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)

$\underset{s}{\mathrm{AgCl}}\underset{s}{A{g}^{+}}+\underset{s}{C{l}^{-}}$ Solubility $=1.435g/litre$ Solubility $\text{=}\frac{1.435}{143.5}mole/litre$ $143.5=$ molecular weight Solubility product $={(s)}^2$ $={(0.01)}^2={10}^{-4}$