Question

Conductivity of saturated solution of silver phosphate is $9\times {10}^{-6}$ $S{m}^{-1}$ and equivalent conductivity $1.5\times {10}^{-4}$ $S{m}^2$/equivalent. Find $K_{sp}$ of $A{g}_{3}P{O}_4$.

• A. $3.49\times {10}^8(mol{L}^{-1}{)}^4$
• B. $3.49\times {10}^{-8}(mol{L}^{-1}{)}^4$
• C. $6.98\times {10}^8(mol{L}^{-1}{)}^4$
• D. $6.98\times {10}^{-8}(mol{L}^{-1}{)}^4$

Answer 1

The correct option is A $3.49\times {10}^8(mol{L}^{-1}{)}^4$

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, $C=\frac{conductivity}{{\lambda }^o},k$
$\Rightarrow C=\frac{9\times {10}^{-6}S{m}^{-1}eq}{1.5\times {10}^{-4}S{m}^{2}eq}=0.06m^{-3}mol$
$\Rightarrow C=0.06\times {10}^{3}d{m}^{-3}mol$
$A{g}_{3}P{O}_4\rightleftharpoons 3A{g}^{+}+P{O}_4^{3-}$
$\Rightarrow K_{sp}=\left[A{g}^{+}{]}^3\right[P{O}_4{]}^{3-}$
$\Rightarrow K_{sp}=\left(3C{)}^3\right(C)=27C^4$
$=27\times [0.06\times {10}^3{]}^4$
$\Rightarrow K_{sp}=3.49\times {10}^8(mol{L}^{-1}{)}^4$