Question

The solubility product of silver bromide is $5.0\times {10}^{-13}$ at 298k. Calculate its solubility at this temperature.

• A. $7.07\times {10}^{-7}mol{L}^{-1}$
• B. $7.08\times {10}^{-10}mol{L}^{-1}$
• C. $7.06\times {10}^{-5}mol{L}^{-1}$
• D. $7.07\times {10}^{-18}mol{L}^{-1}$

Answer 1

The correct option is A $7.07\times {10}^{-7}mol{L}^{-1}$

$k_{sp}=5.0\times {10}^{-13}$
$AgBr\rightleftharpoons A{g}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$
Let solubility of AgBr is s mole/litre
$\therefore \left[A{g}^{+}\right]=s,\left[B{r}^{-}\right]=s$
$5\times {10}^{-13}=s\times s$
$s=\sqrt{5\times {10}^{-13}}$
$=\sqrt{50\times {10}^{-14}}=7.07\times {10}^{-7}mol{L}^{-1}$