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Ionic Equilibrium question

2004 · Shift 0 · Q7
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Ionic Equilibrium question

2004 · Shift 0 · Q7

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
The molar solubility (in ol L-1) of a sparingly soluble salt MX4MX_4MX4​ is "s". The corresponding solubility product is Ksp. 's' is given in term of Ksp by the relation :
  1. A
    s = (256 Ksp)1/5
  2. B
    s = (128 Ksp)1/4
  3. C
    s = ( Ksp / 128)1/4
  4. D
    s = (Ksp / 256)1/5
View written solutionFree

Correct answer: D

  1. Write the dissociation of the salt

For the sparingly soluble salt MX4MX_4MX4​:

MX4(s)⇌M4++4X−MX_4(s) \rightleftharpoons M^{4+} + 4X^-MX4​(s)⇌M4++4X−

If the molar solubility is s mol L−1s\,\text{mol L}^{-1}smol L−1, then at equilibrium:

  • [M4+]=s[M^{4+}] = s[M4+]=s
  • [X−]=4s[X^-] = 4s[X−]=4s
  1. Write the solubility product expression

Ksp=[M4+][X−]4K_{sp} = [M^{4+}][X^-]^4Ksp​=[M4+][X−]4

Substitute the equilibrium concentrations:

Ksp=(s)(4s)4K_{sp} = (s)(4s)^4Ksp​=(s)(4s)4

  1. Simplify

Ksp=s⋅44s4K_{sp} = s \cdot 4^4 s^4Ksp​=s⋅44s4

Ksp=256s5K_{sp} = 256 s^5Ksp​=256s5

  1. Solve for sss

s5=Ksp256s^5 = \frac{K_{sp}}{256}s5=256Ksp​​

s=(Ksp256)1/5s = \left(\frac{K_{sp}}{256}\right)^{1/5}s=(256Ksp​​)1/5

  1. Match with the options

This corresponds to:

Option D:

s=(Ksp256)1/5s = \left(\frac{K_{sp}}{256}\right)^{1/5}s=(256Ksp​​)1/5

  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So, the answer agrees with the stored correct answer.

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