Question

The conductivity of a saturated solution of $A{g}_{3}P{O}_4$ is $9\times {10}^{-6}S{m}^{-1}$ and its equivalent conductivity is $1.50\times {10}^{-6}S{m}^{-1}equivalen{t}^{-1}$. The $K_{sp}$ of $A{g}_{3}P{O}_4$ is:

• A. $4.32\times {10}^{-18}$
• B. $1.8\times {10}^{-9}$
• C. $8.64\times {10}^{-13}$
• D. none of these

Answer 1

Answer: (D)

none of these

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

$s$ - -

- $3s$ $s$

$K_{sp}=\left(35{)}^3\right(s)=27s^4$

as $s=\frac{K}{{\wedge }_{eq}}$ $K$= specific conductance; ${\wedge }_{eq}$= Equivalent conductance

$\Rightarrow K_{sp}=27{\left(\frac{K}{{\wedge }_{eq}}\right)}^4$

$=27{\left(\frac{9\times {10}^{-6}s{m}^{-1}}{1.50\times {10}^{-6}s{m}^{-1}equivalen{t}^{-1}}\right)}^4$

$K_{sp}=34992$