Chemical And Ionic Equilibrium Result Question 5
5. If solubility product of $\mathrm{Zr}_3\left(\mathrm{PO}4\right)4$ is denoted by $K{\text {sp }}$ and its molar solubility is denoted by $S$, then which of the following relation between $S$ and $K{\text {sp }}$ is correct?
(2019 Main, 8 April I)
(a) $S=\left(\frac{K_{\mathrm{sp}}}{144}\right)^{1 / 6}$
(b) $S=\left(\frac{K_{\mathrm{sp}}}{6912}\right)^{1 / 7}$
(c) $S=\left(\frac{K_{\mathrm{sp}}}{929}\right)^{1 / 9}$
(d) $S=\left(\frac{K_{\mathrm{sp}}}{216}\right)^{1 / 7}$
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Answer:
Correct Answer: 5. ( b )
Solution:
- Key Idea: The concentration of a substance in a saturated solution is defined as its solubility $(S)$.
For $A_x B_y \rightleftharpoons x A^{y+}+y B^{x-} ; K_{s p}=\left[A^{y+}\right]^x\left[B^{x-}\right]^y$
For, $\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4$,
$\begin{aligned} & \mathrm{Zr}_3\left(\mathrm{PO}_4\right)4(s) \rightleftharpoons \underset{3 S \mathrm{M}}{3 \mathrm{Zr}^{4+}}(a q)+ \underset{4 S \mathrm{M}}{4 \mathrm{PO}4^{3-}}(a q) \\ & K{\mathrm{sp}}=\left[\mathrm{Zr}^{4+}\right]^3\left[\mathrm{PO}4^{3-}\right]^4 \\ & \quad K{\mathrm{sp}}=(3 S)^3(4 S)^4=6912 S^7 \text { or } S=\left(\frac{K{\mathrm{sp}}}{6912}\right)^{\frac{1}{7}} \end{aligned}$
Thus, the relation between molar solubility $(S)$ and solubility product $\left(K_{\mathrm{sp}}\right)$ will be
$S=\left(\frac{K_{\mathrm{sp}}}{6912}\right)^{1 / 7}$
BETA
