Question

Calculate the ionization constant of acetic acid:

pH of 0.01 M acetic acid: 3.50

pH of 1.00 M acetic acid: 2.34

Answer 1

Solution-

Given:

0.01 M acetic acid pH = 3.5

1.00 M acetic acid pH = 2.34

First we will discuss 0.01 M acetic acid

pH = 3.5

We know that

pH = -log[H3O+]

-log[H3O+] = 3.5

Taking Antilog of both side

[H3O+] = x = 0.000316

Dissociation of CH3COOH is as follows

CH3COOH(aq) + H2O(l→CH3COO^-(aq) + H3O⁺(aq)

We use ICE chart

For 0.01M Acetic acid :

     CH3COOH (aq) + H2O ( l ) --> CH3COO- (aq) + H3O + (aq)

I           0.01 M                                             0                        0

C            -x                                                 + x                     +x

E         ( 0.01-x)                                              x                   x

Ionization constant (Ka) expression is as follows

Ka = [ CH3COO- (aq)][ H3O + (aq)] / [ CH3COOH (aq) ]

Ka = x * x/(0.01-x)

     = (0.000316) *(0.000316)/0.009684

Ka = 1.0 x 10^^-5

Answer ionization constant of 0.01 M acetic acid (ka) = 1.0 x 10^^-5

Let’s calculate Ionization constant of 1.00 M acetic acid pH =3.4

pH = 3.4

We know that

pH = -log[H3O+]

-log[H3O+] = 3.4

Taking Antilog of both side

[H3O+] = x = 0.000398

Ice chart for 1.0 M Acetic acid :

CH3COOH (aq) + H2O ( l ) --> CH3COO- (aq) + H3O + (aq)

I           1.0 M                                             0                        0

C            -x                                                 + x                     +x

E         (1-x)                                              x             x

Ionization constant (Ka) expression is as follows

Ka = [ CH3COO- (aq)][ H3O + (aq)] / [ CH3COOH (aq) ]

Ka = x * x/(0.01-x) = (0.000398 * 0.000398)/(1-0.000398)

Ka = (0.000398 * 0.000398)/0.9996 = 1.58 x 10 ^^-7

Answer ionization constant of 1.0 M acetic acid (ka) = 1.58 x 10 ^^-7