Solubility product of $\mathrm{AgBr}$ is $4.9 \times 10^{-13}$. What is its solubility?

Solubility product of $\mathrm{AgBr}$ is $4.9 \times 10^{-13}$. What is its solubility?
  1. $2.4 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}$
  2. $3.2 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}$
  3. $4.9 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}$
  4. $7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}$

Solution

$\begin{aligned} & \mathrm{K}_{\mathrm{sp}}(\mathrm{AgBr})=\mathrm{S}^2 \\ & 4.9 \times 10^{-13}=\mathrm{S}^2 \\ & \mathrm{~S}=\sqrt{4.9 \times 10^{-13}} \\ & \mathrm{~S}=7.0 \times 10^{-7} \mathrm{~mol} \mathrm{dm}^{-3}\end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 2)

Practice more Ionic Equilibria questions on Aicharya