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Ionic Equilibrium question

2022 · Shift 2 · Q1
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  5. /2022 · Shift 2 · Q1

Ionic Equilibrium question

2022 · Shift 2 · Q1

JEE AdvancedChemistryIonic EquilibriumNumerical+3 / −1
Concentration of H2SO4\mathrm{H}_{2} \mathrm{SO}_{4}H2​SO4​ and Na2SO4\mathrm{Na}_{2} \mathrm{SO}_{4}Na2​SO4​ in a solution is 1M1 \mathrm{M}1M and 1.8×10−2M1.8 \times 10^{-2} \mathrm{M}1.8×10−2M, respectively. Molar solubility of PbSO4\mathrm{PbSO}_{4}PbSO4​ in the same solution is X×10−YM\mathrm{X} \times 10^{-\mathrm{Y}} \mathrm{M}X×10−YM(expressed in scientific notation). The value of YYY is ‾\underline{\hspace{2cm}}​. [Given: Solubility product of PbSO4(Ksp)=1.6×10−8\mathrm{PbSO}_{4}\left(K_{s p}\right)=1.6 \times 10^{-8}PbSO4​(Ksp​)=1.6×10−8. For H2SO4,Kal\mathrm{H}_{2} \mathrm{SO}_{4}, K_{a l}H2​SO4​,Kal​ is very large and Ka2=1.2×10−2]\left.K_{a 2}=1.2 \times 10^{-2}\right]Ka2​=1.2×10−2]
Numerical answer
View written solutionFree

Correct answer: 7

  1. Relevant equilibria

We need the sulfate ion concentration in the given solution, because for

PbSO4(s)⇌Pb2++SO42−,\mathrm{PbSO_4(s)} \rightleftharpoons \mathrm{Pb^{2+}} + \mathrm{SO_4^{2-}},PbSO4​(s)⇌Pb2++SO42−​,

after adding solid PbSO4\mathrm{PbSO_4}PbSO4​,

Ksp=[Pb2+][SO42−]=1.6×10−8.K_{sp}=[\mathrm{Pb^{2+}}][\mathrm{SO_4^{2-}}]=1.6\times 10^{-8}.Ksp​=[Pb2+][SO42−​]=1.6×10−8.

If the molar solubility is sss, then

[Pb2+]=s,[\mathrm{Pb^{2+}}]=s,[Pb2+]=s,

and since the medium already contains a large sulfate concentration, we can take

[SO42−]≈initial sulfate concentration in solution.[\mathrm{SO_4^{2-}}] \approx \text{initial sulfate concentration in solution}.[SO42−​]≈initial sulfate concentration in solution.

So first find [SO42−][\mathrm{SO_4^{2-}}][SO42−​] in the mixture of 1 M1\,\mathrm M1M H2SO4\mathrm{H_2SO_4}H2​SO4​ and 1.8×10−2 M1.8\times 10^{-2}\,\mathrm M1.8×10−2M Na2SO4\mathrm{Na_2SO_4}Na2​SO4​.

  1. Second dissociation of sulfuric acid

First dissociation is complete:

H2SO4→H++HSO4−\mathrm{H_2SO_4} \to \mathrm{H^+} + \mathrm{HSO_4^-}H2​SO4​→H++HSO4−​

So from 1 M1\,\mathrm M1M H2SO4\mathrm{H_2SO_4}H2​SO4​ initially we get

  • [H+]=1[\mathrm{H^+}] = 1[H+]=1
  • [HSO4−]=1[\mathrm{HSO_4^-}] = 1[HSO4−​]=1

Also, Na2SO4\mathrm{Na_2SO_4}Na2​SO4​ fully dissociates and contributes

[SO42−]=1.8×10−2.[\mathrm{SO_4^{2-}}] = 1.8\times 10^{-2}.[SO42−​]=1.8×10−2.

