• # question_answer Two sparingly soluble salts AB and $X{{Y}_{2}}$ have the same solubility product. Which salt will be more soluble? Explain.

Suppose solubility of $AB=a\,\text{mol}\,{{L}^{-1}}$. Then $AB\rightleftharpoons \,{{A}^{+}}+{{B}^{-}},$${{K}_{sp}}\,=[{{A}^{+}}]\,[{{B}^{-}}]\,=a\times a={{a}^{2}}$ $\therefore \,\,\,a=\sqrt{{{K}_{sp}}}$ Suppose solubility of salt $X{{Y}_{2}}=b\,\,mol\,{{L}^{-1}}$. Then $X{{Y}_{2}}\,\rightleftharpoons \,{{X}^{2+}}\,+2\,{{Y}^{-}},$${{K}_{sp}}\,=[{{X}^{2+}}]\,{{[{{Y}^{-}}]}^{2}}=b\,{{(2\,b)}^{2}}$ i.e., $4\,{{b}^{3}}\,={{K}_{sp}}$ or $b={{({{K}_{sp}}/4)}^{1/3}}$ Obviously $b>a$ (as ${{K}_{sp}}$ have values with negative powers of 10). Hence, salt $X{{Y}_{2}}$ is more soluble.