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.
Solution and Explanation
(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, Ka1 = \(\frac{[H^+][HS^-]}{[H_2S]}\) = 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, Ka1 = \(\frac{[H^+][HS^-]}{[H_2S]}\)
Ka1 = \(\frac{[y](0.1+y)}{(0.1-y)}\)
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, Ka2 =\(\frac{[H^+][S_2^-]}{[HS^-]}\)
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 ⇒ Ka1 = 1.74 × 10-5
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