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JEE Main 201419 AprilOnline
Exam
JEE Main
Year
2014
Date
19 April
Mode
Online
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2014 · 19 April · Online · Chemistry · Ionic Equilibrium

Zirconium phosphate [ Zr _ 3 ( PO _ 4 ) _ 4 ] dissociates into...

Zirconium phosphate [Zr3(PO4)4][ {Zr}_{3}{( {PO}_{4} )}_{4} ] dissociates into three zirconium cations of charge +4+ 4 and four phosphate anions of charge −3- 3. If molar solubility of zirconium phosphate is denoted by ss and its solubility product by Ksp{K}_{sp} then which of the following relationship between ss and Ksp{K}_{sp} is correct ?

Options

  1. AS={Ksp/6912}7\text{S} = {\left\{{\text{K}}_{\text{sp}} / 6 9 1 2\right\}}^{7}
  2. BS={Ksp/144}1/7\text{S} = {\left\{{\text{K}}_{\text{sp}} / 1 4 4\right\}}^{1 / 7}
  3. CS={Ksp/(6912)1/7}\text{S} = \left\{{\text{K}}_{\text{sp}} / {\left(6 9 1 2\right)}^{1 / 7}\right\}
  4. DS=(Ksp/6912)1/7\text{S} = {\left({\text{K}}_{sp} / 6 9 1 2\right)}^{1 / 7}

Correct answer

D. S=(Ksp/6912)1/7\text{S} = {\left({\text{K}}_{sp} / 6 9 1 2\right)}^{1 / 7}

Step-by-step solution

In the saturated solution of

Zr3(PO4)4+aq⇌3Zr4+3S molar+4PO43−4S molar{\text{Zr}}_{3} {\left({\text{PO}}_{4}\right)}_{4} + \text{aq} ⇌ 3 \mathop{{\text{Zr}}^{4 +}}_{3 \text{S molar}} + 4 \mathop{{\text{PO}}_{4}^{3 -}}_{4 \text{S molar}}

Ksp=[Zr4+]3[PO43−]4=(3S)3(4S)4{\text{K}}_{\text{sp}} = {\left[{\text{Zr}}^{4 +}\right]}^{3} {\left[{\text{PO}}_{4}^{3 -}\right]}^{4} = {\left(3 \text{S}\right)}^{3} {\left(4 \text{S}\right)}^{4}

=27×256×s7=6912S7= 2 7 × 2 5 6 × {\text{s}}^{7} = 6 9 1 2 {\text{S}}^{7}

S=(Ksp6912)1/7\text{S} = {\left(\dfrac{{\text{K}}_{\text{sp}}}{6 9 1 2}\right)}^{1 / 7}
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