Need help w/Chemistry question about titrations and pH!

Question

A volume of 25.0 mL of 0.100 M CH3CO2H(aq) is titrated with 0.100 M NaOH(aq). What is the pH after the addition of 12.5 mL of NaOH?

Ka for CH3CO2H = 1.8e10-5

Answer 1

CH3CO2H + NaOH ==> CH3CO2Na + H2O

During the titration, a BUFFER is formed consisting of CH3CO2H and CH3CO2^-

We can use the Henderson Hasselbalch equation to find the pH, but first we will need to find

[CH3CO2H] and [CH3CO2^-]

ICE Table:

CH3CO2H + NaOH ==> CH3CO2Na + H2O

0.0025 mol.........0.00125mol.......0 mol.............Initial

-0.00125..........-.0.00125.........+0.00125........Change

0.00125................0...................0.00125..........Equilibrium

HH equation:

pH = pKa + log [salt]/[acid]

pKa = -log Ka = -log 1.8x10-5 = 4.74

pH = 4.74 + log [0.00125]/[0.00125]

pH = 4.74