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NEET 2020
Exam
NEET
Year
2020
Subject
Chemistry
Chapter
Ionic Equilibrium

NEET 2020 · Chemistry · Ionic Equilibrium

The solubility product for a salt of the type AB is 4 × 10 ^ - 8 ....

The solubility product for a salt of the type ABAB is 4×10−8.4 \times {10}^{- 8} . What is the molarity of its saturated solution?

Options

  1. A2×10−4mol/L2 \times {10}^{- 4}mol / L
  2. B16×10−16mol/L16 \times {10}^{- 16}mol / L
  3. C2×10−16mol/L2 \times {10}^{- 16}mol / L
  4. D4×10−4mol/L4 \times {10}^{- 4}mol / L

Correct answer

A. 2×10−4mol/L2 \times {10}^{- 4}mol / L

Step-by-step solution

Ksp{K}_{sp} for AB=s2=4×10−8AB = {s}^{2} = 4 \times {10}^{- 8}
∴∴ Molarity of solution = solubility =Ksp= \sqrt{{K}_{sp}}
=4×10−8= \sqrt{4 \times {10}^{- 8}}
=2×10−4mol/L= 2 \times {10}^{- 4}mol / L
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