Salt Hydrolysis

Table of Content

Salt of a Strong Acid and a Weak Base
- Relation Between Hydrolysis Constant and Degree of Hydrolysis
Salt of a Weak Acid and a Strong Base
Salt of a Weak Acid and a Weak Base
Hydrolysis of Amphiprotic Anion
Hydrolysis at a Glance
Related Resources

Pure water is a weak electrolyte. Weak because it does not completely disassociates into hydrogen and hydroxide but exist in equilibrium with these two ions. It is neutral in nature, i.e., H⁺ion concentration is exactly equal to OH^- ion concentration

[H⁺] = [OH^-]

When this condition is disturbed by decreasing the concentration of either of the two ions, the neutral nature changes into acidic or basic. When [H⁺] > [OH^-], the water becomes acidic and when [H⁺] < [OH^-], the water acquires basic nature. This is exactly the change which occurs during the phenomenon known as salt hydrolysis. It is defined as a reaction in which the cation or anion or both of a salt react with water to produce acidity or alkalinity.

Salts are strong electrolytes. When dissolved in water, they dissociate almost completely into ions. In some salts, cations are more reactive in comparison to anions and these react with water to produce H⁺ ions. Thus, the solution acquires acidic nature.

M⁺ + H₂O $\rightleftharpoons$ MOH + H⁺

Weak base

In other salts, anions may be more reactive in comparison to cations and these react with water to produce OH^- ions. Thus, the solution becomes basic.

A^- + H₂O $\rightleftharpoons$ HA + OH^-

Weak acid

The process of salt hydrolysis is actually the reverse of neutralization.

Salt + Water $\rightleftharpoons$ Acid + Base

If acid is stronger than base, the solution is acidic and in case base is stronger than acid, the solution is alkaline. When both the acid and the base are either strong or weak, the solution is generally neutral in nature.

As the nature of the cation or the anion of the salt determines whether its solution will be acidic or basic, it is proper to divide the salts into four categories.

Salt of a strong acid and a weak base.
Examples: FeCl₃, CuCl₂, AlCl₃, NH₄Cl, CuSO₄, etc.
Salt of a strong base and a weak acid.
Examples: CH₃COONa, NaCN, NaHCO₃, Na₂CO₃, etc.
Salt of a weak acid and a weak base.
Examples: CH₃COONH₄, (NH₄)₂CO₃, NH₄HCO₃, etc.
Salt of a strong acid and a strong base.
Examples: NaCl, K₂SO₄, NaNO₃, NaBr, etc.

Refer to the following video for salt hydrolysis

Salt of a Strong Acid and a Weak Base

The solution of such a salt is acidic in nature. The cation of the salt which has come from weak base is reactive. It reacts with water to form a weak base and H⁺ ions.

B⁺ + H₂O $\rightleftharpoons$ BOH + H⁺

Weak base

Consider, for example, NH₄Cl. It ionises in water completely into NH₄ and CF ions. ions react with water to form a weak base (NH₄OH) and H⁺ ions.

NH⁺₄ + H₂O $\rightleftharpoons$ NH₄OH + H⁺

C(1-x) Cx Cx

Thus, hydrogen ion concentration increases and the solution becomes acidic.

Applying law of mass action,

$K_h = \frac{[Hx ][NH_4OH]}{[NH_4^+ ]} = \frac{(x^2 C)}{(1-x)}$ ...... (i)

where C is the concentration of salt and x the degree of hydrolysis.

Other equilibria which exist in solution are

NH₄OH $\rightleftharpoons$ NH⁺₄ + OH^-,

$K_b = \frac{[NH^+_4][OH^-]}{[NH_4OH]}$ .... (ii)

H₂O $\rightleftharpoons$ H⁺ + OH^-,

K_b = [H⁺][H^-] ..... (iii)

From eqs. (II) and (iii)

K_w/K_b =[H⁺ ][NH⁴ OH]/[NH₄⁺ ] = K_h .... (iv)

[H⁺] = [H⁺ ][NH₄⁺]/[NH₄OH] = K_w/K_b ×[NH₄⁺ ]/[NH₄ OH]

log [H⁺] = log K_w - log K_b + log[salt]/[base]

