Question

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

[Given :
${\lambda }^0\left(A{g}^{+}\right)=63.0Sc{m}^{2}mo{l}^{-1}$
${\lambda }^0\left(C{l}^{-}\right)=67.0Sc{m}^{2}mo{l}^{-1}$]

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

Answer 1

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

At infinite dilution, solution can be considered to be of zero concentration or infinite dilution.

For a sparingle soluble salt at saturated solution is considered to be at a infinite dilution condition. Thus,
${\Lambda }_m^0={\Lambda }_m$

$\text{Solubility}=\text{Molarity}$

${\Lambda }_m^0=\frac{\kappa \times 1000}{\text{Solubilty}\left(S\right)}$

$\text{Solubilty}\left(S\right)=\frac{\kappa \times 1000}{{\Lambda }_m^0}$

${\Lambda }_m^0={\lambda }_{+}^0+{\lambda }_{-}^0$

${\Lambda }_{AgCl}^o={\lambda }_{A{g}^{+}}^o+{\lambda }_{C{l}^{-}}^o$

$\text{Solubility,}S=\frac{1000\times \kappa }{{\Lambda }_{AgCl}^o}$

$=\frac{1000\times 2.6\times {10}^{-6}}{{\lambda }_{A{g}^{+}}^0+{\lambda }_{Cl}^0}$

$=\frac{2.6\times {10}^{-3}}{63+67}$

$S=2\times {10}^{-5}mol{L}^{-1}$
For $AgCl$,
$AgCl\rightleftharpoons A{g}^{+}+C{l}^{-}$
$K_{sp}=S^2$
$K_{sp}=(2\times {10}^{-5}{)}^2{M}^2$
$K_{sp}=4.0\times {10}^{-10}{M}^2$