Chemistry - Electrochemistry Question with Solution | TestHub

Suppose the solubility of $\mathrm{AgBr}$ in $10^{-7}{\mathrm{MAgNO}}_3$ is $\mathrm{mol}/\mathrm{L}$ $$\begin{array}{rl}& \underset{s\mathrm{mol}/\mathrm{L}}{\mathrm{AgBr}}\rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{Br}}^{-}\\ & {\mathrm{AgNO}}_3\rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{NO}}_3^{-}\\ & 10^{-7}\mathrm{M}10^{-7}\mathrm{M}\times 10^{-7}\mathrm{M}\\ & \text{ Total }\left[{\mathrm{Ag}}^{+}\right]=\left(s+10^{-7}\right)\mathrm{M}\\ & K_{\mathrm{sp}}\text{ of }\mathrm{AgBr}=\left[{\mathrm{Ag}}^{+}\right]\left[{\mathrm{Br}}^{-}\right]\\ & 12\times 10^{-14}=\left(s+10^{-7}\right)(s)=s^2+10^{-7}s\\ & s^2+10^{-7}s-12\times 10^{-14}=0\\ & \end{array}$$ On solving $s=3\times 10^{-7}\mathrm{M}$ Hence, $\left[{\mathrm{Br}}^{-}\right]=3\times 10^{-7}\mathrm{M}=3\times 10^{-7}\times 10^3{\mathrm{m}}^3=3\times 10^{-4}{\mathrm{m}}^3$ $$\left[{\mathrm{Ag}}^{+}\right]=\left(3\times 10^{-7}+10^{-7}\right)\mathrm{M}=4\times 10^{-7}\times 10^3{\mathrm{m}}^3=4\times 10^{-4}{\mathrm{m}}^3$$ $$\left[{\mathrm{NO}}_3^{-}\right]=10^{-7}\mathrm{M}=10^{-7}\times 10^{+3}=1\times 10^{-4}{\mathrm{m}}^3$$ $$\because \lambda =\frac{k}{C}\text{ or }k=\lambda \times C$$ $$\begin{array}{rl}\therefore k_{{\mathrm{Br}}^{-}}& =3\times 10^{-4}\times 8\times 10^{-3}{\mathrm{Sm}}^{-1}=24\times 10^{-7}{\mathrm{Sm}}^{-1}\\ k_{{\mathrm{Ag}}^{+}}& =4\times 10^{-4}\times 6\times 10^{-3}{\mathrm{Sm}}^{-1}=24\times 10^{-7}{\mathrm{Sm}}^{-1}\\ k_{{\mathrm{NO}}_3^{-}}& =1\times 10^{-4}\times 7\times 10^{-3}=7\times 10^{-7}{\mathrm{Sm}}^{-1}\\ k_{\text{Total }}& =k_{{\mathrm{Br}}^{-}}+k_{{\mathrm{Ag}}^{+}}+k_{{\mathrm{NO}}_3^{-}}\\ & =(24\times 10^{-7}+24\times 10^{-7}+7\times 10^{-7}{\mathrm{Sm}}^{-1}\text{ ) }\\ & =55\times 10^{-7}{\mathrm{Sm}}^{-1}\\ & =55{\mathbf{Sm}}^{-1}\text{ (in term of }10^{-7}{\mathrm{Sm}}^{-1})\end{array}$$