Question

When 1.88g of AgBr(s) is added to a ${10}^{-3}M$ aqueous solution of KBr, the concentration of Ag is $5\times {10}^{-10}$M. If the same amount of AgBr(s) is added to a ${10}^{-2}$ M aqueous solution of $Ag{NO}_F$ 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

Answer: (D)

$5\times {10}^{-11}M$

n of $AgBr=\frac{1.88}{188}=0.01mol$

$AgBr\left(s\right)\rightleftharpoons A{g}^{+}+B{r}^{-}$

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

$\therefore k{s}_p=\left[A{g}^{+}\right]\left[B{r}^{-}\right]$

$=5\times {10}^{-10}\times {10}^{-3}$

$=5\times {10}^{-13}(mol/lt{)}^2$

$\therefore$ when $\left[AgN{U}_3\right]={10}^{-2}M$

$\left[A{g}^{+}\right]={10}^{-2}M$

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

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