Class 11 Chemistry - Chapter Equilibrium NCERT Solutions | The first ionization constant of H2S is

Welcome to the NCERT Solutions for Class 11th Chemistry - Chapter Equilibrium. This page offers a step-by-step solution to the specific question from Exercise 1, Question 45: the first ionization constant of h2s is 9 1 times....
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 S2– 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.

             H2↔  H+  +  HS-

Ci          0.1         0         0

C      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,       H2↔  H+  +  HS-

Ci          0.1          0         0

C       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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