Calculate the `H^+` ion concentration in 0.1 M `CH_3COOH` if the degree of dissociation of the acid is 0.0132

Explanation

Acetic acid dissociates as : `CH_(3)CO OH + H_(2)O hArr CH_(3)CO O^(-) + H_(2)O^(+)`
If C moles/litre is the initial concentration and `alpha` is the degree of dissociation, we can write
`{:(,CH_(3)CO OH,+H_(2)O,hArr,CH_(3)CO O^(-)+,H_(3)O^(+)),("Initial conc.",C,,,,),("Conc. at eqm.",C-C alpha,,,C alpha,C alpha),(,,,,,):}`
Thus, `[H_(3)O^(+)]=C xx alpha = 0.1 xx 0.0132 = 1.32xx10^(-3) M`
`:. pH = - log [H_(3)O^(+)]=- log (1.32xx10^(-3)) = - log 1.32 - log 10^(-3) = - 0.1206 + 3 = 2.8794 `.