Question

The solubility of a salt of weak acid (AB) at pH= 3 is $Y\times {10}^{-3}mol{L}^{-1}$. The value of Y is.
(Given that the value of solubility product of AB $\left(K_{sp}\right)=2\times {10}^{-10}$ and the value of ionization constant of $HB\left(K_a\right)=1\times {10}^{-8})$

Answer 1

$AB\rightleftharpoons \underset{S}{A^{+}}+\underset{(S-x)}{B^{-}}$
$2\times {10}^{-10}=S(S-x)....\left(1\right)$
$\underset{(S-x)}{B^{-}}+\underset{{10}^{-3}}{H^{+}}\rightleftharpoons \underset{x}{HB}$
$\frac{1}{{10}^{-8}}=\frac{x}{(S-x)\times {10}^{-3}}$
$\frac{x}{S-x}={10}^5....\left(2\right)$
Multiply equation (1) and (2).
$S.x=2\times {10}^{-5}$
From Eq. (1)
$S^2-Sx=2\times {10}^{-10}$
$S^2-2\times {10}^{-5}=2\times {10}^{-10}$
$S^2=2\times {10}^{-5}+2\times {10}^{-10}\cong 2\times {10}^{-5}$
$S=4.47\times {10}^{-3}$
y = 4.47