining the Extend of Hydrolysis

Complete Hydrolysis

If the cation/anion and water are stronger than their conjugate pairs, B⁺ is more acidic than the conjugate hydronium ion, water is more basic than conjugate BOH, A^– is more basic than the conjugate hydroxide ion, and water is more acidic than conjugate HA.

Example: Ion that undergoes complete hydrolysis.

$\begin{array}{l} P H_{4}^{+} + H_{2} O \to P H_{3} + H_{3} O^{+} \end{array}$
form acid solution.
NH₂^–, H^–form a basic solution.
$\begin{array}{l} H^{-} + H_{2} O \to H_{2} + O H^{-} \end{array}$
form basic solution.

No Hydrolysis

If the cation/anion and water are weaker than their conjugate pairs, B⁺ is less acidic than the conjugate hydronium ion and conjugate BOH is more basic than water, A^– is less basic than the conjugate hydroxide ion, and water is less acidic than conjugate HA.

Example: Ion that does not undergo complete hydrolysis.

Cations of strong bases: Alkali and alkaline metal ions.
Anions of strong acids: Chloride, nitrate, sulphate, phosphate chlorate ions.

⇒ Na⁺ + H₂O NaOH + H₃O⁺

⇒ Cl^– + H₂O HCl + OH^–

Limited Hydrolysis

When the cations or anions are not so strong compared to their conjugate pairs, hydrolysis relative to strength will take place, and accordingly, the solution may be acidic or basic.

Ions undergoing limited hydrolysis: Cations of weak bases result in an acidic solution. Anions of weak acids yield basic solutions.

NH₄⁺, Al³⁺ CH₃COO^–, CN^–. C₂O₄^2-, PO₄^3-

\begin{array}{l} \Rightarrow N H_{4}^{+} + H_{2} O ⇌ N H_{4} O H + H_{3} O^{+} \end{array}

\begin{array}{l} \Rightarrow C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-} \end{array}

What Are Salts?

Acid and base react to form the salt. Salts are of four types.

Salt of the strong acid and strong base – NaCl
Salt of the strong acid and weak base – NH₄Cl, FeCl₃, CuCl₂, AlCl₃, CuSO₄
Salt of the weak acid and strong base – CH₃COONa, NaCN, NaHCO₃, Na₂CO₃
Salt of the weak acid and weak base – CH₃COONH_4,(NH₄)₂CO₃, NH₄HCO₃

a) pH Aqueous Solutions of Salt of a Strong Acid and Strong Base

Anions and cations of the strong acid and bases do not undergo hydrolysis with water, so the solution will be neutral, with pH = 7.

b) pH Aqueous Solutions of Salt of a Strong Acid and Weak Base

Examples: NH₄Cl, FeCl₃, CuCl₂, AlCl₃, CuSO₄

Only the cation of the weak base undergoes hydrolysis to make the solution acidic.

Say, ‘h’ be the degree of hydrolysis.

⇆ NH₄Cl NH₄⁺ + Cl^–

⇆ NH₄⁺ + H₂O NH₄OH + H⁺

At the start, concentration, mole/l C 0 0

At equilibrium C(1 – h) Ch Ch

\begin{array}{l} Hydrolysis constant = K_{h} = \frac{[N H_{4} O H] [H^{+}]}{[N H_{4}^{+}]} \frac{C h + C h}{C (1 - h)} = \frac{c h^{2}}{(1 - h)} = c h^{2} \end{array}

h is small, so (1 – h) = 1

\begin{array}{l} h = \sqrt{\frac{K h}{c}} [H^{+}] = c h = \sqrt{c K h} \end{array}

For the equilibrium,

\begin{array}{l} N H_{4} O H ⇌ N H_{4}^{+} + O H^{-} Equilibrium constant of the base = K b = \frac{[N H_{4}^{+}] [O H^{-}]}{[N H_{4} O H]} \end{array}

For the equilibrium,

\begin{array}{l} H_{2} O ⇌ H^{+} + O H^{-} Ionic product of water = K w = [H^{+}] [O H^{-}] = 10^{- 14} \end{array}

\begin{array}{l} \frac{[N H_{4} O H] [H^{+}]}{[N H_{4}^{+}]} \times \frac{[N H_{4}^{+}] [O H^{-}]}{[N H_{4} O H]} = [H^{+}] [O H^{-}] \end{array}

