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MHT CET 202624 AprilMorning Shift
Exam
MHT CET
Year
2026
Date
24 April
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

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

The solubility of AgBr at 298 K is 2×10^ -7 \ kg/dm^3 . Find the...

The solubility of AgBr at 298 K is 2×10−7 kg/dm32\times10^{-7}\ \text{kg/dm}^3. Find the solubility product of AgBr at same temperature. [Molar mass of AgBr=188\text{AgBr}=188 g/mol]

Options

  1. A1.981×10−121.981\times10^{-12}
  2. B1.632×10−121.632\times10^{-12}
  3. C1.425×10−121.425\times10^{-12}
  4. D1.124×10−121.124\times10^{-12}

Correct answer

D. 1.124×10−121.124\times10^{-12}

Step-by-step solution

Given solubility of AgBr =2×10−7 kg/dm3= 2 \times 10^{-7} \text{ kg/dm}^3

Since 1 kg=103 g1 \text{ kg} = 10^3 \text{ g} and 1 dm3=1 L1 \text{ dm}^3 = 1 \text{ L}, the solubility in g/L\text{g/L} is:

s=2×10−7×103 g/L=2×10−4 g/Ls = 2 \times 10^{-7} \times 10^3 \text{ g/L} = 2 \times 10^{-4} \text{ g/L}

Molar mass of AgBr =188 g/mol= 188 \text{ g/mol}

Molar solubility (S)inmol/L) in \text{mol/L} is:

S=2×10−4188 mol/L≈1.06×10−6 mol/LS = \dfrac{2 \times 10^{-4}}{188} \text{ mol/L} \approx 1.06 \times 10^{-6} \text{ mol/L}

The dissociation of AgBr is given by:

AgBr(s)⇌Ag+(aq)+Br−(aq)\text{AgBr(s)} \rightleftharpoons \text{Ag}^+(\text{aq}) + \text{Br}^-(\text{aq})

The solubility product KspK_{sp} is:

Ksp=[Ag+][Br−]=S2K_{sp} = [\text{Ag}^+][\text{Br}^-] = S^2

Ksp=(1.06×10−6)2=1.1236×10−12≈1.124×10−12K_{sp} = (1.06 \times 10^{-6})^2 = 1.1236 \times 10^{-12} \approx 1.124 \times 10^{-12}

Answer: 1.124×10−121.124\times10^{-12}
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