Question

If thesolubility product of $Z{r}_3(P{O}_4{)}_4$ is denoted by $K_{sp}$, and its molar solubility is denoted by $S$, then which of the following relation between $S$ and $K_{sp}$ is correct?

• A. $S=(\frac{K_{sp}}{6912}{)}^{\frac{1}{7}}$
• B. $S=(\frac{K_{sp}}{144}{)}^{\frac{1}{6}}$
• C. $S=(\frac{K_{sp}}{929}{)}^{\frac{1}{9}}$
• D. $S=(\frac{K_{sp}}{216}{)}^{\frac{1}{7}}$

Answer 1

The correct option is A $S=(\frac{K_{sp}}{6912}{)}^{\frac{1}{7}}$

We know,
$Z{r}_3\left(P{O}_4{)}_4\right(aq)\leftrightharpoons 3Z{r}^{4+}(aq)+4P{O}_4^{3-}(aq)$
$\Rightarrow K_{sp}=\left[Z{r}^{4+}{]}^3\right[P{O}_4^{3-}{]}^4$
$\Rightarrow K_{sp}=\left(3S{)}^3\right(4S{)}^4=6912S^7$
$\Rightarrow S=(\frac{K_{sp}}{6912}{)}^{\frac{1}{7}}$