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Ionic Equilibrium question

2021 · 18 Mar · Shift 2 · Q21
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  5. /2021 · 18 Mar · Shift 2 · Q21

Ionic Equilibrium question

2021 · 18 Mar · Shift 2 · Q21

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
The solubility of CdSO4CdSO_4CdSO4​ in water is 8.0 ×\times× 10 −-− 4 mol L −-− 1. Its solubility in 0.01 M H2SO4H_2SO_4H2​SO4​ solution is ‾×\underline{\hspace{2cm}}\times​× 10 −-− 6 mol L −-− 1. (Round off to the Nearest Integer). (Assume that solubility is much less than 0.01 M)
Numerical answer
View written solutionFree

Correct answer: 64

  1. Write the dissolution equilibrium

For cadmium sulfate,

CdSO4(s)⇌Cd2++SO42−CdSO_4(s) \rightleftharpoons Cd^{2+} + SO_4^{2-}CdSO4​(s)⇌Cd2++SO42−​

If its solubility in pure water is s=8.0×10−4 mol L−1s = 8.0 \times 10^{-4}\,\text{mol L}^{-1}s=8.0×10−4mol L−1, then

[Cd2+]=s,[SO42−]=s[Cd^{2+}] = s, \qquad [SO_4^{2-}] = s[Cd2+]=s,[SO42−​]=s

So,

Ksp=s2=(8.0×10−4)2=64×10−8=6.4×10−7K_{sp} = s^2 = (8.0 \times 10^{-4})^2 = 64 \times 10^{-8} = 6.4 \times 10^{-7}Ksp​=s2=(8.0×10−4)2=64×10−8=6.4×10−7
  1. Consider solubility in 0.01 M H2SO40.01\,M\,H_2SO_40.01MH2​SO4​

Assume sulfuric acid is the source of common ion SO42−SO_4^{2-}SO42−​ and, as stated, solubility is much smaller than 0.010.010.01 M. Then

[SO42−]≈0.01[SO_4^{2-}] \approx 0.01[SO42−​]≈0.01

Let the new solubility be s′s's′. Then

[Cd2+]=s′[Cd^{2+}] = s'[Cd2+]=s′

and

Ksp=[Cd2+][SO42−]=s′(0.01)K_{sp} = [Cd^{2+}][SO_4^{2-}] = s'(0.01)Ksp​=[Cd2+][SO42−​]=s′(0.01)

Thus,

s′=Ksp0.01=6.4×10−710−2=6.4×10−5 mol L−1s' = \frac{K_{sp}}{0.01} = \frac{6.4 \times 10^{-7}}{10^{-2}} = 6.4 \times 10^{-5}\,\text{mol L}^{-1}s′=0.01Ksp​​=10−26.4×10−7​=6.4×10−5mol L−1
  1. Match with the required format

We need to write solubility as

‾×10−6 mol L−1\underline{\hspace{1cm}} \times 10^{-6}\,\text{mol L}^{-1}​×10−6mol L−1

Now,

6.4×10−5=64×10−66.4 \times 10^{-5} = 64 \times 10^{-6}6.4×10−5=64×10−6

So the required integer is

64\boxed{64}64​
  1. Comparison with stored answer

Stored correct answer = 646464.

Our derived answer matches it.

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