Question

Calculate the mass of $KBr$ (in g) that needs to be added to 1 litre of a $0.05M$ solution of $AgN{O}_3$ so that the precipitation of $AgBr$ just starts.
Solubility product $\left(K_{sp}\right)$ of $AgBr$ is $3.5\times {10}^{-13}$.
Take molecular mass of $KBr$ to be $119gmo{l}^{-1}$

• A. $2.5\times {10}^{-11}g$
• B. $6.92\times {10}^{-7}g$
• C. $8.33\times {10}^{-10}g$
• D. $4.56\times {10}^{-5}g$

Answer 1

The correct option is C $8.33\times {10}^{-10}g$

From, $AgN{O}_3$,
the concentration $\left[A{g}^{+}\right]=0.05M$
$K_{sp}\left(AgBr\right)=3.5\times {10}^{-13},\left[A{g}^{+}\right]=0.05M$.
Molecular mass of $KBr=120gmo{l}^{-1}$

The equilibrium between undissociated $AgBr$ and it's ions $A{g}^{+}$ and $B{r}^{-}$ can be represented as:

$AgBr\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$
Precipitation starts when ionic product $Q_{sp}$ just exceeds solubility product $K_{sp}$.
For the limiting case,
$K_{sp}=Q_{sp}$

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

putting the values,
$\Rightarrow \left[B{r}^{-}\right]=\frac{K_{sp}}{\left[A{g}^{+}\right]}=\frac{3.5\times {10}^{-13}}{0.05}=7\times {10}^{-12}$
For $1L$ solution,
Concentration=Moles

$\therefore n_{B{r}^{-}}=7\times {10}^{-12}$
Since,
$KBr\left(s\right)\to K^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$

No. of moles of $B{r}^{-}$= No of moles of $KBr$
$\therefore n_{KBr}=7\times {10}^{-12}$

Precipitation starts when $7\times {10}^{-12}$ moles of $KBr$ is added to 1L of $AgN{O}_3$ solution.


Mass of $KBr=\text{number of moles}\times \text{molecular mass}$

putting the values,
$=7\times {10}^{-12}mol\times 119gmo{l}^{-1}=8.33\times {10}^{-10}g$