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Ionic Equilibrium question

2021 · 25 Feb · Shift 2 · Q3
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  5. /2021 · 25 Feb · Shift 2 · Q3

Ionic Equilibrium question

2021 · 25 Feb · Shift 2 · Q3

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
The solubility of Ca(OH)2Ca(OH)_2Ca(OH)2​ in water is : [Given : The solubility product of Ca(OH)2Ca(OH)_2Ca(OH)2​ in water = 5.5 ×\times× 10 −-− 6]
  1. A
    1.11 ×\times× 10 −-− 6
  2. B
    1.11 ×\times× 10 −-− 2
  3. C
    1.77 ×\times× 10 −-− 6
  4. D
    1.77 ×\times× 10 −-− 2
View written solutionFree

Correct answer: B

  1. Write the dissolution equilibrium

For calcium hydroxide:

Ca(OH)2(s)⇌Ca2++2OH−Ca(OH)_2 (s) \rightleftharpoons Ca^{2+} + 2OH^-Ca(OH)2​(s)⇌Ca2++2OH−

If the solubility is sss mol L−1^{-1}−1, then at equilibrium:

[Ca2+]=s,[OH−]=2s[Ca^{2+}] = s, \qquad [OH^-] = 2s[Ca2+]=s,[OH−]=2s

  1. Write the solubility product expression

Ksp=[Ca2+][OH−]2K_{sp} = [Ca^{2+}][OH^-]^2Ksp​=[Ca2+][OH−]2

Substitute the equilibrium concentrations:

Ksp=s(2s)2=4s3K_{sp} = s(2s)^2 = 4s^3Ksp​=s(2s)2=4s3

Given:

Ksp=5.5×10−6K_{sp} = 5.5 \times 10^{-6}Ksp​=5.5×10−6

So,

4s3=5.5×10−64s^3 = 5.5 \times 10^{-6}4s3=5.5×10−6

s3=5.5×10−64=1.375×10−6s^3 = \frac{5.5 \times 10^{-6}}{4} = 1.375 \times 10^{-6}s3=45.5×10−6​=1.375×10−6

  1. Calculate sss

s=1.375×10−63s = \sqrt[3]{1.375 \times 10^{-6}}s=31.375×10−6​

Now,

1.375=(1.1)31.375 = (1.1)^31.375=(1.1)3

and

10−6=(10−2)310^{-6} = (10^{-2})^310−6=(10−2)3

Therefore,

s=1.1×10−2 mol L−1s = 1.1 \times 10^{-2} \text{ mol L}^{-1}s=1.1×10−2 mol L−1

  1. Match with the given options

s=1.1×10−2≈1.11×10−2s = 1.1 \times 10^{-2} \approx 1.11 \times 10^{-2}s=1.1×10−2≈1.11×10−2

So the correct option is:

B: 1.11×10−21.11 \times 10^{-2}1.11×10−2

  1. Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

They agree.

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