Chemistry - Electrochemistry Question with Solution | TestHub

The concentrations of the ions can be calculated as follows,

$\left[{\mathrm{Ag}}^{+}\right]={10}^{-5}\mathrm{M}$

$\left[{\mathrm{NO}}_3^{-}\right]={10}^{-5}\mathrm{M}$

$\left[{\mathrm{Br}}^{-}\right]=\frac{\mathrm{Ksp}}{\left[{\mathrm{Ag}}^{+}\right]}=4.9\times {10}^{-8}\mathrm{M}$

${\mathrm{\Lambda }}_{\mathrm{m}}=\frac{\mathrm{k}}{1000\times \mathrm{M}}=\frac{\mathrm{Specific} \mathrm{conductance}}{1000\times \mathrm{Molarity}}$

For ${\mathrm{Ag}}^{+}$

$6\times {10}^{-3}=\frac{{\mathrm{k}}_{{\mathrm{Ag}}^{+}}}{1000\times {10}^{-5}}$

${\mathrm{k}}_{\mathrm{Ag}+}=6\times {10}^{-5}{\mathrm{Sm}}^{-1}$

$\Rightarrow 6000\times {10}^{-8}{\mathrm{Sm}}^{-1}$

for ${\mathrm{Br}}^{-}$

$8\times {10}^{-3}=\frac{{\mathrm{k}}_{{\mathrm{Br}}^{-}}}{1000\times 4.9\times {10}^{-8}}$

${\mathrm{k}}_{\mathrm{Br}-}=39.2\times {10}^{-8}{\mathrm{Sm}}^{-1}$

for ${\mathrm{NO}}_3^{-}$

$7\times {10}^{-3}=\frac{{\mathrm{k}}_{{\mathrm{NO}}_3^{-}}}{1000\times {10}^{-5}}$

${\mathrm{k}}_{{\mathrm{NO}}_3^{-}}=7\times {10}^{-5}{\mathrm{Sm}}^{-1}$

$=7000\times {10}^{-8}{\mathrm{Sm}}^{-1}$

Conductivity of solution

$\Rightarrow (6000+7000+39.2)\times {10}^{-8}$

$\Rightarrow 13039.2\times {10}^{-8} \mathrm{S} {\mathrm{m}}^{-1}$