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
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
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