The answer is 4.74.
Solution:
The product of the volume and concentration of each solution gives the number of moles of the acid HA and the base OH- present before the neutralization:
0.0400L (0.10mol/L HA) = 0.00400 moles HA
0.0200L (0.10mol/L OH-) = 0.00200 moles OH-
We can see that the 0.00200 moles of the strong base OH- consumes 0.00200 moles of the weak acid HA:
HA + OH- → A- + H2O
initial 0.00400 0.00200 0
change -0.00200 -0.00200 +0.00200
equilibrium 0.00200 0 0.00200
The concentration of HA is equal to the concentration of A- after the reaction:
[HA] = [A-] = 0.00200mol / 0.0400L+0.0200L = 0.03333 M
The equilibrium-constant expression for HA is
Ka = [H+][A-] / [HA]
Ka = [H+] = 1.8x10^-5
We can now solve for pH:
pH = -log [H+] = -log(1.8x10^-5) = 4.74
