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
Jul 28, 2016

Answer:

#pK_a=6.52#

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^-#

#color(red)(I)" "1" "mol" " 0" "mol" " 0" " "mol#

#color(red)(C)" "-alpha" "mol" " alpha" "mol" " alpha" " "mol#

#color(red)(E)" "1-alpha" "mol" " alpha" "mol" " alpha" " "mol#

Where #alpha# is the degree of dissociation of #HA#

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 #H_2O# is irrelevant as it acts as sovent.

Now the acid dissociation constant

#K_a=([H_3O^+][A^-])/[HA]^2#

#=((calpha)*(calpha))/(c(1-alpha))#

#=(calpha^2)/(1-alpha)#

Given #c=6.03xx10^-6M and alpha=20%=0.2#

#K_a=(6.03xx10^-6*(0.2)^2)/(1-0.2)#

#=(6.03xx10^-6xx0.04)/0.8#

#3.015xx10^-7#

The #pK_a# of the weak acid

#pK_a=-logK_a=-log(3.015xx10^-7)#

#=7-log3.015=6.52#