Question

The solubility of $ AgCl $ is 1.435 g/L What will be the solubility product?

• A. $ {{10}^{-4}} $
• B. 0.01
• C. 0.001
• D. 0.1

Answer 1

Answer: (A)

$ \underset{s}{\mathop{Ag}}\,Cl\underset{s}{\mathop{A{{g}^{+}}}}\,+\underset{s}{\mathop{C{{l}^{-}}}}\, $ Solubility =1.435 g/L Solubility $ =\frac{1.435}{143.5}mol/L $ 143.5 = molecular weight Solubility product $ ={{(s)}^{2}} $ $ ={{(0.01)}^{2}}={{10}^{-4}} $