Q2 / 160 · TS EAMCET 2015 · Chemistry

If the solubility of $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$ in water is $x$ mol L $\mathrm{L}^{-1}$, its solubility product in $\operatorname{mol}^5 \mathrm{~L}^{-5}$ is

  • A
    $6 x^5$
  • B
    $36 x^5$
  • C
    $64 x^5$
  • Correct answer
    D
    $108 x^5$

Solution

$\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2 \rightleftharpoons 3 \mathrm{Ca}^{2+}+2 \mathrm{PO}_4^{3-}$ $$ \begin{aligned} K_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^3\left[\mathrm{PO}_4^{3-}\right]^2 & =[3 x]^3[2 \mathrm{x}]^2=\left[27 \mathrm{x}^3\right]\left[4 \mathrm{x}^2\right] \\ & =108 x^5 \end{aligned} $$

Hence, the correct answer is option: 'D'.

Relevant Questions