Discussion Forum

1)

The solubility of salt of a weak acid (AB) at pH 3 is Y× 10^-3 mol L^-1. The value of Y is.........

[Given that the value of solubility product of AB (K_sp)=2× 10^-10 and the value of ionization constant of HB (K_a)=1× 10^-8]

Answer:

Option B

Explanation:

Let Solubility of AB in the buffer of pH 3=x

      $AB(s)\rightleftharpoons A^{+}(aq)+B^{-}(aq)$    K₁ =K_sp

     $B^{-}(aq)+H^{+}(aq)\rightleftharpoons HB^{}(aq)$   $K_{2}=\frac{1}{K_{a}}$

---

  $AB(s)+H^{+}(aq)\rightleftharpoons HB (aq)+A^{+}(aq)$

                                                           x              x

---

$K_{3}=\frac{K_{sp}}{k_{a}}$

 $K_{3}=\frac{[HB][A^{+}]}{[H^{+}]}=\frac{K_{sp}}{K_{a}}$

  $\therefore$  $\frac{x^{2}}{(10^{-3})}=\frac{2\times 10^{-10}}{1\times 10^{-8}}$

$\therefore$    x= 4.47 × 10^-3 M = Y× 10^-3 M

$\therefore$    y = 4.47