SATHEE: Equilibrium Question 840

Equilibrium Question 840

Question: Zirconium phosphate $[Z r_{3} {(P O_{4})}_{4}]$ dissociates into three zirconium cations of charge + 4 and four phosphate anions of charge $- 3$ . If molar solubility of zirconium phosphate is denoted by S and its solubility product by K then which of the following relationship between S and $K_{s p}$ is correc'

Options:

A) $S = K_{s p} / {(6912)}^{1 / 7}$

B) $S = {K_{s p} / 144}^{1 / 7}$

C) $S = (K_{s p} / 6912)^{1 / 7}$

D) $S = {K_{s p} / 6912}^{7}$

Show Answer

Answer:

Correct Answer: C

Solution:

$[Z r_{3} {(P O_{4})}_{4}] \to \underset{3 S}{3 Z r^{4 +}}, + \underset{4 S}{4 P O_{4}^{3 -}},$

$K_{s p} = {(3 S)}^{3} {(4 S)}^{4}$

$= 27 S^{3} \times 256 S^{4}$

$= 6912 S^{7} .$

$∴ S = {(\frac{K_{s p}}{6912})}^{1 / 7}$

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Equilibrium Question 840

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