Question

Decinormal solution of acetic acid ionised to an extent of 1.3%. ph of this solution will be

Answer 1

The dissociation of acetic acid is:

CH3COOH   CH3COO- + H+

c – cα  cα cα

Concentration = 0.1 M

Degree of ionization, α = 1.3 %  

c α = (0.1) * (1.3* 10^-2)  

[H +]=1.3 * 10^-3

pH can be determined as:

-log (H^+) = pH = - log (1.3 * 10^-3) = 2.87

Thus pH of solution is 2.87