-pH = -pK_w + pK_b + log[salt]/[base]

pK_w - pH = pK_b + log[salt]/[base]

$pOH = pK_b + log\frac{[salt]}{[base]}$

Relation Between Hydrolysis Constant and Degree of Hydrolysis

The extent to which hydrolysis proceeds is expressed as the degree of hydrolysisand is defined as the fraction of one mole of the salt that is hydrolysed when the equilibrium has been attained. It is generally expressed as h or x.

h = (Amount of salt hydrolysed)/(Total salt taken)

Considering again eq. (i),

K_h = x²C/(1-x) or K_h = h²C/(1-h)

When h is very small (1-h) → 1,

H₂ = K_h × 1/c

or h = √(K_h/C) = √(K_w/K_b $\times$ C)

[H⁺] = h × C = √(C $\times$ K_h)/K_b

log [H⁺] = 1/2 log K_w + 1 1/2log C - 1/2log K_b

$pH = \frac{1}{2}pKw - \frac{1}{2} log C - \frac{1}{2} pKb = 7 - \frac{1}{2} pKb - \frac{1}{2}log C$

Salt of a Weak Acid and a Strong Base

The solution of such a salt is basic in nature. The anion of the salt is reactive. It reacts with water to form a weak acid and OH^- ions.

A^- + H₂O; $\rightleftharpoons$ HA + OH^-

Weak acid

Consider, for example, the salt CH₃COONa. It ionises in water completely to give CH₃COO^- and Na⁺ ions. CH₃COO^- ions react with water to form a weak acid, CH₃COOH and OH^- ions.

CH₃COO^- + H₂O $\rightleftharpoons$ CH₃COOH + OH^-

C(1-x) Cx Cx

Thus, OH^- ion concentration increases, the solution becomes alkaline.

Applying law of mass action,

K_h = [CH₃COOH][OH^-]/[CH₃CO^-] = (Cx×Cx)/C(1-x) = (Cx²)/(1-x) ) ...... (i)

Other equations present in the solution are:

CH₃COOH $\rightleftharpoons$ CH₃COO^- + H⁺,

K_a = [CH₃COO^-][H⁺]/[CH₃COOH] ...... (ii)

H₂O $\rightleftharpoons$ H⁺ + OH^-,

K_w = [H⁺][OH^-] ....... (iii)

From eqs. (ii) and (iii),

log [OH^-] = log K_w - log K_a + log[salt]/[acid]

-pOH = -pK_w + pK_a + log[salt]/[acid]

pK_w - pOH = pK_a + log[salt]/[acid]

$pH = pK_a + log\frac{[salt]}{[base]}$

Considering eq. (i) again,

K_h = cx²/(1-x) or K_h = Ch²/(1-h)

When h is very small, (1-h) → 1

or h² = K_h/C

or h = √K_h/C

[OH^-] = h × C = √(CK_h) = √(C $\times$ K_w/K_a)

[H⁺] = K_w/[OH^-]

= K_w/√(C*K_w/K_a) = √(K_a $\times$ K_w)/K_c

-log [H⁺] = -1/2log K_w - 1/2log K_a + 1/2log C

pH = 1/2pK_w + 1/2pK_a + 1/2log C

= 7 + 1/2pK_a + 1/2log C.

$pH = 7 - \frac{1}{2} pK_a - \frac{1}{2}log C$

Salt of a Weak Acid and a Weak Base

Maximum hydrolysis occurs in the case of such a salt as both the cation and anion are reactive and react with water to produce H⁺ and OFT ions. The solution is generally neutral but it can be either slightly acidic or slightly alkaline if both the reactions take place with slightly different rates. Consider, for example, the salt CH₃COONH₄. It gives CH₃COO^- and ions in solution. Both react with water.

salt-of-a-weak-acid-and-a-weak-base

Other equilibria which exist in solution are:

CH₃COOH $\rightleftharpoons$ CH₃COO^- + H⁺,

K_a = [CH³COO^-][H⁺]/[CH₃COOH] ..... (i)

NH₄OH $\rightleftharpoons$ NH⁺₄ + OH^-,

Kb = [NH⁺₄] [OH^-]/[NH₄OH] ..... (ii)

H₂O $\rightleftharpoons$ H⁺ + OH^-

K_w = [H⁺][OH^-] ..... (iii)

From Eqs. (i), (ii) and (iii),

K_h = Kw/K_a.K_b = [CH₃COOH][NH₄OH]/[CH₃COO^-][NH⁺₄] .... (iv)

Let C be the concentration and h be the degree of hydrolysis

K_h = h²/(1-h)²

When h is small, (1-h) → 1.

