what is the ph of the resulting mixture when 10.00ml of 0.400 mol/l KOH is added to 15.00 ml of 0.300 mol/l acetic acid?
acetic acid is HAc
KOH + HAc ==> KAc + H2O
millimols KOH = 10 x 0.4 = 4
mmols HAc = 15 x 0.3 = 4.5
mmols KAc formed = 4
mmols HAc left unreacted = 0.5
Substitue into the Henderson-Hasselbalch equation and solve for pH. You will need to look up the pKa for acetic acid. I think it's about 4.76.