Question

At 298 K, the conductivity of a saturated solution of $AgCl$ in water is $2.6\times {10}^{-6}Sc{m}^{-1}$. What is its solubility product at 298K?

$\left[Given:{\lambda }^{\infty }\left(A{g}^{+}\right)=63.0Sc{m}^{2}mo{l}^{-1},{\lambda }^{\infty }\left(C{i}^{-}\right)=67.0Sc{m}^{2}mo{l}^{-1}\right]$

• A. $2.0\times {10}^{-5}{M}^2$
• B. $4.0\times {10}^{-20}{M}^2$
• C. $4.0\times {10}^{-16}{M}^2$
• D. $2\times {10}^{-8}{M}^2$

Answer 1

The correct option is B $4.0\times {10}^{-20}{M}^2$

$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$

$1-\alpha$ $\alpha$ $\alpha$

$\alpha =\frac{\lambda }{{\lambda }_{\infty }}$

$\therefore \alpha =\frac{2.6\times {10}^{-6}}{63+67}$

$\alpha =\frac{2.6}{130}\times {10}^{-6}$

$\therefore \alpha =2\times {10}^{-8}=\left[A{g}^{+}\right]=\left[C{l}^{-}\right]$

$\therefore K_{sp}=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

$={\alpha }^2$

$=(2\times {10}^{-8}{)}^2$

$=4\times {10}^{-16}c{m}^2$

$=4\times {10}^{-20}{M}^2$

Hence, option b is correct.