Question

The specific conductance of a saturated solution of $AgCl$ is $K{\Omega }^{-1}c{m}^{-1}$. The amount of $Ag$ and $C{l}^{-}$ are $x$ and $y$ respectively. The solubility product of $AgCl$ is:

• A. $\frac{1000K}{x+y}$
• B. $\frac{(1000K{)}^2}{(x+y{)}^2}$
• C. $\frac{1000\times 143.5\times K}{x+y}$
• D. none of these

Answer 1

Answer: (B)

$\frac{(1000K{)}^2}{(x+y{)}^2}$

${\lambda }_{\infty }^{o}AgCl={\lambda }_{\infty }^{o}Ag+{\lambda }_{\infty }^{o}C{l}^{-}$

$=x+y$

${\lambda }_{\infty }^{o}AgCl=\frac{1000K}{S}$

$S=\frac{1000K}{x+y}$

$Ag\to A{g}^{+}+C{l}^{-}$

$S$ $S$

$K_{sp}=S^2=\frac{(1000K{)}^2}{(x+y{)}^2}$