The dissociation constant of acetic acid is `8 xx 10^(-5)` ta `25^(@)C`. Find the `pH` of
i. `M//10` ii. `M//100` solution of acetic acid.

Answer

B, C

Explanation

i. For weak acid: `(C = 0.1M)`
`K_(a) = (C alpha^(2))/(1-alpha) = (0.1alpha^(2))/(1-alpha)`
`:. alpha + 1.8 xx 10^(-4)alpha - 1.8 xx 10^(-4) = 0`
`:. alpha = 0.0129 = 1.29 xx 10^(-2)`
Note: `alpha = 0.0129` which is less than `0.05`, so the term `(1-alpha) ~~1` can be taken. Moreover if `K_(a) lt 10^(-5)` and 'C' is fairly high `(0.1M or 0.01M)` then `(1-alpha)` can be taken as `1`.
`:. Calpha = [H^(o+)] = 0.0129 xx 0.1 = 1.29 xx 10^(-3)`
`pH = log [H^(o+)] =- log (1.29 xx 10^(-3))= 2.89`
Alternatively: `rArr (K_(a) = 1.8 xx 10^(-5), pK_(a) = 4.7447 ~~ 4.74)`
`pH_(W_(A)) = (1)/(2) (pK_(a) - log C)`
`= (1)/(2) (4.74 - log 10^(-1)) = 2.87`.
ii. `pH_(W_(A)) = (1)/(2) (pK_(a) - logC)`
`= (1)/(2) (4.74 - log 10^(-1)) = 3.37`.
Alternatively: `K_(a) = (C alpha^(2))/(1-alpha) = (0.01alpha^(2))/(1-alpha)`
`:. alpha + 1.8 xx 10^(-3) alpha - 1.8 xx 10^(-3) = 0`
`:. alpha = 1.29 xx 10^(-2)`
`[H^(o+)] = C alpha = 1.29 xx 10^(-2) xx 0.01 = 1.29 xx 10^(-4)`.
`pH =- log 1(1.29 xx 10^(-4)) = 3.89`.