Ionic Equilibrium

The $\mathrm{pH}$ of a salt solution formed from a weak acid and a weak base can be determined using the relationship between their dissociation constants. Specifically, the formula for the $\mathrm{pH}$ of the solution is given by :

$ \mathrm{pH} = \frac{1}{2} (\mathrm{p} K_w + \mathrm{p} K_a - \mathrm{p} K_b) $

where :

$\mathrm{p} K_w$ is the ionization constant of water, which is 14 at $25^{\circ} \mathrm{C}$,

$\mathrm{p} K_a$ is the acid dissociation constant, and

$\mathrm{p} K_b$ is the base dissociation constant.

Given that :

$\mathrm{p} K_a = 4$,

$\mathrm{p} K_b = 5$,

substitute these values into the formula :

$ \mathrm{pH} = \frac{1}{2} (14 + 4 - 5) $

Simplify the expression inside the parentheses :

$ \mathrm{pH} = \frac{1}{2} (14 + 4 - 5) = \frac{1}{2} (13) = 6.5 $

Thus, the $\mathrm{pH}$ of the salt solution at $25^{\circ} \mathrm{C}$ is:

$ \boxed{6.5} $

$A X_2$ is ionised as follows:

$\underset{S \mathrm{~mol} \mathrm{~L}^{-1}}{A X_2} \rightleftharpoons \underset{S}{A^{2+}}+\underset{2 S}{2 X^{-}}$

Solubility product of $A X_2$

$K_{\text {sp }}=\left[A^{2+}\right]\left[X^{-}\right]^2=[S] \times[2 S]^2=4 S^3$

$\begin{aligned} & \because K_{\text {sp }} \text { of } A X_2=3.2 \times 10^{-11} \\ & \therefore \quad 3.2 \times 10^{-11}=4 S^3 \end{aligned}$

or, $\quad S^3=0.8 \times 10^{-11}=8 \times 10^{-12}$

or, $\quad S=\sqrt[3]{8 \times 10^{-12}}$

$\begin{aligned} & =2 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \\ \text { Solubility } & =2 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \end{aligned}$