Question 45 of 69advanced🔧 ApplyNumerical5 marks

The first ionization constant of H2_2S is 9.1×10−89.1 \times 10^{-8}. Calculate the concentration of HS−^{-} ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H2_2S is 1.2×10−131.2 \times 10^{-13}, calculate the concentration of S2−^{2-} under both conditions.

Correct Answer

(i) [HS−]=9.54×10−5[HS^-] = 9.54 \times 10^{-5} M

(ii) [HS−]=9.1×10−8[HS^-] = 9.1 \times 10^{-8} M

(iii) [S2−]=1.2×10−13[S^{2-}] = 1.2 \times 10^{-13} M (without HCl) and [S2−]=1.092×10−19[S^{2-}] = 1.092 \times 10^{-19} M (with HCl)

Exercise: EXERCISES | Q: 6.45 | (Page 45)
For More Understanding

Explanation

To find the concentration of HS−HS^- in a 0.10.1 M H2SH_2S solution, we use the first ionization constant Ka1K_{a1}. When HClHCl is added, the common ion H3O+H_3O^+ suppresses the ionization of H2SH_2S due to the common ion effect. For the S2−S^{2-} concentration, we use the second dissociation constant Ka2K_{a2} and the overall equilibrium constant K=Ka1×Ka2K = K_{a1} \times K_{a2}.

Solution Steps

  1. Part 1: [HS−][HS^-] in 0.1 M H2SH_2S solution

    H2S+H2O⇌H3O++HS−H_2S + H_2O \rightleftharpoons H_3O^+ + HS^- Ka1=[H3O+][HS−][H2S]=9.1×10−8K_{a1} = \frac{[H_3O^+][HS^-]}{[H_2S]} = 9.1 \times 10^{-8}

    Let [H3O+]=[HS−]=x[H_3O^+] = [HS^-] = x and [H2S]≈0.1[H_2S] \approx 0.1 M.

    x20.1=9.1×10−8\frac{x^2}{0.1} = 9.1 \times 10^{-8} x2=9.1×10−9x^2 = 9.1 \times 10^{-9}

    x=9.1×10−9=9.54×10−5x = \sqrt{9.1 \times 10^{-9}} = 9.54 \times 10^{-5} M So, [HS−]=9.54×10−5[HS^-] = 9.54 \times 10^{-5} M.

  2. Part 2: [HS−][HS^-] in 0.1 M H2SH_2S with 0.1 M HCl Due to the common ion effect, HClHCl completely dissociates to provide [H3O+]=0.1[H_3O^+] = 0.1 M.

    Ka1=(0.1)[HS−]0.1=[HS−]K_{a1} = \frac{(0.1)[HS^-]}{0.1} = [HS^-]

    [HS−]=9.1×10−8[HS^-] = 9.1 \times 10^{-8} M. The concentration is drastically reduced.

  3. Part 3: [S2−][S^{2-}] without HCl For the second dissociation: HS−⇌H3O++S2−HS^- \rightleftharpoons H_3O^+ + S^{2-}

    Ka2=[H3O+][S2−][HS−]=1.2×10−13K_{a2} = \frac{[H_3O^+][S^{2-}]}{[HS^-]} = 1.2 \times 10^{-13}

    Since [H3O+]≈[HS−][H_3O^+] \approx [HS^-] from the first dissociation, [S2−]=Ka2=1.2×10−13[S^{2-}] = K_{a2} = 1.2 \times 10^{-13} M.

  4. Part 4: [S2−][S^{2-}] with 0.1 M HCl Overall equilibrium: H2S⇌2H3O++S2−H_2S \rightleftharpoons 2H_3O^+ + S^{2-}

    K=Ka1×Ka2=(9.1×10−8)×(1.2×10−13)=1.092×10−20K = K_{a1} \times K_{a2} = (9.1 \times 10^{-8}) \times (1.2 \times 10^{-13}) = 1.092 \times 10^{-20} K=[H3O+]2[S2−][H2S]K = \frac{[H_3O^+]^2[S^{2-}]}{[H_2S]} 1.092×10−20=(0.1)2[S2−]0.11.092 \times 10^{-20} = \frac{(0.1)^2[S^{2-}]}{0.1} 1.092×10−20=0.1×[S2−]1.092 \times 10^{-20} = 0.1 \times [S^{2-}]

    [S2−]=1.092×10−19[S^{2-}] = 1.092 \times 10^{-19} M.

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