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JEE MAIN 2019
08-04-2019 S1
Question
If solubility product of $\mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4$ is denoted by $\mathrm{K}_{\mathrm{sp}}$ and its molar solubility is denoted by S , then which of the following relation between S and $\mathrm{K}_{\mathrm{sp}}$ is correct?
Select the correct option:
A
$\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{929}\right)^{\frac{1}{9}}$
B
$S=\left(\frac{K_{s p}}{216}\right)^{\frac{1}{7}}$
C
$S=\left(\frac{K_{s p}}{144}\right)^{\frac{1}{6}}$
D
$S=\left(\frac{K_{s p}}{6912}\right)^{\frac{1}{7}}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \mathrm{Zr}_3\left(\mathrm{PO}_4\right)_4 \rightleftharpoons 3 \mathrm{Zr}_{3 \mathrm{~S}}^{+4}+4 \mathrm{PO}_4^{3-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Zr}^{+4}\right]^3\left[\mathrm{PO}_4^{3-}\right]^4=(3 \mathrm{~S})^3(4 \mathrm{~S})^4 \\ & \mathrm{~K}_{\mathrm{sp}}=6912 \mathrm{~S}^7 \\ & \mathrm{~S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}\right)^{1 / 7}\end{aligned}$
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