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
$\operatorname{AgBr}(s) \rightleftharpoons Ag ^{+}(a q)+B r^{-}(a q)$
Molar solubility $(s)$ of $AgBr =\sqrt{4.9 \times 10^{-13}}$
$=7.0 \times 10^{-7}$ moldm $^{-3}$