Kh x Kb = Kw ;

\begin{array}{l} K h = \frac{K w}{K b} \end{array}

\begin{array}{l} S o [H^{+}] = c h = \sqrt{c K h} = \sqrt{c \frac{K w}{K b}} = - \frac{1}{2} [\log K w - \log K b + \log c] \end{array}

pH aqueous solutions of salt of a strong acid and weak base

\begin{array}{l} p H = \frac{1}{2} [p K w - p K b - \log c] \end{array}

c) pH Aqueous Solutions of Salt of a Weak Acid and Strong Base

Examples: CH₃COONa, NaCN, NaHCO₃, Na₂CO₃

\begin{array}{l} C H_{3} C O O N a ⇌ N a^{-} + C H_{3} C O O^{-} \end{array}

$\begin{array}{l} C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-} \end{array}$

At the start, concentration, mole/l C 0 0

At equilibrium C(1-h) Ch Ch

\begin{array}{l} Hydrolysis constant = K h = \frac{[C H_{3} C O O H] [O H^{-}]}{[C H_{3} C O O^{-}]} = \frac{(c h + c h)}{c (1 + h)} = \frac{c h^{2}}{(1 - h)} = C H^{2} \end{array}

h is small, so (1-h)=1

\begin{array}{l} h = \sqrt{\frac{K h}{c}}; [O H^{-}] = c h = \sqrt{c K h} \end{array}

For the equilibrium,

\begin{array}{l} C H_{3} C O O H ⇌ C H_{3} C O O^{-} + H^{+} Equlibrium constant of acid, K a = \frac{[C H_{3} C O O^{-}] [H^{+}]}{C H_{3} C O O H} \end{array}

For the equilibrium,

\begin{array}{l} H_{2} O^{+} + O H^{-} Ionic product of water = K w = [H^{+}] [O H^{-}] = 10^{- 14} \end{array}

\begin{array}{l} \frac{[C H_{3} C O O H] [O H^{-}]}{[C H_{3} C O O^{-}]} \times \frac{[C H_{3} C O O^{-}] [H^{+}]}{[C H_{3} C O O H]} = [H^{+}] [O H^{-}] \end{array}

Kh x Ka = Kw ; Kh = Kw/Ka

So,

\begin{array}{l} [0 H^{-}] = c h = \sqrt{c K h} = \sqrt{c \frac{K w}{K a}} \end{array}

\begin{array}{l} = - \log \sqrt{c \frac{K w}{K a}} \\ = - \frac{1}{2} [\log K w - \log K a + \log c] \\ p O H = \frac{1}{2} [p K w - p K a - \log c] \end{array}

\begin{array}{l} p H = 14 - p O H = 14 - \frac{1}{2} [p K w - p K a - \log c] \\ = 14 + \frac{1}{2} [- p K w + p K a + \log c] \end{array}

$\begin{array}{l} p H = \frac{1}{2} [28 - p K w + p K a + \log c] \\ = \frac{1}{2} [2 p K w - p K w + p K a + \log c] \\ = \frac{1}{2} [p K w + p K a + \log c] \end{array}$

pH aqueous solution of salt, weak acid and strong base,

\begin{array}{l} p H = \frac{1}{2} [p K w + p K b + \log c] \end{array}

d) pH Aqueous Solutions of Salt of a Weak Acid and Weak Base

Examples: CH₃COONH_4,(NH₄)₂CO₃, NH₄HCO₃

\begin{array}{l} C H_{3} C O O N H_{4} ⇌ N H_{4}^{+} + C H_{3} C O O^{-} \end{array}

\begin{array}{l} C H_{3} C O O^{-} + N H_{4}^{+} + H_{2} O ⇌ C H_{3} C O O H + N H_{4} O H \end{array}

At the start, concentration, mole/l C C 0 0

At equilibrium C(1-h) C(1-h) Ch Ch

\begin{array}{l} Hydrolysis constant = K h = \frac{[C H_{3} C O O H] [N H_{4} O H]}{[C H_{3} C O O^{-}] [N H_{4}^{+}]} = \frac{(c h + c h)}{c (1 - h) c (1 - h)} = \frac{h^{2}}{((1 - h)^{2})}; \end{array}

\begin{array}{l} K h = \frac{h^{2}}{1 - h}; \frac{h}{1 - h} = \sqrt{K h} \end{array}

