Question

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

• A. $2.4\times 10^{-7}$ mol dm ${}^{-3}$
• B. $7.0\times 10^{-7}$ mol dm ${}^{-3}$
• C. $4.9\times 10^{-7}$ mol dm ${}^{-3}$
• D. $3.2\times 10^{-7}$ mol dm ${}^{-3}$

Answer 1

Answer: (C)

The equilibrium reaction of $AgBr$ is
$\mathrm{AgBr}(s)\rightleftharpoons A{g}^{+}(aq)+B{r}^{-}(aq)$
Molar solubility $(s)$ of $AgBr=\sqrt{4.9\times 10^{-13}}$
$=7.0\times 10^{-7}$ moldm ${}^{-3}$