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Ionic Equilibrium question

2018 · 15 Apr · Shift 1 · Q25
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  5. /2018 · 15 Apr · Shift 1 · Q25

Ionic Equilibrium question

2018 · 15 Apr · Shift 1 · Q25

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution (Ksp of PbCl2PbCl_2PbCl2​ = 3.2 ×\times× 10-8 atomic mass of Pb = 207 u ) is :
  1. A
    0.36 L
  2. B
    17.98 L
  3. C
    0.18 L
  4. D
    1.798 L
View written solutionFree

Correct answer: C

  1. Write the dissolution equilibrium
PbCl2(s)⇌Pb2++2Cl−PbCl_2(s) \rightleftharpoons Pb^{2+} + 2Cl^-PbCl2​(s)⇌Pb2++2Cl−

If the solubility of PbCl2PbCl_2PbCl2​ in water is s mol L−1s\,\text{mol L}^{-1}smol L−1, then

[Pb2+]=s,[Cl−]=2s[Pb^{2+}] = s, \qquad [Cl^-] = 2s[Pb2+]=s,[Cl−]=2s
  1. Use the solubility product expression
Ksp=[Pb2+][Cl−]2=s(2s)2=4s3K_{sp} = [Pb^{2+}][Cl^-]^2 = s(2s)^2 = 4s^3Ksp​=[Pb2+][Cl−]2=s(2s)2=4s3

Given:

Ksp=3.2×10−8K_{sp} = 3.2 \times 10^{-8}Ksp​=3.2×10−8

So,

4s3=3.2×10−84s^3 = 3.2 \times 10^{-8}4s3=3.2×10−8 s3=3.2×10−84=0.8×10−8=8×10−9s^3 = \frac{3.2 \times 10^{-8}}{4} = 0.8 \times 10^{-8} = 8 \times 10^{-9}s3=43.2×10−8​=0.8×10−8=8×10−9 s=8×10−93=2×10−3 mol L−1s = \sqrt[3]{8 \times 10^{-9}} = 2 \times 10^{-3}\,\text{mol L}^{-1}s=38×10−9​=2×10−3mol L−1
  1. Find molar mass of PbCl2PbCl_2PbCl2​
M(PbCl2)=207+2(35.5)=278 g mol−1M(PbCl_2) = 207 + 2(35.5) = 278\,\text{g mol}^{-1}M(PbCl2​)=207+2(35.5)=278g mol−1
  1. Find moles in 0.1 g of PbCl2PbCl_2PbCl2​
n=0.1278≈3.597×10−4 moln = \frac{0.1}{278} \approx 3.597 \times 10^{-4}\,\text{mol}n=2780.1​≈3.597×10−4mol
  1. Calculate minimum volume for a saturated solution

At saturation, concentration of dissolved PbCl2PbCl_2PbCl2​ is s=2×10−3 mol L−1s = 2 \times 10^{-3}\,\text{mol L}^{-1}s=2×10−3mol L−1.

Hence,

V=ns=3.597×10−42×10−3V = \frac{n}{s} = \frac{3.597 \times 10^{-4}}{2 \times 10^{-3}}V=sn​=2×10−33.597×10−4​ V≈0.17985 LV \approx 0.17985\,\text{L}V≈0.17985L V≈0.18 LV \approx 0.18\,\text{L}V≈0.18L
  1. Match with options

The correct option is:

C: 0.18 L\boxed{\text{C: } 0.18\,\text{L}}C: 0.18L​
  1. Comparison with stored answer

Stored correct answer: C

Our derived answer is also C, so they agree.

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