For the equilibrium,

\begin{array}{l} C H_{3} C O O H ⇌ C H_{3} C O O^{-} + H^{+} \end{array}

\begin{array}{l} Equilibrium constant of acid = K a = \frac{[C H_{3} C O O^{-}] [H^{+}]}{[C H_{3} C O O H]} [H^{+}] = K a \times \frac{[C H_{3} C O O H]}{[C H_{C} O O^{-}]} \end{array}

\begin{array}{l} [H^{+}] = K a \frac{c h}{c (1 - h)} = K a \frac{h}{1 - h} = K a \sqrt{k h} \end{array}

For the equilibrium,

\begin{array}{l} N H_{4} O H ⇌ N H_{4}^{+} + O H^{-} Equilibrium constant of the base = K b = \frac{[N H_{4}^{-}] [O H^{-}]}{N H_{4} O H} \end{array}

For the equilibrium,

\begin{array}{l} H_{2} O ⇌ H^{+} + O H^{-} Ionic product of water = K w = [H^{+}] [O H^{-}] = 10^{- 14} \end{array}

\begin{array}{l} \frac{[C H_{3} C O O H] [N H_{4} O H]}{[C H_{3} C O O^{-}] [N H_{4}^{+}]} \times \frac{[C H_{3} C O O^{-}] [H^{+}]}{[C H_{3} C O O H]} \times \frac{[N H_{4}^{-}] [O H^{-}]}{N H_{4} O H} = [H^{+}] [O H^{-}] \end{array}

Kh x Ka x Kb = Kw ;

\begin{array}{l} K h = \frac{K w}{K a K b} \end{array}

\begin{array}{l} [H^{+}] = K a \sqrt{K h} = K a \frac{K w}{K a K b} = \sqrt{\frac{K w K a}{K b}} \end{array}

$\begin{array}{l} p H = - \log [H^{+}] \\ = - \log [\sqrt{\frac{K w K a}{K b}}] \\ = - \frac{1}{2} [\log K w + \log K a - \log K b] \\ = \frac{1}{2} [p K w + p K a - p K b] \end{array}$

pH aqueous solutions of salt of a weak acid and strong base

\begin{array}{l} p H = \frac{1}{2} [p K w + p K b - p K b] \end{array}

e) pH of Polyprotic Acids and Their Salts

1) pH of polyprotic acids:

Acids that can ionise to give two or more hydrogen ions are polyprotic acids.

Examples: Sulphuric, phosphoric, carbonic and oxalic acid.

These acids ionise in steps. But the ionisation may stop with the first ionisation unless it is required.

Due to common ions produced in the earlier steps.

Orthophosphoric acid can ionise in three steps, two yielding three hydrogen ions as follows.

$\begin{array}{l} H_{3} P O_{4} ⇌ H^{+} + H_{2} P O_{4}^{-} K_{a_{1}} = \frac{[H^{+}] [H_{2} P O]_{4}^{1}}{[H_{3} P O_{4}]} \end{array}$

\begin{array}{l} H_{2} P O_{4}^{-} ⇌ H^{+} + H P O_{4}^{-} K_{a_{2}} = \frac{[H^{+}] [H P O_{4}^{2 -}]}{[H_{2} P O_{4}^{-}]} \end{array}

\begin{array}{l} H_{2} P O_{4}^{-} ⇌ H^{+} P O_{4}^{-} K a 3 = \frac{[H^{+}] [H P O_{4}^{3 -}]}{[H P O_{4}^{2 -}]} \end{array}

For all acids, K_a₁ >> K_a2 >> K_a3….

So, the pH of the acids is calculated on the first ionisation constant only.

For weak acids,

\begin{array}{l} K_{a_{1}} = \frac{[H^{+}] [H_{2} P O_{4}^{3 -}]}{[H_{3} P O_{4}]} = \frac{C h - C h}{C (1 - h)} = \frac{c h^{2}}{(1 - h)}, h <<< 1, (1 - α) = 1 \end{array}

\begin{array}{l} ∴ K_{a_{1}} = C h^{2} h = \sqrt{\frac{K a 1}{c}} [H^{+}] o f i o n = C h \end{array}

and

\begin{array}{l} p H = - \log \sqrt{C K a 1} = \frac{1}{2} (- \log K a 1 - \log C); p H = \frac{1}{2} (p K a l - \log C) \end{array}

