Question

The solubility product of $AgBr$ solution is $4\times 10^{-13}$ moles/litre. The $A{g}^{+}$ ion concentration in it is $1\times 10^{-6}$ moles/litre. The $B{r}^{-}$ ion concentration is:

• A. $1\times 10^{-6}$
• B. $2\times 10^{-6}$
• C. $2\times 10^{-7}$
• D. $4\times 10^{-7}$

Answer 1

Answer: (D)

The solubility product of $AgBr\left[A{g}^{+}\right]\left[B{r}^{-}\right]$
$=4\times 10^{-13}$ moles/litre
The $A{g}^{+}$ion concentration $=1\times 10^{-6}mols/litre$
$\therefore B{r}^{-}=\frac{\text{ solubility product of }AgBr}{\left[A{g}^{+}\right]}$
$=\frac{4\times 10^{-13}}{1\times 10^{-6}}$
$=4\times 10^{-7}\text{ moles/litre }$