Question

$1\times 10^{-5}MAgN{O}_3$ is added to $1L$ of saturated solution of $AgBr$. The conductivity of this solution at $298K$ is ___
[Given : $K_{\text{SP }}(AgBr)=4.9\times 10^{-13}$ at $298K$
${\lambda }_{A{g}^{+}}^0=6\times 10^{-3}S{m}^{2}mo{l}^{-1}$
${\lambda }_{B{r}^{-}}^0=8\times 10^{-3}S{m}^{2}mo{l}^{-1}$
${\lambda }_{N{O}_3^{-}}^0=7\times 10^{-3}S{m}^{2}mo{l}^{-1}$ ]

Answer 1

Answer: 14

$\left[A{g}^{+}\right]=10^{-5}$
$\left[N{O}_3^{-}\right]=10^{-5}$
$[B{r}^{-}]=\frac{Ksp}{[A{g}^{+}]}=4.9\times 10^{-8}$
${\Lambda }_m=\frac{k}{1000\times M}$
For $A{g}^{+}$
$6\times 10^{-3}=\frac{K_{A{g}^{+}}}{1000\times 10^{-5}}$
$K_{Ag+}=6\times 10^{-5}$
$\Rightarrow 6000\times 10^{-8}$
$\text{ for }B{r}^{-}$
$8\times 10^{-3}=\frac{K_{B{r}^{-}}}{1000\times 4.9\times 10^{-8}}$
$K_{Br-}=39.2\times 10^{-8}$
$\text{ for }N{O}_3^{-}$
$7\times 10^{-3}=\frac{K_{N{O}_3^{-}}}{1000\times 10^{-5}}$
$K_{N{O}_3}=7\times 10^{-5}$
$=7000\times 10^{-8}$
Conductivity of solution
$\Rightarrow (6000+7000+39.2)\times 10^{-8}$
$\Rightarrow 13039.2\times 10^{-8}S{m}^{-1}$