2) pH of salts of polyprotic acids:K₃PO₄, Na₂CO₃, FeCl₃, (NH₄)C₂O₄

Salts will be completely ionised

\begin{array}{l} K h 1 = \frac{[O H^{-}] [H P O_{4}^{2 -}]}{[P O_{4}^{3 -}]} \end{array}

Phosphate hydrolyzes as follows:

\begin{array}{l} P O_{4}^{3 -} + H_{2} O ⇌ [H P O_{4}^{2 -}] + [O H^{-}] \end{array}

\begin{array}{l} K h 2 = \frac{[O H^{-}] [H_{2} P O_{4}^{-}]}{[H P O_{4}^{2 -}]} \end{array}

\begin{array}{l} [H P O_{4}^{2 -}] + [H_{2} O] ⇌ [H_{2} P O_{4}^{-}] + [O H^{-}] \end{array}

\begin{array}{l} K h 3 = \frac{[O H^{-}] [H_{3} P O_{4}]}{[H_{2} P O_{4}^{-}]} \end{array}

\begin{array}{l} [H_{2} P O_{4}^{-}] + [H_{2} O] ⇌ [H_{3} P O_{4}] + [O H] \end{array}

On hydrolysis, [PO₄^3-] + H₂O [HPO₄^2-] + [OH^–]

At equilibrium,

moles / l C (1-h) Ch Ch

\begin{array}{l} K h 1 = \frac{(C h + C h)}{C (1 - h)} = \frac{C h^{2}}{(1 - h)}, h <<< 1, (1 - h) = 1 \end{array}

\begin{array}{l} K h 1 = C h^{2}; h = \sqrt{\frac{K h 1}{C}}; [H^{+}] = \frac{K w}{O H^{[-]}} = \frac{K w}{\sqrt{C K h 1}} \end{array}

\begin{array}{l} H P O_{4}^{-} H^{+} + P O_{4}^{-} \end{array}

\begin{array}{l} K a 3 = \frac{[H^{+}] [P O_{4}^{3}]}{[H P O_{4}^{2 -}]} \end{array}

\begin{array}{l} K h 1 \times K a 3 = \frac{[O H^{-}] [H P O_{4}^{2 -}]}{[P O_{4}^{3 -}]} \frac{[H^{+}] [P O_{4}^{3 -}]}{[O H^{-}]} = [H^{+}] [O H^{-}] = K w \end{array}

Hence,

\begin{array}{l} K h 1 = \frac{K w}{K a 3} \end{array}

\begin{array}{l} [H^{+}] = \frac{K w}{\sqrt{C K h 1}} = \sqrt{\frac{K w K a 3}{C}}; p H = - \log \sqrt{\frac{K w K a 3}{C}} = - \frac{1}{2} (\log K w + \log K a 3 - \log C) \end{array}

$\begin{array}{l} p H = - \frac{1}{2} (p K w + p K a 3 + \log C) \end{array}$

The pH of salts of polyprotic acids,

\begin{array}{l} K_{3} P O_{4} = - \frac{1}{2} (p K w + p K a 3 + \log C) \end{array}

Hydrolysis of Amphoteric Anion

Examples: NaHCO3, NaHS

Amphoteric anions can ionise to give hydrogen ions and hydrolyze to give hydroxide ions.

⇒ Ionisation

\begin{array}{l} H C O_{3}^{-} + H_{2} O ⇌ C O_{3}^{2 -} + H_{3} O^{+} with K a 1 \end{array}

⇒ Hydrolysis

\begin{array}{l} H C O_{3}^{-} + H_{2} O ⇌ H_{2} C O_{3} + O H^{-} with K a 2 \end{array}

\begin{array}{l} p H_{(H C O_{3 -})} = \frac{p K a 1 + p K a 2}{2} \end{array}

What Is Hydrolysis?

Factors Determining the Extend of Hydrolysis

Complete Hydrolysis

No Hydrolysis

Limited Hydrolysis

What Are Salts?

a) pH Aqueous Solutions of Salt of a Strong Acid and Strong Base

b) pH Aqueous Solutions of Salt of a Strong Acid and Weak Base

c) pH Aqueous Solutions of Salt of a Weak Acid and Strong Base

d) pH Aqueous Solutions of Salt of a Weak Acid and Weak Base

e) pH of Polyprotic Acids and Their Salts

Hydrolysis of Amphoteric Anion

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