JEE PYQ: Ionic Equilibrium Question 21
Question 21 - 2019 (08 Apr Shift 1)
Type: MCQ
If solubility product of $\text{Zr}_3(\text{PO}4)4$ 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) $S = \left(\frac{K_{sp}}{144}\right)^{1/6}$
(b) $S = \left(\frac{K_{sp}}{6912}\right)^{1/7}$
(c) $S = \left(\frac{K_{sp}}{929}\right)^{1/9}$
(d) $S = \left(\frac{K_{sp}}{216}\right)^{1/7}$
Show Answer
Answer: (a)
Solution
$\text{Zr}_3(\text{PO}_4)_4 \rightleftharpoons 3\text{Zr}^{4+} + 4\text{PO}_4^{3-}$
$K_{sp} = [\text{Zr}^{4+}]^3 [\text{PO}_4^{3-}]^4 = (3S)^3(4S)^4 = 27 \times 256 \times S^7 = 6912 S^7$
$S = \left(\frac{K_{sp}}{6912}\right)^{1/7}$
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