Class 11 Chemistry - Chapter Equilibrium NCERT Solutions | The ionization constant of acetic acid i

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 46: the ionization constant of acetic acid is 1 74 x 1....
Question 46

The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Answer

1) CH3COOH        CH3COO-  + H+     Ka = 1.74x10-5

2) H2O + H2O       H3O+   +   OH-        Kw  = 1.0x10-14

Since Ka > Kw

         CH3COOH  + H2O       CH3COO-  +    H3O+

Ci  =  0.05                                     0                   0

       0.05- 0.05α                     0.05α             0.05α  

ka =  0.05α x   0.05α  /   0.05- 0.05α  

= 0.05α x   0.05α  /   0.05 (1-α)

= 0.05α2  /  (1-α)

⇒ 1.74x10-5  =  0.05α2  /  (1-α)

⇒ 1.74x10-5   -  1.74x10-5 α  =  0.05α2 

⇒ 0.05α2  +    1.74x10-5 α  -  1.74x10-5 =0

D =  b2 - 4ac

   = (1.74x10-5 )2 - 4 (0.05) (1.74x10-5)

   = 3.02x10-25  + 0.348x10-5

Method 2:

Degree of Dissociation,

CH3COOH      CH3COO- + H+

Thus, concentration of CH3COO- = c.α 

= 0.05x1.86x10-2

= 0.93x10-2

=.00093M

Since [oAc-] = [H+]

[H+] = .00093 = 0.093x10-2

pH = -log[H+]

     =  -log (0.93x10-2)

∴ pH = 3.03

Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.

 

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