(a) pH of aspirin solution
Let's write the chemical equation as
mmmmmmmmlHAm+mH2O⇌H3O+m+mlA-
I/mol⋅dm-3:mm0.05mmmmmmmml0mmmmmll0
C/mol⋅dm-3:mml-xmmmmmmmm+xmlmmml+x
E/mol⋅dm-3:m0.05 -lxmmmmmmmlxmmxmmmx
Ka=[H3O+][A-][HA]=x20.05 -lx=3.27×10-4
Check for negligibility
0.053.27×10-4=153<400
∴ x is not less than 5 % of the initial concentration of [HA].
We cannot ignore it in comparison with 0.05, so we must solve a quadratic.
Then
x20.05−x=3.27×10-4
x2=3.27×10-4(0.05−x)=1.635×10-5−3.27×10-4x
x2+3.27×10-4x−1.635×10-5=0
x=1.68×10-5
[H3O+]=xlmol/L=1.68×10-5lmol/L
pH=-log[H3O+]=-log(1.68×10-5)=4.77
(b) [H3O+] at pH 4
[H3O+]=10-pHlmol/L=1.00×10-4lmol/L
(c) Concentration of A- in the buffer
We can now use the Henderson-Hasselbalch equation to calculate the [A-].
pH=pKa+log([A-][HA])
4.00=−log(3.27×10-4)+log([A-]0.05)=3.49+log([A-]0.05)
log([A-]0.05)=4.00 - 3.49=0.51
[A-]0.05=100.51=3.24
[A-]=0.05×3.24=0.16
The concentration of A- in the buffer is 0.16 mol/L.