K_h = h²

h = √Kh = √K_w/K_a*K_b

[H⁺] K_a × h

= K_a × √K_w/K_a*K_b

= √K_w $\times$ K_a/K_b

-log [H⁺] = -1/2log K_a - 1/2log K_w + 1/2log K_b

pH = 1/2pK_a + 1/2pK_w - 1/2pK_b

= 7 + 1/2pK_a - 1/2pK_b

$\fn_phv pH = 7 + \frac{1}{2} pK_a - \frac{1}{2}log C$

When pK_a = pK_b, pH = 7, i.e., solution will be neutral in nature.

When pK_a > pK_b. The solution will be alkaline as the acid will be slightly weaker than base and pH value will be more than 7.

In case pK_a < pK_b, the solution will be acidic as the acid is relatively stronger than base and pH will be less than 7.

Salt of a Strong Acid and a Strong Base

Such a salt, say NaCl, does not undergo hydrolysis as both the ions are not reactive. The solution is thus, neutral in nature.

Hydrolysis of Amphiprotic Anion

Let us consider hydrolysis of amphiprotic anion only, i.e., when counter cation is not hydrolysed, example of some salts of this category are NaHCO₃, NaHS, Na₂HPO₄, NaH₂PO₄.

amphiprotic-anions

Here, H₂PO^-₄ and HPO^2-₄ are amphiprotic anions. pH after their hydrolysis can be calculated as,

pH of H₂PO^-₄ in aqueous medium = (pk_a1 + pk_a2)/2

pH of H₂PO^2-₄ in aqueous medium = (pk_a2 + pk_a3)/2

Here, H₂PO^2-₄ is conjugate base of H₂PO^-₄ and H₃PO₄ is conjugate acid of H₂PO^-₄.

Similarly, PO^3-_{4 i}s conjugate base of HPO^-2₄ and HPO^-₄ is conjugate acid of PO^3-₄ .

(iv) Let us consider amphiprotic bicarbonate anion.

hydrolysis-in-aqueous-medium

pH HCO^-₃of ion after hydrolysis in aqueous medium = (pk_a1 + pk_a2)/2

(v) Let us consider the hydrolysis of amphiprotic anion along with cation, e.g., NH₄HCO₃, NH₄HS.

In above examples both cations and anions are derived from weak base and weak acids respectively hence, both will undergo hydrolysis in aqueous medium.

When these salts are dissolved in water, [H₃O⁺] concentration can be determined as,

[H₃O⁺] = √k_a1[k_w/k_b + k_a2]

pH = -log = √k_a1[k_w/k_b + k_a2]

Hydrolysis at a Glance

Salt

Nature

Degree

Hydrolysis Constant

NaCl

(Strong acid + Strong Base)

2. Ch₃COONa (Weak acid + Strong base)

3. NH₄Cl

(Strong acid + Weak base)

4. CH₃COONH₄ (Weak acid + Weak base)

Neutral

Base

Acidic

No Hydrolysis

h = √kw/Cka

h = √kw/Ckb

h = √kw/(ka + kb)

K_h = kw/ka

K_h = kw/Ckb

K_h = kw/(ka + kb)

pH=1/2[pkw + pka + logC]

pH=1/2[pkw- pkb - logC]

pH=1/2[pkw + pka - pkb]

In the case of salt of weak acid and weak base, nature of medium after hydrolysis is decided in the following manner:

(i) If K_a = K_b, the medium will be neutral.

(ii) If K_a > K_b, the medium will be acidic.

(iii) If K_a < K_b, the medium will be basic.

The degree of hydrolysis of salts of weak acids and weak bases is unaffected by dilution because there is no concentration term in the expression of degree ofhydrolysis.

Note : Degree of hydrolysis always increases with increase in temperature because at elevated temperature increase in K_w is greater as compared to K_a and K_b.

Related Resources

To read more, Buy study materials of Chemical Equilibrium comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here.