Class 11 Chemistry - Chapter Equilibrium NCERT Solutions | The first ionization constant of H2S is
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.
(i) To calculate the concentration of HS- ion:
Case I (in the absence of HCl):
Let the concentration of HS- be x M.
H2S ↔ H+ + HS-
Ci 0.1 0 0
Cf 0.1-x x x
Then Ka1 = [H+ ] [ HS-] / H2S
9.1 × 10–8 = xx / 0.1-x
(9.1 × 10–8) (0.1-x) = x2
Taking 0.1 - x M ; 0.1M, we have
(9.1 × 10–8) (0.1) = x2
9.1 x 10-9 = x2
Case II (in the presence of HCl):
In the presence of 0.1 M of HCl, let [HS-] be y M.
Then, H2S ↔ H+ + HS-
Ci 0.1 0 0
Cf 0.1-y y y
Also, HCI ↔ H+ + CI-
0.1 0.1
Now, Ka1 = [H+ ] [ HS-] / H2S
Ka1 = [y] [0.1+y] / [0.1-y]
9.1 × 10–8 = y x 0.1 / 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- ↔ H+ + S2-
HS- = 9.54 x 10-5 M (From first ionization, case I)
Let S2- be X.
Also, [H+] = 9.54 x 10-5 M (From first ionization, case I)
Ka2 = (9.54 x 10-5) (X) / (9.54 x 10-5)
1.2x10-13 = X = S2-
Case II (in the presence of 0.1 M HCl):
Again, let the concentration of HS- be X' M.
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