Question

The molar solubility, $s$ of $B{a}_3(P{O}_4{)}_2$ in terms of $K_{sp}$ is:

• A. $s=(K_{sp}{)}^{1/2}$
• B. $s=(K_{sp}{)}^{1/5}$
• C. $s={\left(\frac{K_{sp}}{27}\right)}^{1/5}$
• D. $s={\left(\frac{K_{sp}}{108}\right)}^{1/5}$

Answer 1

The correct option is D $s={\left(\frac{K_{sp}}{108}\right)}^{1/5}$

The dissociation equilibrium for $B{a}_3(P{O}_4{)}_2$ is $B{a}_3(P{O}_4{)}_2\rightleftharpoons 3B{a}^{2+}+2P{O}_4^{3-}$

The expression for the solubility product is $K_{SP}=\left[B{a}^{2+}{]}^3\right[P{O}_4^{3-}{]}^2$

But, $\left[B{a}_3\right(P{O}_4{)}_2]=s,[B{a}^{2+}]=3s,[P{O}_4^{3-}]=2s$

Substitute values in the above expression,

$K_{sp}=\left[B{a}^{2+}{]}^3\right[P{O}_4^{3-}{]}^2=\left(3s{)}^3\right(2s{)}^2=108s^5$

Hence, $s={\left(\frac{K_{sp}}{108}\right)}^{1/5}$.