Question

The first ionization constant of H₂S is 9.1 × 10^–8. Calculate the concentration of HS^– ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H₂S is 1.2 × 10^–13, calculate the concentration of S₂^– under both conditions.

Answer 1

(i) To calculate the concentration of HS– ion: Case I (in the absence of HCl): Let the concentration of HS– be x M.
       H₂S \(\leftrightarrow\) H⁺ + HS^-
Cƒ    0.1       0        0
Cƒ   0.1 - x    x        x
Then, K_a1 = \(\frac{[H^+][HS^-]}{[H_2S]}\) = 9.1 × 10^-8 = \(\frac{(x)(x)}{0.1-x}\) = (9.1 × 10 ^-8)(0.1 -x) = x²
Taking 0.1 - x M ; 0.1 M, we have (9.1 × 10 ^-8)(0.1) = x²
x = \(\sqrt{9.1\times10^{-9}}\) = 9.54 × 10^-5 M
⇒ [HS^-] = 9.54 × 10^-5 M
Case II (in the presence of HCl) : In the presence of 0.1 M of HCl, let [HS^-] be y M.
Then, H₂S\(\leftrightarrow\)HS^- + H⁺
Cƒ      0.1      0        0
Cƒ    0.1 - y   y        y
Also, HCl \(\leftrightarrow\) H⁺ + Cl^- 
                     0.1    0.1
Now, K_a1 = \(\frac{[H^+][HS^-]}{[H_2S]}\)
K_a1 = \(\frac{[y](0.1+y)}{(0.1-y)}\)
9.1 × 10 ^-8 = \(\frac{y\times0.1}{0.1}\) (∵ 0.1 - y ; 0.1M) (and 0.1 + y; 0.1M)
9.1 × 10^-8 = y ⇒ [HS^-] = 9.1 × 10^-8
(ii) To calculate the concentration of [S₂^-]: Case I (in the absence of 0.1 M HCl): HS^-\(\leftrightarrow\)H⁺ + S₂^-
[HS^-] = 9.54 × 10^-5 M (from first ionization, case : I)
Let [S₂^-]be X.
Also, [H⁺] = 9.54 × 10^-5M (from first ionization, case II)
Ka₂ = \(\frac{[H^+][S_2^-]}{[HS^-]}\)
Ka₂ = \(\frac{(9.54 × 10 ^{-5})(X)}{9.54 × 10 ^{-5}}\)
1.2 × 10^-13 = X = [S₂^-]
Case II (in the presence of 0.1 M HCl): Again, let the concentration of HS^- be X' M.
[HS^-] = 9.1 × 10^-8 M (from first ionization, case II)
[H⁺] = 0.1M (from HCl, case II)
[S₂^-] = X'
Then, K_a2 =\(\frac{[H^+][S_2^-]}{[HS^-]}\)
1.20 × 10 ^-13 = \(\frac{(0.1)(X')}{9.1 × 10^{ - 8}}\) = 10.92 × 10^-21 = 0.1 X' = \(\frac{10.92 × 10^{-21}}{0.1}\) = X'
X' = \(\frac{10.92 × 10^{-20}}{0.1}\) = 1.092 × 10^-19 M ⇒ K_a1 = 1.74 × 10^-5