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Ionic Equilibrium question

2023 · 8 Apr · Shift 2 · Q16
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  5. /2023 · 8 Apr · Shift 2 · Q16

Ionic Equilibrium question

2023 · 8 Apr · Shift 2 · Q16

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
The solubility product of BaSO4\mathrm{BaSO}_{4}BaSO4​ is 1×10−101 \times 10^{-10}1×10−10 at 298 K298 \mathrm{~K}298 K. The solubility of BaSO4\mathrm{BaSO}_{4}BaSO4​ in 0.1 M K2SO4(aq)0.1 ~\mathrm{M} ~\mathrm{K}_{2} \mathrm{SO}_{4}(\mathrm{aq})0.1 M K2​SO4​(aq) solution is ‾\underline{\hspace{2cm}}​×10−9 g L−1\times 10^{-9} \mathrm{~g} \mathrm{~L}^{-1}×10−9 g L−1(nearest integer). Given: Molar mass of BaSO4\mathrm{BaSO}_{4}BaSO4​ is 233 g mol−1233 \mathrm{~g} \mathrm{~mol}^{-1}233 g mol−1
Numerical answer
View written solutionFree

Correct answer: 233

  1. Write the dissolution equilibrium
BaSO4(s)⇌Ba2+(aq)+SO42−(aq)\mathrm{BaSO_4(s)} \rightleftharpoons \mathrm{Ba^{2+}(aq)} + \mathrm{SO_4^{2-}(aq)}BaSO4​(s)⇌Ba2+(aq)+SO42−​(aq)

For this equilibrium,

Ksp=[Ba2+][SO42−]=1×10−10K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{SO_4^{2-}}] = 1 \times 10^{-10}Ksp​=[Ba2+][SO42−​]=1×10−10
  1. Use the common ion effect

The solution already contains 0.1 M0.1\,\mathrm{M}0.1M K2SO4\mathrm{K_2SO_4}K2​SO4​, so initially

[SO42−]≈0.1[\mathrm{SO_4^{2-}}] \approx 0.1[SO42−​]≈0.1

Let the solubility of BaSO4\mathrm{BaSO_4}BaSO4​ in this solution be s mol L−1s\,\mathrm{mol\,L^{-1}}smolL−1. Then,

[Ba2+]=s,[SO42−]=0.1+s[\mathrm{Ba^{2+}}] = s, \qquad [\mathrm{SO_4^{2-}}] = 0.1 + s[Ba2+]=s,[SO42−​]=0.1+s

Since sss will be very small compared to 0.10.10.1, we take

0.1+s≈0.10.1 + s \approx 0.10.1+s≈0.1

Hence,

Ksp=s(0.1)=1×10−10K_{sp} = s(0.1) = 1 \times 10^{-10}Ksp​=s(0.1)=1×10−10

So,

s=1×10−100.1=1×10−9 mol L−1s = \frac{1 \times 10^{-10}}{0.1} = 1 \times 10^{-9}\,\mathrm{mol\,L^{-1}}s=0.11×10−10​=1×10−9molL−1
  1. Convert molar solubility to g L−1^{-1}−1

Given molar mass of BaSO4\mathrm{BaSO_4}BaSO4​ is 233 g mol−1233\,\mathrm{g\,mol^{-1}}233gmol−1.

Solubility in g L−1=s×233\text{Solubility in g L}^{-1} = s \times 233Solubility in g L−1=s×233 =1×10−9×233= 1 \times 10^{-9} \times 233=1×10−9×233 =233×10−9 g L−1= 233 \times 10^{-9}\,\mathrm{g\,L^{-1}}=233×10−9gL−1

Thus, the required number is

233\boxed{233}233​
  1. Comparison with stored correct answer

Stored correct answer = 233233233

Our derived answer = 233233233

So they agree.

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