Question

The conductivity of saturated solution of $BaS{O}_4$ is $3.06\times {10}^{-6}$ mho ${cm}^{-1}$ and its equivalent conductance 1.53 mho ${cm}^2{eq}^{-1}$. The $K_{sp}$ for $BaS{O}_4$ will be:

• A. ${10}^{3}M$
• B. $4\times {10}^{6}M$
• C. ${10}^{-3}M$
• D. $4\times {10}^{-6}M$

Answer 1

The correct option is A ${10}^{3}M$

$\begin{array}{cc}\Lambda eq=1.53mhoc{m}^{2}e{q}^{-1}& K=3.6\times {10}^{-6}mhoc{m}^{-1}\\ \Lambda eq=\frac{K}{C}& \\ C=\frac{K}{\Lambda eq}\Rightarrow \frac{3.06\times {10}^{-6}c{m}^{-1}mho}{1.53\times mhoc{m}^{2}e{q}^{-1}}& \\ \Rightarrow 2.00\times {10}^{-6}c{m}^{-3}eq.\Rightarrow 2.00\times {10}^{-6}\times 1000d{m}^{-3}e{q}^{-1}& \\ C=2.00\times {10}^{-3}Le{q}^{-1}& N=nM4n-factor\\ \Rightarrow 2\times {10}^{-3}N& \\ forBaS{O}_{4}n-factor\to 2& \\ C=\frac{2\times {10}^{-3}}{2}M\Rightarrow 1\times {10}^{-3}M& \\ Ksp=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]& \\ \Rightarrow \left(M{o}^{-3}\right)\left(1\times {10}^{-3}\right)& \\ Ksp={10}^3& \end{array}$