5.5 Dibasic and polyprotic weak acids

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

A dibasic acid is the one which has the capacity of giving two protons per molecule. e.g. ${H_2}S{O_4}$

Suppose ${H_2}A$ is a weak dibasic acid which dissociates as

$${H_2}A \rightleftharpoons {{\text{H}}^ + }{\text{ + }}H{A^ - }$$

$C(1 - \alpha )$ $C{\alpha _1} + C{\alpha _1}{\alpha _2}$ $C{\alpha _1}(1 - {\alpha _2})$

$$H{A^ - } \rightleftharpoons {{\text{H}}^ + }{\text{ + }}{A^{2 - }}$$

$C{\alpha _1}(1 - {\alpha _2})$ $C{\alpha _1} + C{\alpha _1}{\alpha _2}$ $C{\alpha _1}{\alpha _2}$

Thus,

$${K_{{a_1}}}{\text{ = }}\frac{{\left[ {{H^ + }} \right]\left[ {H{A^ - }} \right]}}{{\left[ {{H_2}A} \right]}}{\text{ = }}\frac{{(C{\alpha _1}{\alpha _2} + C{\alpha _1})(C{\alpha _1}(1 - {\alpha _2}))}}{{C(1 - {\alpha _1})}}{\text{ }}$$

$${K_{{a_2}}}{\text{ = }}\frac{{\left[ {{H^ + }} \right]\left[ {{A^2}^ - } \right]}}{{\left[ {HA} \right]}}{\text{ = }}\frac{{(C{\alpha _1}{\alpha _2} + C{\alpha _1})(C{\alpha _1}{\alpha _2})}}{{C{\alpha _1}(1 - {\alpha _2})}}{\text{ }}$$

Here too, equations can be solved to get the values for ${{\alpha _1}}$ and ${{\alpha _2}}$ using which ${\left[ {{H^ + }} \right]}$ and hence ${\text{pH}}$ can therefore be calculated.