Chemistry - Electrochemistry Question with Solution | TestHub

Saturated solution corresponding to infinite dilution

${\wedge }_{\text{m}}^{\text{o}}$for salt$={\text{λ}}^{°}{}_{\text{M}{\left({\text{NH}}_3\right)}_4{\text{Br}}_2^{+}}+{\text{λ}}^{°}{}_{{\text{H}}_2{\text{PO}}_2{}^{-}}=100\text{S}{\text{m}}^2{\text{mole}}^{-1}+50\text{S}{\text{m}}^{\text{2}}{\text{mol}}^{-1}=150\text{S}{\text{m}}^{\text{2}}{\text{mol}}^{-1}$

$\mathrm{\kappa }$$=\frac{1}{\rho }=\frac{1}{200}{\text{Ω}}^{-1}{\text{cm}}^{-1}\text{=}\frac{1}{2}{\text{Ω}}^{-1}{\text{m}}^{-1}$

${\wedge }_{\text{m}}^{\text{o}}\left(\text{salt}\right)=\frac{\mathrm{\kappa }}{\mathrm{s}}=\frac{\frac{1}{2}{\mathrm{\Omega }}^{-1}{\text{m}}^{-1}}{\text{s moles}{\text{m}}^{-3}}$

$\text{s}\left({\text{moles m}}^{-3}\right)=\frac{1}{2\times 150}=\frac{1}{300}=3·33\times 10^{-3}$

$\text{s}\left({\text{moles L}}^{-1}\right)=3·33\times 10^{-6}$

${\text{K}}_{\text{sp}}={\left(3·33\times 10^{-6}\right)}^2$

$=1.11\times {10}^{-11}{\text{M}}^2$