Question

The solubility product of silver bromide is $5.0\times {10}^{-13}$. The quantity of potassium bromide (molar mass is taken as $120$ g $mo{l}^{-1}$) to be added to $1$ liter of $0.05$M solution of silver nitrate to start the precipitate of AgBr is:

• A. $5.0\times {10}^{-8}$ g
• B. $1.2\times {10}^{-10}$ g
• C. $1.2\times {10}^{-9}$ g
• D. $6.2\times {10}^{-5}$ g

Answer 1

Answer: (C)

$1.2\times {10}^{-9}$ g

$K_{sp}$ of $AgBr=\left[A{g}^{+}\right]\left[B{r}^{-}\right]=5\times {10}^{-13}$

$\left[A{g}^{+}\right]=0.05M$

$\left[B{r}^{-}\right]=\frac{5.0\times {10}^{-13}}{0.05}$

$=1\times {10}^{-11}M$

Moles of $KBr=1\times {10}^{-11}\times 1$

Weight of $KBr=1\times {10}^{-11}\times 120$

$=1.2\times {10}^{-9}gm$