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Hydrolysis, Salts and Types

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

\(\begin{array}{l}BA\rightleftharpoons B^{+} + A^{-}\end{array} \)

Cation hydrolysis

\(\begin{array}{l}B^{+}+2H_2O\rightleftharpoons BOH + H_{3}O^{+}\end{array} \)

Anion hydrolysis

\(\begin{array}{l}A^{-}+H_2O\rightleftharpoons HA + OH^{-}\end{array} \)

Hydrolysis of salt

\(\begin{array}{l}Salt + Water\rightleftharpoons Acid + Base\end{array} \)

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.

  • \(\begin{array}{l}PH_{4}^{+} + H_2O\rightarrow PH_3 + H_3O^{+}\end{array} \)
    form acid solution.
  • NH2, Hform a basic solution.
  • \(\begin{array}{l}H^{-} + H_2O\rightarrow H_2 + OH^{-}\end{array} \)
    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. C2O42-, PO43-

\(\begin{array}{l}\Rightarrow NH_{4}^{+}+H_2O\rightleftharpoons NH_4OH+H_3O^{+}\end{array} \)

\(\begin{array}{l}\Rightarrow CH_3COO^{-}+H_2O\rightleftharpoons CH_3COOH + OH^{-}\end{array} \)

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 – CH3COONH4, (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

\(\begin{array}{l}\text{Hydrolysis constant}=K_h=\frac{[NH_4OH][H^{+}]}{[NH_{4}^{+}]} \frac{Ch+Ch}{C(1-h)}=\frac{ch^{2}}{(1-h)} = ch^{2}\end{array} \)

h is small, so (1 – h) = 1

\(\begin{array}{l}h = \sqrt{\frac{Kh}{c}} \ [H^{+}] = ch = \sqrt{cKh}\end{array} \)

For the equilibrium,

\(\begin{array}{l}NH_4OH \rightleftharpoons NH_{4}^{+} + OH^{-}\ \text{Equilibrium constant of the base} = Kb = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_4OH]}\end{array} \)

For the equilibrium,

\(\begin{array}{l}H_2O \rightleftharpoons H^{+} + OH^{-}\ \text{Ionic product of water} = Kw = [H^{+}][OH^{-}]=10^{-14}\end{array} \)

\(\begin{array}{l}\frac{[NH_{4}OH][H^{+}]}{[NH_{4}^{+}]} \times \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{4}OH]} = [H^{+}][OH^{-}]\end{array} \)

Kh x Kb = Kw ;

\(\begin{array}{l}Kh= \frac{Kw}{Kb}\end{array} \)

\(\begin{array}{l}So\; [H^{+}]= ch=\sqrt{cKh} = \sqrt{c \frac {Kw}{Kb}} = – \frac{1}{2}[\log Kw – \log Kb + \log c]\end{array} \)

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

\(\begin{array}{l}pH = \frac{1}{2}[pKw-pKb-\log c]\end{array} \)

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

Examples: CH3COONa, NaCN, NaHCO3, Na2CO3

\(\begin{array}{l}CH_3COONa\rightleftharpoons Na^{-} + CH_3COO^{-}\end{array} \)

\(\begin{array}{l}CH_3COO^{-} + H_2O\rightleftharpoons CH_3COOH + OH^{-}\end{array} \)

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

At equilibrium C(1-h) Ch Ch

\(\begin{array}{l}\text{Hydrolysis constant} = Kh = \frac{[CH_3COOH][OH^{-}]}{[CH_3COO^{-}]} = \frac{(ch+ch)}{c(1+h)} = \frac{ch^{2}}{(1-h)} = CH^{2}\end{array} \)

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

\(\begin{array}{l}h = \sqrt{\frac{Kh}{c}}; [OH^{-}] = ch = \sqrt{cKh}\end{array} \)

For the equilibrium,

\(\begin{array}{l}CH_3COOH\rightleftharpoons CH_3COO^{-} + H^{+}\ \text{Equlibrium constant of acid}, Ka=\frac{[CH_3COO^{-}][H^{+}]}{CH_3COOH}\end{array} \)

For the equilibrium,

\(\begin{array}{l}H_{2}O^{+} + OH^{-}\ \text{Ionic product of water} = Kw = [H^{+}][OH^{-}]=10^{-14}\end{array} \)

\(\begin{array}{l}\frac{[CH_{3}COOH][OH^{-}]}{[CH_{3}COO^{-}]} \times \frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]} = [H^{+}][OH^{-}]\end{array} \)

Kh x Ka = Kw ; Kh = Kw/Ka

So,

\(\begin{array}{l}[0H^{-}] = ch = \sqrt{cKh} = \sqrt {c\frac{Kw}{Ka}}\end{array} \)