Now consider:

HSO4−⇌H++SO42−,Ka2=1.2×10−2\mathrm{HSO_4^-} \rightleftharpoons \mathrm{H^+} + \mathrm{SO_4^{2-}}, \qquad K_{a2}=1.2\times 10^{-2}HSO4−​⇌H++SO42−​,Ka2​=1.2×10−2

Let xxx M of HSO4−\mathrm{HSO_4^-}HSO4−​ dissociate. Then,

[HSO4−]=1−x,[\mathrm{HSO_4^-}] = 1-x,[HSO4−​]=1−x, [H+]=1+x,[\mathrm{H^+}] = 1+x,[H+]=1+x, [SO42−]=1.8×10−2+x.[\mathrm{SO_4^{2-}}] = 1.8\times 10^{-2}+x.[SO42−​]=1.8×10−2+x.

Apply Ka2K_{a2}Ka2​:

Ka2=[H+][SO42−][HSO4−]=(1+x)(1.8×10−2+x)1−x=1.2×10−2K_{a2}=\frac{[\mathrm{H^+}][\mathrm{SO_4^{2-}}]}{[\mathrm{HSO_4^-}]} =\frac{(1+x)(1.8\times10^{-2}+x)}{1-x}=1.2\times10^{-2}Ka2​=[HSO4−​][H+][SO42−​]​=1−x(1+x)(1.8×10−2+x)​=1.2×10−2

Since xxx is small compared with 1, use 1+x≈11+x\approx11+x≈1 and 1−x≈11-x\approx11−x≈1:

1.8×10−2+x≈1.2×10−21.8\times10^{-2}+x \approx 1.2\times10^{-2}1.8×10−2+x≈1.2×10−2

which gives a negative xxx, impossible. This means the already present sulfate from Na2SO4\mathrm{Na_2SO_4}Na2​SO4​ suppresses dissociation strongly, so the equilibrium actually shifts left, and we should directly use the equilibrium relation with the dominant acid concentration.

Because the solution already has about 1 M1\,\mathrm M1M H+\mathrm{H^+}H+, for the pair HSO4−/SO42−\mathrm{HSO_4^-}/\mathrm{SO_4^{2-}}HSO4−​/SO42−​:

[SO42−][HSO4−]=Ka2[H+]approx1.2×10−21=1.2×10−2\frac{[\mathrm{SO_4^{2-}}]}{[\mathrm{HSO_4^-}]}=\frac{K_{a2}}{[\mathrm{H^+}]} approx \frac{1.2\times10^{-2}}{1}=1.2\times10^{-2}[HSO4−​][SO42−​]​=[H+]Ka2​​approx11.2×10−2​=1.2×10−2

Thus with total sulfur from sulfuric acid mostly as HSO4−+SO42−≈1\mathrm{HSO_4^-}+\mathrm{SO_4^{2-}}\approx 1HSO4−​+SO42−​≈1, we get sulfate from acid approximately

[SO42−]from acid≈1.2×10−2.[\mathrm{SO_4^{2-}}]_{\text{from acid}} \approx 1.2\times10^{-2}.[SO42−​]from acid​≈1.2×10−2.

Adding sulfate from sodium sulfate:

[SO42−]≈1.2×10−2+1.8×10−2=3.0×10−2 M.[\mathrm{SO_4^{2-}}] \approx 1.2\times10^{-2}+1.8\times10^{-2}=3.0\times10^{-2}\,\mathrm M.[SO42−​]≈1.2×10−2+1.8×10−2=3.0×10−2M.
  1. Solubility of PbSO4\mathrm{PbSO_4}PbSO4​

Now,

Ksp=s[SO42−]K_{sp}=s[\mathrm{SO_4^{2-}}]Ksp​=s[SO42−​]

so

s=1.6×10−83.0×10−2s=\frac{1.6\times10^{-8}}{3.0\times10^{-2}}s=3.0×10−21.6×10−8​ s≈5.33×10−7 Ms\approx 5.33\times10^{-7}\,\mathrm Ms≈5.33×10−7M

Thus in scientific notation,

s=X×10−Ys = X\times 10^{-Y}s=X×10−Y

with

Y=7.Y=7.Y=7.
  1. Comparison with stored answer

My derived value is

Y=7Y=7Y=7

whereas the stored correct answer is 666.

So I do not agree with the stored answer.

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