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Ionic Equilibrium question

2018 · Shift 0 · Q19
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Ionic Equilibrium question

2018 · Shift 0 · Q19

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
An aqueous solution contains 0.10 M H2SH_2SH2​S and 0.20 M HClHClHCl. If the equilibrium constants for the formation of HS−HS^-HS− from H2SH_2SH2​S is 1.0 ×\times× 10–7 and that of S2−S^{2-}S2− from HS−HS^-HS− ions is 1.2 ×\times× 10–13 then the concentration of S2−S^{2-}S2− ions in aqueous solution is :
  1. A
    5 ×\times× 10–19
  2. B
    5 ×\times× 10–8
  3. C
    3 ×\times× 10–20
  4. D
    6 ×\times× 10–21
View written solutionFree

Correct answer: C

  1. Write the equilibria and constants

For diprotic acid H2SH_2SH2​S:

H2S⇌H++HS−,Ka1=1.0×10−7H_2S \rightleftharpoons H^+ + HS^- , \qquad K_{a1}=1.0\times 10^{-7}H2​S⇌H++HS−,Ka1​=1.0×10−7

HS−⇌H++S2−,Ka2=1.2×10−13HS^- \rightleftharpoons H^+ + S^{2-} , \qquad K_{a2}=1.2\times 10^{-13}HS−⇌H++S2−,Ka2​=1.2×10−13

Given:

  • [H2S]=0.10 M[H_2S]=0.10\,M[H2​S]=0.10M
  • [HCl]=0.20 M[HCl]=0.20\,M[HCl]=0.20M

Since HClHClHCl is a strong acid, it dissociates completely, so approximately

[H+]≈0.20 M[H^+] \approx 0.20\,M[H+]≈0.20M

Because of the common ion effect, dissociation of H2SH_2SH2​S is strongly suppressed.

  1. Find [HS−][HS^-][HS−] using Ka1K_{a1}Ka1​

From

Ka1=[H+][HS−][H2S]K_{a1} = \frac{[H^+][HS^-]}{[H_2S]}Ka1​=[H2​S][H+][HS−]​

So,

[HS−]=Ka1[H2S][H+][HS^-] = \frac{K_{a1}[H_2S]}{[H^+]}[HS−]=[H+]Ka1​[H2​S]​

Substitute values:

[HS−]=(1.0×10−7)(0.10)0.20[HS^-] = \frac{(1.0\times 10^{-7})(0.10)}{0.20}[HS−]=0.20(1.0×10−7)(0.10)​

[HS−]=5.0×10−8 M[HS^-] = 5.0\times 10^{-8}\,M[HS−]=5.0×10−8M

  1. Find [S2−][S^{2-}][S2−] using Ka2K_{a2}Ka2​

From

Ka2=[H+][S2−][HS−]K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]}Ka2​=[HS−][H+][S2−]​

Thus,

[S2−]=Ka2[HS−][H+][S^{2-}] = \frac{K_{a2}[HS^-]}{[H^+]}[S2−]=[H+]Ka2​[HS−]​

Substitute [HS−]=5.0×10−8[HS^-]=5.0\times 10^{-8}[HS−]=5.0×10−8:

[S2−]=(1.2×10−13)(5.0×10−8)0.20[S^{2-}] = \frac{(1.2\times 10^{-13})(5.0\times 10^{-8})}{0.20}[S2−]=0.20(1.2×10−13)(5.0×10−8)​

[S2−]=3.0×10−20 M[S^{2-}] = 3.0\times 10^{-20}\,M[S2−]=3.0×10−20M

  1. Alternative direct method

Using both dissociation constants together:

Ka1Ka2=[H+]2[S2−][H2S]K_{a1}K_{a2} = \frac{[H^+]^2[S^{2-}]}{[H_2S]}Ka1​Ka2​=[H2​S][H+]2[S2−]​

Hence,

[S2−]=Ka1Ka2[H2S][H+]2[S^{2-}] = \frac{K_{a1}K_{a2}[H_2S]}{[H^+]^2}[S2−]=[H+]2Ka1​Ka2​[H2​S]​

[S2−]=(1.0×10−7)(1.2×10−13)(0.10)(0.20)2[S^{2-}] = \frac{(1.0\times 10^{-7})(1.2\times 10^{-13})(0.10)}{(0.20)^2}[S2−]=(0.20)2(1.0×10−7)(1.2×10−13)(0.10)​

[S2−]=3.0×10−20 M[S^{2-}] = 3.0\times 10^{-20}\,M[S2−]=3.0×10−20M

This matches the previous result.

  1. Evaluate options
  • A: 5×10−195 \times 10^{-19}5×10−19 ❌
  • B: 5×10−85 \times 10^{-8}5×10−8 ❌ (this is [HS−][HS^-][HS−], not [S2−][S^{2-}][S2−])
  • C: 3×10−203 \times 10^{-20}3×10−20 ✅
  • D: 6×10−216 \times 10^{-21}6×10−21 ❌

Therefore, the correct option is C.

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