question_answer

Two sparingly soluble salts AB and \[X{{Y}_{2}}\] have the same solubility product. Which salt will be more soluble? Explain.

Answer:

                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.