Chemistry > Ionic Equilibrium > 6.0 Salt Hydrolysis
Ionic Equilibrium
1.0 Reversible Reaction
2.0 $pH$ Scale
3.0 Arrehenius Theory of Electrolyte Ionization (Dissociation)
4.0 Ionization of Water
5.0 Determination of $pH$ of acids and bases
5.1 Strong acid
5.2 Weak acid
5.3 Strong Acid + Weak Acid
5.4 Two weak acids
5.5 Dibasic and polyprotic weak acids
6.0 Salt Hydrolysis
6.1 Salts of weak acid + strong base
6.2 Salt of weak base + strong acid
6.3 Salt of weak acid and weak base
7.0 Buffer Solution
8.0 Solubility and Solubility Product
6.3 Salt of weak acid and weak base
5.2 Weak acid
5.3 Strong Acid + Weak Acid
5.4 Two weak acids
5.5 Dibasic and polyprotic weak acids
6.2 Salt of weak base + strong acid
6.3 Salt of weak acid and weak base
$${\text{N}}{{\text{H}}_4}{\text{OH + }}C{H_3}COOH{\text{ }} \to {\text{ }}C{H_3}COO{\text{N}}{{\text{H}}_4}{\text{ + }}{{\text{H}}_2}O$$
The salt would dissolve as $$C{H_3}COO{\text{N}}{{\text{H}}_4}{\text{ }} \to {\text{ }}N{H_4}^ + {\text{ + }}C{H_3}CO{O^ - }{\text{ }}$$
and the hydrolysis is shown as
$$C{H_3}CO{O^ - }{\text{ + }}N{H_4}^ + {\text{ + }}{H_2}O \rightleftharpoons {\text{N}}{{\text{H}}_4}{\text{OH}}{\text{ + }}C{H_3}COOH$$
${C(1 - \alpha )}$ ${C(1 - \alpha )}$ ${C\alpha }$ ${C\alpha }$
The hyrolysis constant is given as
$${K_h} = \frac{{\left[ {C{H_3}COOH} \right]\left[ {{\text{N}}{{\text{H}}_4}{\text{OH}}} \right]}}{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {N{H_4}^ + } \right]}}$$
and $${K_h} = \frac{{Kw}}{{{K_a}{K_b}}} = \frac{{C\alpha \times C\alpha }}{{C(1 - \alpha )C(1 - \alpha )}}$$
$$\frac{\alpha }{{1 - \alpha }} = \sqrt {\frac{{Kw}}{{{K_a}{K_b}}}} $$
The acetic acid formed would partially decompose to give ${C{H_3}CO{O^ - }}$ and ${{H^ + }}$. But, because of common ion effect ( which is due to unhydrolyzsed ${C{H_3}CO{O^ - }}$) it is possible to neglect the acetate ion coming from $C{H_3}COOH$.
Therefore,
$$C{H_3}COOH \rightleftharpoons {C{H_3}CO{O^ - }}{\text{ + }}{H^ + }$$
${C\alpha }$ ${C(1 - \alpha )}$
$${K_a} = \frac{{C(1 - \alpha )\left[ {{H^ + }} \right]}}{{C\alpha }}$$
Neglecting $\alpha $,$${K_a} = \frac{{\left[ {{H^ + }} \right]}}{\alpha }$$
$$\left[ {{H^ + }} \right] = {K_a}\alpha = {K_a}\sqrt {\frac{{Kw}}{{{K_a}{K_b}}}} = \sqrt {\frac{{Kw{K_a}}}{{{K_b}}}} {\text{ }}$$
NOTE : If ${K_a}{\text{ > }}{K_b}$, the solution would be acidic,
${K_a}{\text{ < }}{K_b}$, the solution would be basic, and
${K_a}{\text{ = }}{K_b}$, the solution would be neutral.
Example 4. When $0.2M{\text{ }}C{H_3}COOH$ is neutralised with $0.2M{\text{ }}NaOH$ in $0.5L$ of water, the resolving solution is slightly alkaline. Calculate $pH$ of resulting solution. ${K_a}(C{H_3}COOH) = 1.8 \times {10^{ - 5}}$
Solution: $$C{H_3}CO{O^ - }{\text{ + }}{H_2}O \rightleftharpoons C{H_3}COOH{\text{ + }}O{H^ - }$$
${C(1 - \alpha )}$ ${C\alpha }$ ${C\alpha }$
where $\alpha $ is the degree of hydrolysis.
$${K_h} = \frac{{C{\alpha ^2}}}{{1 - \alpha }}$$
For $\alpha \ll 1$, ${K_h} = C{\alpha ^2}$
$${K_h} = \frac{{{K_w}}}{{{K_a}}} = \frac{{1 \times {{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}} = 5.5 \times {10^{ - 10}}$$
So, $${K_h} = C{\alpha ^2} = 5.5 \times {10^{ - 10}}$$
or, $${\alpha ^2} = 27.5 \times {10^{ - 10}}$$$$\alpha = 5.25 \times {10^{ - 5}}$$
$$\therefore \left[ {{H^ + }} \right] = \frac{{{K_w}}}{{\left[ {O{H^ - }} \right]}} = \frac{{{{10}^{ - 14}}}}{{C\alpha }} = \frac{{C\alpha }}{{1.05 \times {{10}^{ - 5}}}} = 0.952 \times {10^{ - 9}}M$$
$$pH = - \log \left[ {{H^ + }} \right] = - \log (0.952 \times {10^{ - 9}}) = 8.98$$
Example 5. Calculate amount of ammonium chloride required to dissolve in $500{\text{ }}ml$ water to have $pH = 4.5$ (${{K_b}}$ for $N{H_4}OH = 1.8 \times {10^{ - 5}}$)
Solution. $$\left[ {{H^ + }} \right] = {10^{ - pH}} = {10^{ - 4.5}} = 3.162 \times {10^{ - 5}}M$$
$$N{H_4}^ + {\text{ + }}{H_2}O \rightleftharpoons {\text{N}}{{\text{H}}_4}{\text{OH}}{\text{ + }}{{H^ + }}$$
${C(1 - \alpha )}$ ${C\alpha }$ ${C\alpha }$
${K_h} = C{\alpha ^2}$, for $\alpha \ll 1$
$${K_h} = \frac{{{K_w}}}{{{K_a}}} = \frac{{1 \times {{10}^{ - 14}}}}{{1.8 \times {{10}^{ - 5}}}} = 5.5 \times {10^{ - 10}}$$
$$\therefore \alpha = \frac{{{K_h}}}{{C\alpha }} = \frac{{{K_h}}}{{\left[ {{H^ + }} \right]}} = \frac{{5.5 \times {{10}^{ - 10}}}}{{3.162 \times {{10}^{ - 5}}}} = 1.74 \times {10^{ - 5}}$$
$$C = \frac{{\left[ {{H^ + }} \right]}}{\alpha } = \frac{{3.162 \times {{10}^{ - 5}}}}{{1.74 \times {{10}^{ - 5}}}} = 1.8{\text{ }}mol{L^{ - 1}}$$
i.e. $1.8/2 = 0.9$ mole in $500{\text{ ml}}$ water
Mass in grams is $$ = 0.9 \times 53.5 = 48.15{\text{ }}g$$
