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Ionic Equilibrium question

2005 · Shift 0 · Q6
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Ionic Equilibrium question

2005 · Shift 0 · Q6

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
The solubility product of a salt having general formula MX2MX_2MX2​, in water is: 4 ×\times× 10-12 . The concentration of M2+M^{2+}M2+ ions in the aqueous solution of the salt is :
  1. A
    2.0 ×\times× 10-6 M
  2. B
    4.0 ×\times× 10-10 M
  3. C
    1.0 ×\times× 10-4 M
  4. D
    1.6 ×\times× 10-4 M
View written solutionFree

Correct answer: C

  1. Write the dissolution equilibrium

For a salt of formula MX2MX_2MX2​:

MX2(s)⇌M2+(aq)+2X−(aq)MX_2(s) \rightleftharpoons M^{2+}(aq) + 2X^{-}(aq)MX2​(s)⇌M2+(aq)+2X−(aq)

  1. Assume molar solubility = sss

If the solubility of MX2MX_2MX2​ in water is sss mol L−1^{-1}−1, then at equilibrium:

[M2+]=s[M^{2+}] = s[M2+]=s [X−]=2s[X^-] = 2s[X−]=2s

  1. Write the solubility product expression

Ksp=[M2+][X−]2K_{sp} = [M^{2+}][X^-]^2Ksp​=[M2+][X−]2

Substitute the equilibrium concentrations:

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

Given:

Ksp=4×10−12K_{sp} = 4 \times 10^{-12}Ksp​=4×10−12

So,

4s3=4×10−124s^3 = 4 \times 10^{-12}4s3=4×10−12

s3=10−12s^3 = 10^{-12}s3=10−12

s=10−4s = 10^{-4}s=10−4

Thus,

[M2+]=s=1.0×10−4 M[M^{2+}] = s = 1.0 \times 10^{-4}\,\text{M}[M2+]=s=1.0×10−4M

  1. Check options
  • A: 2.0×10−62.0 \times 10^{-6}2.0×10−6 M — incorrect
  • B: 4.0×10−104.0 \times 10^{-10}4.0×10−10 M — incorrect
  • C: 1.0×10−41.0 \times 10^{-4}1.0×10−4 M — correct
  • D: 1.6×10−41.6 \times 10^{-4}1.6×10−4 M — incorrect

Therefore, the correct option is C.

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