Question

Determine the [H₃O⁺] and the pH of a 0.250 solution of Carbonic acid; Ka1 = 4.3 x 10^-7; Ka2 = 5.6 x 10^-11

Answer 1

H2CO3 + H2OHCO3- +H3O+ (ka1=4.3*10^-7)

HCO3- +H2OCO32- + H3O+ (ka2=5.6*10^-11)

1) ICE table for first reaction

                       [H2CO3]               [HCO3-]                [H3O+]

initial concentration        0.250                        0                            0

change                              -X                         +X                            +X

equilibrium conc              0.250-X                    X                                  X

ka1=[HCO3-] [H3O+]/ [H2CO3]

4.3*10^-7=X^2/(0.250-X)    (ignoring x wrt 0.250 as it is very small ,less weak acid dissociation takes place)

4.3*10^-7=X^2/(0.250)

X^2=1.075*10^-7=10.75*10^-8

X=sqrt(10.75*10^-8)=3.28*10^-4=[H3O+]=[HCO3]

again, further dissociation of HCO3 takes place,

2)HCO3- +H2OCO32- + H3O+ (ka2=5.6*10^-11)

ICE table

                                [ HCO3-]                      [CO32-]             [H3O+]

initial                          0.250                            0                    3.28*10^-4

change    -y    + y +y

eqm                           0.250-y    +y       3.28*10^-4+y

ka2= [CO32-][H3O+]/[H2CO3]=( 3.28*10^-4+y)(y)/(0.250-y)            

Again y is very small as ka2 is very small ,very less dissociation takes place

ka2=( 3.28*10^-4)(y)/(0.250)

5.6*10^-11=( 3.28*10^-4)(y)/(0.250)

0.35*10^-11=( 3.28*10^-4)(y)

y=0.11*10^-7 =[H3O+]

[H3O+]=0.11*10^-7=1.1 *10^-8

pH=-log[H3O+]=-log (1.1 *10^-8)=-0.041+8=7.96

pH=7.96