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MHT CET 202624 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
24 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

MHT CET 2026 · 24 April · Evening Shift · Chemistry · Ionic Equilibrium

Solubility of MA_2 type of electrolyte is 0.5 × 10^ -4 \ mol/litre...

Solubility of MA2\text{MA}_2 type of electrolyte is 0.5×10−4 mol/litre0.5 \times 10^{-4}\ \text{mol/litre}. Then find out kspk_{sp} of electrolyte?

Options

  1. A5×10−135 \times 10^{-13}
  2. B5×10−125 \times 10^{-12}
  3. C25×10−1325 \times 10^{-13}
  4. D5×10−145 \times 10^{-14}

Correct answer

A. 5×10−135 \times 10^{-13}

Step-by-step solution

The dissociation of the electrolyte MA2\text{MA}_2 is given by:

MA2⇌\textM2++2\textA−\text{MA}_2 \rightleftharpoons \textM^{2+} + 2\textA^{-}

Let the solubility of the electrolyte be ss. Then the concentrations of the ions are:

[\textM2+]=s[\textM^{2+}] = s

[\textA−]=2s[\textA^{-}] = 2s

The solubility product KspK_{sp} is:

Ksp=[\textM2+][\textA−]2K_{sp} = [\textM^{2+}][\textA^{-}]^2

Ksp=(s)(2s)2=4s3K_{sp} = (s)(2s)^2 = 4s^3

Given s=0.5×10−4 mol/L=5×10−5 mol/Ls = 0.5 \times 10^{-4}\ \text{mol/L} = 5 \times 10^{-5}\ \text{mol/L}, substituting this value:

Ksp=4×(5×10−5)3K_{sp} = 4 \times (5 \times 10^{-5})^3

Ksp=4×125×10−15K_{sp} = 4 \times 125 \times 10^{-15}

Ksp=500×10−15=5×10−13K_{sp} = 500 \times 10^{-15} = 5 \times 10^{-13}

Answer: 5×10−135 \times 10^{-13}
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