\(\begin{array}{l}=-\log \sqrt{c\frac{Kw}{Ka}}\\ = -\frac{1}{2}[\log Kw-\log Ka +\log c]\\ pOH = \frac{1}{2}[pKw – pKa – \log c]\end{array} \)

\(\begin{array}{l}pH = 14-pOH = 14-\frac{1}{2}[pKw – pKa – \log c] \\= 14+\frac{1}{2}[-pKw + pKa+ \log c]\end{array} \)

\(\begin{array}{l}pH = \frac{1}{2}[28-pKw+pKa+\log c] \\= \frac{1}{2}[2pKw – pKw + pKa + \log c]\\ = \frac{1}{2}[pKw +pKa +\log c]\end{array} \)

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

\(\begin{array}{l}pH = \frac{1}{2}[pKw+pKb+\log c]\end{array} \)

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

Examples: CH3COONH4, (NH4)2CO3, NH4HCO3

\(\begin{array}{l}CH_3COO NH_4 \rightleftharpoons NH_{4}^{+} + CH_3COO^{-}\end{array} \)

\(\begin{array}{l}CH_3COO^{-}+NH_{4}^{+}+H_2O \rightleftharpoons CH_3COOH + NH_4OH\end{array} \)

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

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

\(\begin{array}{l}\text{Hydrolysis constant} = Kh = \frac{[CH_3COOH][NH_4OH]}{[CH_3COO^{-}][NH_{4}^{+}]} = \frac{(ch+ch)}{c(1-h)c(1-h)} = \frac{h^{2}}{((1-h)^{2})} ;\end{array} \)

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

For the equilibrium,

\(\begin{array}{l}CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}\end{array} \)

\(\begin{array}{l}\text{Equilibrium constant of acid} = Ka = \frac{[CH_3COO^{-}][H^{+}]}{[CH_3COOH]}[H^{+}] = Ka\times \frac{[CH_3COOH]}{[CH_COO^{-}]}\end{array} \)

\(\begin{array}{l}[H^{+}] = Ka\frac{ch}{c(1-h)} = Ka\frac{h}{1-h} = Ka\sqrt{kh}\end{array} \)

For the equilibrium,

\(\begin{array}{l}NH_4OH\rightleftharpoons NH_{4}^{+}+OH^{-}\ \text{Equilibrium constant of the base} = Kb = \frac{[NH_{4}^{-}][OH^{-}]}{NH_4OH}\end{array} \)

For the equilibrium,

\(\begin{array}{l}H_2O\rightleftharpoons H^{+} + OH^{-} \ \text{Ionic product of water} = Kw = [H^{+}][OH^{-}] = 10^{-14}\end{array} \)

\(\begin{array}{l}\frac{[CH_3COOH][NH_4OH]}{[CH_3COO^{-}][NH_{4}^{+}]}\times \frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]} \times \frac{[NH_{4}^{-}][OH^{-}]}{NH_4OH} = [H^{+}][OH^{-}]\end{array} \)

Kh x Ka x Kb = Kw ;

\(\begin{array}{l}Kh = \frac{Kw}{KaKb}\end{array} \)

\(\begin{array}{l}[H^{+}] = Ka\sqrt{Kh} = Ka\frac{Kw}{KaKb} = \sqrt{\frac{KwKa}{Kb}}\end{array} \)

\(\begin{array}{l}pH = -\log[H^{+}]\\ = -\log[\sqrt{\frac{KwKa}{Kb}}]\\ =-\frac{1}{2}[\log Kw+\log Ka – \log Kb]\\= \frac{1}{2}[pKw+pKa-pKb]\end{array} \)

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

\(\begin{array}{l}pH = \frac{1}{2}[pKw +pKb-pKb]\end{array} \)

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.

\(\begin{array}{l}H_3PO_4\rightleftharpoons H^{+}+H_2PO_{4}^{-}\;\;\;\;K_{a_1} = \frac{[H^{+}][H_2PO]_{4}^{1}}{[H_3PO_4]}\end{array} \)

\(\begin{array}{l}H_2PO_{4}^{-}\rightleftharpoons H^{+}+HPO_{4}^{-}\;\;\;\;K_{a_2} = \frac{[H^{+}][HPO_{4}^{2-}]}{[H_2PO_{4}^{-}]}\end{array} \)

\(\begin{array}{l}H_2PO_{4}^{-}\rightleftharpoons H^{+}PO_{4}^{-}\;\;\;\;Ka3 = \frac{[H^{+}][HPO_{4}^{3-}]}{[HPO_{4}^{2-}]}\end{array} \)

For all acids, Ka1 >> Ka2 >> Ka3….

