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JEE Main 20208 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2020 · 8 January · Morning Shift · Chemistry · Ionic Equilibrium

The stoichiometry and solubility product of a salt with the...

The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively:

Options

  1. AX2Y,2×10−9M3{X}_{2}Y, 2 \times {10}^{- 9}{M}^{3}
  2. BXY2,4×10−9M3X{Y}_{2}, 4 \times {10}^{- 9}{M}^{3}
  3. CXY2,1×10−9M3X{Y}_{2}, 1 \times {10}^{- 9}{M}^{3}
  4. DXY,2×10−6M3XY, 2 \times {10}^{- 6}{M}^{3}

Correct answer

B. XY2,4×10−9M3X{Y}_{2}, 4 \times {10}^{- 9}{M}^{3}

Step-by-step solution

XY2(s)→X+(aq)10−3+2Y−(aq)2×10−3X{Y}_{2\left(s\right)} \to \mathop{{X}^{+} {}_{\left(aq\right)}}_{{10}^{- 3}} + \mathop{2{Y}^{-} {}_{\left(aq\right)}}_{2 \times {10}^{- 3}}

Ksp=[X+][Y−]2{K}_{sp} = \left[{X}^{+}\right]{\left[{Y}^{-}\right]}^{2}

Ksp=[10−3][2×10−3]2{K}_{sp} = \left[{10}^{- 3}\right]{\left[2 \times {10}^{- 3}\right]}^{2}

Ksp=4×10−9M3{K}_{sp} = 4 \times {10}^{- 9}{M}^{3}
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