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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

Calculate the solubility of AgCl in mol/dm^3 at 25\ ^ ^ C . Given:...

Calculate the solubility of AgCl in mol/dm3\text{mol/dm}^3 at 25\ ^{}^{\circ}\text{C}. Given: Ksp (AgCl)=1.8×10−10K_{sp}\ (\text{AgCl}) = 1.8 \times 10^{-10}

Options

  1. A1.8×10−5 M1.8 \times 10^{-5}\ \text{M}
  2. B1.34×10−5 M1.34 \times 10^{-5}\ \text{M}
  3. C1.8×10−10 M1.8 \times 10^{-10}\ \text{M}
  4. D3.6×10−5 M3.6 \times 10^{-5}\ \text{M}

Correct answer

B. 1.34×10−5 M1.34 \times 10^{-5}\ \text{M}

Step-by-step solution

The dissociation reaction of AgCl is given by:

AgCl(s)⇌Ag+(aq)+Cl−(aq)\text{AgCl(s)} \rightleftharpoons \text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq})

Let the solubility of AgCl be ss.

The solubility product constant is:

Ksp=[Ag+][Cl−]=s×s=s2K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s \times s = s^2

Given Ksp=1.8×10−10K_{sp} = 1.8 \times 10^{-10}, substituting this value:

s2=1.8×10−10s^2 = 1.8 \times 10^{-10}

s=1.8×10−10s = \sqrt{1.8 \times 10^{-10}}

s=1.34×10−5 mol/dm3s = 1.34 \times 10^{-5}\ \text{mol/dm}^3

Answer: 1.34×10−5 M1.34 \times 10^{-5}\ \text{M}
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