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

For weak acids,

\(\begin{array}{l}K_{a_1} = \frac{[H^{+}][H_{2}PO_{4}^{3-}]}{[H_{3}PO_{4}]} = \frac{Ch-Ch}{C(1-h)} = \frac{ch^{2}}{(1-h)} ,h< < < 1,(1-\alpha) = 1\end{array} \)

\(\begin{array}{l}\therefore K_{a_1} = Ch^2h=\sqrt{\frac{Ka1}{c}}\;\;\;\;[H^{+}]\;of\;ion=Ch\end{array} \)

and

\(\begin{array}{l}pH = -\log\sqrt{CKa1} = \frac{1}{2}(-\log Ka1 – \log C);\;\;\;\; pH = \frac{1}{2}(pKal-\log C)\end{array} \)

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

Salts will be completely ionised

\(\begin{array}{l}Kh1 = \frac{[OH^{-}][HPO_{4}^{2-}]}{[PO_{4}^{3-}]}\end{array} \)

Phosphate hydrolyzes as follows:

\(\begin{array}{l}PO_{4}^{3-} + H_{2}O\rightleftharpoons [HPO_{4}^{2-}] + [OH^{-}]\end{array} \)

\(\begin{array}{l}Kh2 = \frac{[OH^{-}][H_{2}PO_{4}^{-}]}{[HPO_{4}^{2-}]}\end{array} \)

\(\begin{array}{l}[HPO_{4}^{2-}] + [H_{2}O] \rightleftharpoons [H_{2}PO_{4}^{-}]+[OH^{-}]\end{array} \)

\(\begin{array}{l}Kh3 = \frac{[OH^{-}][H_{3}PO_{4}]}{[H_{2}PO_{4}^{-}]}\end{array} \)

\(\begin{array}{l}[H_{2}PO_{4}^{-}] + [H_{2}O] \rightleftharpoons [H_{3}PO_{4}]+[OH]\end{array} \)

On hydrolysis, [PO43-] + H2O [HPO42-] + [OH]

At equilibrium,

moles / l C (1-h) Ch Ch

\(\begin{array}{l}Kh1 = \frac{(Ch+Ch)}{C(1-h)} = \frac{Ch^{2}}{(1-h)} ,h< < < 1, (1-h) = 1\end{array} \)

\(\begin{array}{l}Kh1 = Ch^{2} ; h= \sqrt {\frac{Kh1}{C}} ; [H^{+}] = \frac{Kw}{OH^{[-]}} = \frac{Kw}{\sqrt{CKh1}}\end{array} \)

\(\begin{array}{l}HPO_{4}^{-}H^{+} + PO_{4}^{-}\end{array} \)
\(\begin{array}{l}Ka3 = \frac{[H^{+}][PO_{4}^{3}]}{[HPO_{4}^{2-}]}\end{array} \)

\(\begin{array}{l}Kh1 \times Ka3 = \frac{[OH^{-}][HPO_{4}^{2-}]}{[PO_{4}^{3-}]} \frac{[H^{+}][PO_{4}^{3-}]}{[OH^{-}]} = [H^{+}][OH^{-}] = Kw\end{array} \)

Hence,

\(\begin{array}{l}Kh1 = \frac{Kw}{Ka3}\end{array} \)

\(\begin{array}{l}[H^{+}] = \frac{Kw}{\sqrt{CKh1}} = \sqrt{\frac{KwKa3}{C}} ; pH = -\log\sqrt{\frac{KwKa3}{C}} = -\frac{1}{2}(\log Kw+\log Ka3 – \log C)\end{array} \)

\(\begin{array}{l}pH = -\frac{1}{2}(pKw+pKa3+\log C)\end{array} \)

The pH of salts of polyprotic acids,

\(\begin{array}{l}K_3PO_4= – \frac{1}{2}(pKw+pKa3+\log C)\end{array} \)

Hydrolysis of Amphoteric Anion

Examples: NaHCO3, NaHS

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

⇒ Ionisation

\(\begin{array}{l}HCO_{3}^{-} + H_2O \rightleftharpoons CO_{3}^{2-} + H_3O^{+}\ \text{with}\ Ka1\end{array} \)

⇒ Hydrolysis

\(\begin{array}{l}HCO_{3}^{-} + H_2O \rightleftharpoons H_2CO_3 + OH^{-}\ \text{with}\ Ka2\end{array} \)

\(\begin{array}{l}pH_{(HCO_{3-})} = \frac{pKa1+pKa2}{2}\end{array} \)
Test Your Knowledge On Hydrolysis Salts And Types!

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