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Ionic Equilibrium question

2013 · Shift 2 · Q6
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  5. /2013 · Shift 2 · Q6

Ionic Equilibrium question

2013 · Shift 2 · Q6

JEE AdvancedChemistryIonic EquilibriumMultiple correct+3 / −0.75
The Ksp of Ag2CrO4Ag_2CrO_4Ag2​CrO4​ is 1.1 ×\times× 10-12 at 298 K. The solubility (in mol/L) of Ag2CrO4Ag_2CrO_4Ag2​CrO4​ in a 0.1 M AgNO3AgNO_3AgNO3​ solution is
  1. A
    1.1 ×\times× 10-11
  2. B
    1.1 ×\times× 10-10
  3. C
    1.1 ×\times× 10-12
  4. D
    1.1 ×\times× 10-9
View written solutionFree

Correct answer: B

  1. Write the dissolution equilibrium

For silver chromate:

Ag2CrO4(s)⇌2Ag++CrO42−Ag_2CrO_4(s) \rightleftharpoons 2Ag^+ + CrO_4^{2-}Ag2​CrO4​(s)⇌2Ag++CrO42−​

So,

Ksp=[Ag+]2[CrO42−]K_{sp} = [Ag^+]^2[CrO_4^{2-}]Ksp​=[Ag+]2[CrO42−​]

Given:

Ksp=1.1×10−12K_{sp} = 1.1 \times 10^{-12}Ksp​=1.1×10−12

  1. Let the solubility in 0.1 M0.1\,M0.1M AgNO3AgNO_3AgNO3​ be sss mol/L

Since AgNO3AgNO_3AgNO3​ is a strong electrolyte, it provides initially:

[Ag+]=0.1 M[Ag^+] = 0.1\,M[Ag+]=0.1M

If sss mol/L of Ag2CrO4Ag_2CrO_4Ag2​CrO4​ dissolves, then:

  • [CrO42−]=s[CrO_4^{2-}] = s[CrO42−​]=s
  • [Ag+]=0.1+2s[Ag^+] = 0.1 + 2s[Ag+]=0.1+2s

Because solubility will be very small in presence of common ion Ag+Ag^+Ag+,

2s≪0.12s \ll 0.12s≪0.1

So we take:

[Ag+]≈0.1[Ag^+] \approx 0.1[Ag+]≈0.1

  1. Substitute into the solubility product expression

Ksp=[Ag+]2[CrO42−]K_{sp} = [Ag^+]^2[CrO_4^{2-}]Ksp​=[Ag+]2[CrO42−​]

1.1×10−12=(0.1)2(s)1.1 \times 10^{-12} = (0.1)^2(s)1.1×10−12=(0.1)2(s)

1.1×10−12=10−2s1.1 \times 10^{-12} = 10^{-2}s1.1×10−12=10−2s

s=1.1×10−1210−2=1.1×10−10 mol/Ls = \frac{1.1 \times 10^{-12}}{10^{-2}} = 1.1 \times 10^{-10}\,\text{mol/L}s=10−21.1×10−12​=1.1×10−10mol/L

  1. Check options
  • A: 1.1×10−111.1 \times 10^{-11}1.1×10−11 ❌
  • B: 1.1×10−101.1 \times 10^{-10}1.1×10−10 ✅
  • C: 1.1×10−121.1 \times 10^{-12}1.1×10−12 ❌
  • D: 1.1×10−91.1 \times 10^{-9}1.1×10−9 ❌

Therefore, the correct option is B.

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