Question

The solubility of $ AgCl $ is $ 1.435gm/litre $ . 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{AgCl}}\,\underset{s}{\mathop{A{{g}^{+}}}}\,+\underset{s}{\mathop{C{{l}^{-}}}}\, $ Solubility $ =1.435\text{ }g/litre $ Solubility $ \text{=}\frac{1.435}{143.5}\text{ }mole/litre $ $ 143.5= $ molecular weight Solubility product $ ={{(s)}^{2}} $ $ ={{(0.01)}^{2}}={{10}^{-4}} $