Question 45

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 S^2– under both conditions.

Answer

(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^-

C_i 0.1 0 0

C_f 0.1-x x x

Then K_a1 = [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^-

C_i 0.1 0 0

C_f 0.1-y y y

Also, HCI ↔ H⁺ + CI^-

0.1 0.1

Now, K_a1 = [H⁺] [ HS^-] / H₂S

K_a1 = [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 [S^2-]:

Case I (in the absence of 0.1 M HCl):

HS^- ↔ H⁺ + S^2-

HS^- = 9.54 x 10^-5 M (From first ionization, case I)

Let S^2-be X.

Also, [H⁺] = 9.54 x 10^-5 M (From first ionization, case I)

K_a2 = (9.54 x 10^-5) (X) / (9.54 x 10^-5)

1.2x10-13 = X = S^2-

Case II (in the presence of 0.1 M HCl):

Again, let the concentration of HS^- be X' M.

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