Question

Solubility of lead chloride is K. Then the number of moles of $ C{{l}^{-}} $ ionsin 1000 ml of saturated lead chloride solution will be

• A. $ {{\left( \frac{K}{4} \right)}^{1/3}} $
• B. $ {{(2K)}^{1/3}} $
• C. $ {{K}^{1/2}} $
• D. $ \frac{K}{2} $

Answer 1

Answer: (A)

: For lead chloride $ (PbC{{l}_{2}}) $ $ {{K}_{sp}}=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}} $ $ {{K}_{sp}}=x.{{(2x)}^{2}} $ $ {{K}_{sp}}=x.4{{x}^{2}} $ $ x={{\left( \frac{K}{4} \right)}^{1/3}} $