A40.00 ml sample of 0.10 m weak acid with ka of 1.8×10−5 is titrated with a 0.10 m strong base. what is the ph after 20.00 ml of base has been added?

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

It would be (at) + cu (s)