Question

decinormal solution of ch3cooh ionized to an extent of 1.3%. ph of the solution

Answer 1

The acetic acid is dissociated to produce acetate ion along with proton and thus the dissociation of acetic acid can be shown as:

The dissociation of acetic acid is:

$CH_3COOH \rightarrow CH3COO- + H+$

$c – cα                          cα                  cα$

The given concentration of acid = 0.1 M

Degree of ionization, α = 1.3 %  

The concentration of proton can be shown as:

$c \alpha  = (0.1) * (1.3* 10^-^2)$

$[H^ +]=1.3 * 10^-^3$

pH is negative of logarithm of proton ion concentration in the reaction and can be determined as:

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

Thus pH of solution is 2.87