Question

The solubility product of $AgCl$ is ${10}^{-4}$ at a certain temperature. What is the molarity of its unsaturated solution?

• A. Greater than $0.10M$
• B. Less than $0.01M$
• C. Equal to $0.01M$
• D. Unpredictable

Answer 1

The correct option is C Equal to $0.01M$

$K_{sp}$ of $AgCl={10}^{-4}$

$AgCl⟶\underset{\left(s\right)}{A{g}^{+}}+\underset{\left(s\right)}{C{l}^{-}}$

$S\times S=K_{sp}$

$\Rightarrow S^2={10}^{-4}$

$\Rightarrow S={10}^{-2}mol/L$

Molarity is equal to $0.01M$