Question

The solubility product of $AgBr$ is $4.9\times {10}^{-9}$. The solubility of AgBr will be:

• A. $7\times {10}^{-4}mole/litre$
• B. $7\times {10}^{-5}g/litre$
• C. $1.316\times {10}^{-2}g/litre$
• D. $1\times {10}^{-3}mole/litre$

Answer 1

Answer: (B)

$7\times {10}^{-5}g/litre$

$AgB{r}_s\to A{g}_{aq}^{-}+B{r}_{aq}^{-}$

$K_{s}p=\left[A{g}^{+}\right]\left[B{r}^{-}\right]$

$4.9\times {10}^{-9}=x\times x$

$4.9\times {10}^{-9}=x^2$

taking square root on both sides

$x=7.0\times {10}^{-5}$

$\therefore$ The solubility of $AgBr=7\times {10}^{-5}moles/lt$