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Ionic Equilibrium question

2022 · 27 Jul · Shift 1 · Q19
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  5. /2022 · 27 Jul · Shift 1 · Q19

Ionic Equilibrium question

2022 · 27 Jul · Shift 1 · Q19

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
At 310 K310 \mathrm{~K}310 K, the solubility of CaF2\mathrm{CaF}_{2}CaF2​ in water is 2.34×10−3 g/100 mL2.34 \times 10^{-3} \mathrm{~g} / 100 \mathrm{~mL}2.34×10−3 g/100 mL. The solubility product of CaF2\mathrm{CaF}_{2}CaF2​ is ‾\underline{\hspace{2cm}}​×10−8( mol/L)3\times 10^{-8}(\mathrm{~mol} / \mathrm{L})^{3}×10−8( mol/L)3. (Give molar mass : CaF2=78 g mol−1\mathrm{CaF}_{2}=78 \mathrm{~g} \mathrm{~mol}^{-1}CaF2​=78 g mol−1)
Numerical answer
View written solutionFree

Correct answer: 0.0108, KSP = 1.08 × 10^-10

  1. Write the dissolution equilibrium
CaF2(s)⇌Ca2+(aq)+2F−(aq)\mathrm{CaF_2(s)} \rightleftharpoons \mathrm{Ca^{2+}(aq)} + 2\mathrm{F^-(aq)}CaF2​(s)⇌Ca2+(aq)+2F−(aq)

If the molar solubility is sss mol/L, then

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

So,

Ksp=[Ca2+][F−]2=s(2s)2=4s3K_{sp}=[\mathrm{Ca^{2+}}][\mathrm{F^-}]^2=s(2s)^2=4s^3Ksp​=[Ca2+][F−]2=s(2s)2=4s3
  1. Convert given solubility into g/L

Given solubility:

2.34×10−3 g per 100 mL2.34\times 10^{-3}\ \text{g per 100 mL}2.34×10−3 g per 100 mL

Since 100 mL=0.1 L100\,\text{mL}=0.1\,\text{L}100mL=0.1L,

solubility in g/L=2.34×10−3×10=2.34×10−2 g/L\text{solubility in g/L} = 2.34\times 10^{-3}\times 10 = 2.34\times 10^{-2}\ \text{g/L}solubility in g/L=2.34×10−3×10=2.34×10−2 g/L
  1. Convert into molar solubility

Molar mass of CaF2=78 g mol−1\mathrm{CaF_2}=78\,\text{g mol}^{-1}CaF2​=78g mol−1

s=2.34×10−278s=\frac{2.34\times 10^{-2}}{78}s=782.34×10−2​ s=3.0×10−4 mol/Ls=3.0\times 10^{-4}\ \text{mol/L}s=3.0×10−4 mol/L
  1. Calculate KspK_{sp}Ksp​
Ksp=4s3=4(3.0×10−4)3K_{sp}=4s^3=4(3.0\times 10^{-4})^3Ksp​=4s3=4(3.0×10−4)3

First,

(3.0×10−4)3=27×10−12=2.7×10−11(3.0\times 10^{-4})^3=27\times 10^{-12}=2.7\times 10^{-11}(3.0×10−4)3=27×10−12=2.7×10−11

Thus,

Ksp=4×2.7×10−11=10.8×10−11=1.08×10−10K_{sp}=4\times 2.7\times 10^{-11}=10.8\times 10^{-11}=1.08\times 10^{-10}Ksp​=4×2.7×10−11=10.8×10−11=1.08×10−10
  1. Match with required format

The question asks:

Ksp=‾×10−8 (mol/L)3K_{sp}=\underline{\hspace{2cm}}\times 10^{-8}\,(\text{mol/L})^3Ksp​=​×10−8(mol/L)3

We have

1.08×10−10=0.0108×10−81.08\times 10^{-10}=0.0108\times 10^{-8}1.08×10−10=0.0108×10−8

So the blank is

0.01080.01080.0108

If an integer is strictly required, this is unusual because the value is not an integer. The actual calculated coefficient is 0.01080.01080.0108.

  1. Final derived answer
Ksp=1.08×10−10 (mol/L)3K_{sp}=1.08\times 10^{-10}\,(\text{mol/L})^3Ksp​=1.08×10−10(mol/L)3

or equivalently, the blank is

0.01080.01080.0108
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