Question

Question 7.45 The first ionization constant of H2­­S is 9.1 × 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 H2S is 1.2 × 10^(–13),
calculate the concentration of S(2–) under both conditions.

Class XI Equilibrium Page 228

Answer 1

Hi

Please see the attached file !


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Answer 2

(i) H2S +H2O <=> H3O+ + HS-
so, Ka = [H3O+][HS-]/[H2S]
use formula ,
[HS- ] = [H3O+] = √{Ka.C}
where Ka is the Ionisation constant and C is the concentration of solution.
so, [HS-] = √{9.1 × 10^-8 × 0.1} = 9.54 × 10^-5 M--------(1)

(ii) in the presence of 0.1M HCl,
then, conc.of Hydronium ion should be 0.1 M
e.g., [H3O+] = 0.1M
given, [H2S] = 0.1 M and Ka = 9.1 × 10^-8
so, Ka = [H3O+][HS-]/[H2S]
9.1 × 10^-8 = 0.1[HS-]/0.1
[HS-] = 9.1 × 10^-8 M

hence, you can see that concentration of HS- decreases in the presence of 0.1 M of HCl due to common ion - effect.

(iii) for second dissociation constant,
HS- + H2O <=> H3O+ + S²-
from equation (1)
[HS-] = C = 9.54 × 10^-5 M
Ka' = [H3O+][S²-]/[HS-]
according to Ostwald's formula,
[H3O+] = [ S-] = √{Ka'.C}
= √{1.2 × 10^-13 × 9.54 × 10^-5 }
= 3.38 × 10^-9 M

(iv) in the presence of 0.1M HCl,
Ka' =[ H3O+][S²-]/[HS-]
1.2 × 10^-13 = 0.1 × [S²-]/9.54 × 10^-5
[S²-] = 1.1448 × 10^-16 M