5.2 Weak acid

3.1 Degree of Ionization (Dissociation)

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

6.1 Salts of weak acid + strong base
6.2 Salt of weak base + strong acid
6.3 Salt of weak acid and weak base

7.1 Acidic buffer
7.2 Basic buffer
7.3 Salt buffer

8.1 Calculating Solubility of a salt

These are the acids which dissociate weakly in water. For example,

$$C{H_3}COOH{\text{ }} \rightleftharpoons {\text{ }}C{H_3}CO{O^ - }{\text{ + }}{{\text{H}}^ + }$$

The equilibrium constant for this is given by $${K_a}{\text{ = }}\frac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}$$

Let initial moles of acetic acid be $C$ and $\alpha $ is the degree of dissociation which is defined as the number of moles of acetic acid which dissociates from one mole of acetic acid, then the number of moles dissociated would be $C\alpha $.

$$C{H_3}COOH{\text{ }} \rightleftharpoons {\text{ }}C{H_3}CO{O^ - }{\text{ + }}{{\text{H}}^ + }$$

Initial $C$ $0$ $0$

At equilibrium $C - C\alpha $ $C\alpha $ $C\alpha $

$${K_a}{\text{ = }}\frac{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]}}{{\left[ {C{H_3}COOH} \right]}}{\text{ = }}\frac{{C\alpha \times C\alpha }}{{C(1 - \alpha )}}{\text{ = }}\frac{{C{\alpha ^2}}}{{1 - \alpha }}$$

Note: If $\alpha \leqslant 0.05$, then $\alpha $ can be ignored in comparison to $1$

Therefore, from above equation, we get

$${K_a} = C{\alpha ^2}$$ or, $$\alpha = \sqrt {\frac{{{K_a}}}{C}} $$

We can thus calculate the concentration of ${H^ + }$ coming from the acid as

$${\left[ {{H^ + }} \right]_A} = C\alpha = C \times \sqrt {\frac{{{K_a}}}{C}} = \sqrt {{K_a} \times C} $$

Example 3. For $10M{\text{ C}}{{\text{H}}_3}COOH$ solution if ${K_a} = {10^{ - 5}}$, then find out

(i) $\alpha $

(ii) $\left[ {{H^ + }} \right]$

(iii) $pH$

Solution: (i) Given that $C = 10M$ and ${K_a} = {10^{ - 5}}$

Degree of ionization is $$\alpha = \sqrt {\frac{{{K_a}}}{C}} = \sqrt {\frac{{{{10}^{ - 5}}}}{{10}}} = {10^{ - 3}}$$

(ii) $${\left[ {{H^ + }} \right]_A} = C\alpha = 10 \times {10^{ - 3}} = {10^{ - 2}}$$

(iii) $$pH = - \log \left[ {{H^ + }} \right] = - \log {10^{ - 2}} = 2$$