Question

When $1.88$ g of $AgBr\left(s\right)$ is added to a ${10}^{-3}$ M aqueous solution of $KBr$, the concentration of $A{g}^{+}$ is $5\times {10}^{-10}$ M. If the same amount of $AgBr\left(s\right)$ is added to a ${10}^{-2}$ M aqueous solution of $AgN{O}_3$, the concentration of $B{r}^{-}$ is:

• A. $9.4\times {10}^{-9}$ M
• B. $5\times {10}^{-10}$ M
• C. $1\times {10}^{-11}$ M
• D. $5\times {10}^{-11}$ M

Answer 1

The correct option is D $5\times {10}^{-11}$ M

$AgBr\leftrightharpoons A{g}^{+}+B{r}^{-}$

$\left[B{r}^{-}\right]=5\times {10}^{-10}+{10}^{-3}={10}^{-3}$

$K_{sp}\left(AgBr\right)=\left[B{r}^{-}\right]\times \left[A{g}^{+}\right]={10}^{-3}\times 5\times {10}^{-10}=5\times {10}^{-13}$

If it is dissolved in $AgN{O}_3$:

$K_{sp}=5\times {10}^{-13}={10}^{-2}\times \left[B{r}^{-}\right]$

$\left[B{r}^{-}\right]=5\times {10}^{-11}M$

Hence option $D$ is correct.