Question

Conductivity of saturated solution of silver phosphate is $9\times {10}^{-6}$ S/m and equivalent conductivity $1.5\times {10}^{-4}$ S m${}^2$/equivalent. Find $K_{sp}$ of $A{g}_{3}P{O}_4$.

Answer 1

Given conductivity of $A{g}_{3}P{O}_4$, $k=9\times {10}^{-6}S{m}^{-1}eq$ of saturated solution

Conductivity of $A{g}_{3}P{O}_4$ at infinite dilution,

${\lambda }^o=1.5\times {10}^{-4}S{m}^{2}eq$

Thus, as concentration, $C=\frac{conductivity}{{\lambda }^o},k$

$\Rightarrow C=\frac{9\times {10}^{-6}S{m}^{-1}eq}{1.5\times {10}^{-4}S{m}^{2}eq}=0.06m^{-3}mol$

$\Rightarrow C=0.06\times {10}^{3}d{m}^{-3}mol$

$A{g}_{3}P{O}_4\rightleftharpoons 3A{g}^{+}+P{O}_4^{3-}$

$\Rightarrow K_{sp}=\left[A{g}^{+}{]}^3\right[P{O}_4{]}^{3-}$

$\Rightarrow K_{sp}=\left(3C{)}^3\right(C)=27C^4$

$=27\times [0.06\times {10}^3{]}^4$

$\Rightarrow K_{sp}=3.49\times {10}^8(mol{L}^{-1}{)}^4$