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JEE Main 20198 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2019 · 8 April · Morning Shift · Chemistry · Ionic Equilibrium

If solubility product of Z r _ 3 ( P O _ 4 ) _ 4 is denoted by K _...

If solubility product of Zr3(PO4)4Z{r}_{3}{( P{O}_{4} )}_{4} is denoted by Ksp{K}_{sp} and its molar solubility is denoted by SS , then which of the following relation between SS and Ksp{K}_{sp} is correct?

Options

  1. AS = {\left(\dfrac{{K}_{sp}}{144}\right)}^{\dfrac{1}{6}}
  2. BS = {\left(\dfrac{{K}_{sp}}{6912}\right)}^{\dfrac{1}{7}}
  3. CS = {\left(\dfrac{{K}_{sp}}{929}\right)}^{\dfrac{1}{9}}
  4. DS = {\left(\dfrac{{K}_{sp}}{216}\right)}^{\dfrac{1}{7}}

Correct answer

B. S = {\left(\dfrac{{K}_{sp}}{6912}\right)}^{\dfrac{1}{7}}

Step-by-step solution

Zr3(PO4)4(s)⇌3Zr+43s(aq)+4PO43−4sZ{r}_{3}{\left(P{O}_{4}\right)}_{4}\left(s\right) ⇌ \mathop{3Z{r}^{+ 4}}_{3s}\left(aq\right) + 4\mathop{P{O}_{4}^{3 -}}_{4s}\left(aq\right)<br/><br />{K}_{sp} = {\left[Z{r}^{+ 4}\right]}^{3}{\left[P{O}_{4}^{3 -}\right]}^{4}
=(3s)3(4s)4= {\left(3s\right)}^{3}{\left(4s\right)}^{4}
Ksp=6912S7{K}_{sp} = 6912{S}^{7}
s=(Ksp/6912)17s = {\left({K}_{sp} / 6912\right)}^{\dfrac{1}{7}}
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