If a 6.03 x 10-6.0 M solution of a weak acid is 20 % ionized, what is the pKa for the acid?
1 Answer
Explanation:
If the formula of a weak acid be reprsented as HA.then its dissociation in aqueous medium can be written as follows:
#HA" "+H_2O""rightleftharpoons""H_3O^+""+" "A^-#
Where
If cM be the initial concentration of the weak acid then the concentrations different species at equilibrium will be as follows
#[HA]=c(1-alpha)M#
#[H_3O^+]=calphaM#
#[A^-]=calphaM#
The concentration of
Now the acid dissociation constant
#K_a=([H_3O^+][A^-])/[HA]^2#
#=((calpha)*(calpha))/(c(1-alpha))#
#=(calpha^2)/(1-alpha)#
Given
#K_a=(6.03xx10^-6*(0.2)^2)/(1-0.2)#
#=(6.03xx10^-6xx0.04)/0.8#
#3.015xx10^-7#
The
#=7-log3.015=6.52#