Class 11 Chemistry - Chapter Equilibrium NCERT Solutions | The ionization constant of propanoic aci

Welcome to the NCERT Solutions for Class 11th Chemistry - Chapter Equilibrium. This page offers a step-by-step solution to the specific question from Excercise ".$ex_no." , Question 59: the ionization constant of propanoic acid is 1 32....
Question 59

The ionization constant of propanoic acid is 1.32 x 10-5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?

Answer

Let the degree of ionization of propanoic acid be α.

Then, representing propionic acid as HA, we have:

HA     +    H2O      ↔      H3O+     +  A-

(0.05-0.0α) ≈ 0.05       0.05α     0.05α

K =  [H3O+]  [A-]  / [HA]

=  (0.05α)(0.05α)  /  0.05

= 0.05 α2

Then, [H3O+]  = 0.05α = 0.05 x 1.63 x 10-2  =  Kb . 15 x 10-4 M

∴ pH  =  3.09

In the presence of 0.1M of HCl, let α' be the degree of ionization.

Then, [H3O+]  = 0.01

[A-]  =  0.05α'

[HA]  =  0.05

Ka  =  0.01 x 0.05α'  / 0.05

⇒ 1.32 x 10-5 =  0.1 α'

α' =  1.32 x 10-3

More Questions From Class 11 Chemistry - Chapter Equilibrium

Write a Comment: