Question

The solubility in water of a sparingly soluble salt $ A{{B}_{2}} $ is $ 1.0\times {{10}^{-5}}mol\text{ }{{L}^{-1}} $ . Its solubility product will be

• A. $ 4\times {{10}^{-15}} $
• B. $ 4\times {{10}^{-10}} $
• C. $ 1\times {{10}^{-15}} $
• D. $ 1\times {{10}^{-10}} $

Answer 1

Answer: (A)

Let the solubility
$=x $ $ x=1.0\times {{10}^{-5}}mol\text{ }{{L}^{-1}} $ (given) $ A{{B}_{2}}\underset{x}{\mathop{{{A}^{2+}}}}\,+\underset{2x}{\mathop{2{{B}^{-}}}}\, $ $ {{K}_{sp}}=[{{A}^{2+}}]{{[{{B}^{-}}]}^{2}} $ $ {{K}_{sp}}=x\times {{(2x)}^{2}} $
$=4{{x}^{3}} $ $ {{K}_{sp}}=4\times {{(1\times {{10}^{-5}})}^{3}} $
$=4\times {{10}^{-15}} $