Consider the titration of 100.0 ml of 0.200 M acetic

We have a titration of 100.0 mL (0.100 L) of 0.200 M acetic acid ${\displaystyle (K_a=1.8\cdot {10}^{-5})}$ with 0.100 M KOH.
(b)
Let us calculate the pH of a solution, after 50.0 mL (0.050 L) of KOH has been added.
1) The total volume of a solution is now ${\displaystyle 100.0mL50.0mL=150.0mL\left(0.150L\right)}$
First, let us calculate the number of moles of acetic acid
$n_{C{H}_{3}COOH}=0.200M\cdot 0.100L=0.02mol$
And the number of moles of KOH
${\displaystyle n_{KOH}=0.100M\cdot 0.050L=0.005mol}$
2) KOH is strong base, and it will dissociate completely into ${\displaystyle K^{+}}$ and ${\displaystyle O{H}^{-}}$. Therefore, we are adding 0.005 mol of ${\displaystyle O{H}_{-}}$ ions. When we add strong base into weak acid solution, the reaction that will occur is
${\displaystyle C{H}_{3}COOH+O{H}^{-}\Rightarrow C{H}_{3}CO{O}^{-}+H_{2}0}$
Therefore, when 0.005 mol of ${\displaystyle O{H}^{-}}$ is added, 0.005 mol of acetic acid is consumed ${\displaystyle \left(C{H}_{3}COOH\right)}$, and 0.005 mol of it's conjugate base is produced ${\displaystyle \left(C{H}_{3}CO{O}^{-}\right)}$. So, after KOH has beed added, we end up with a buffer solution that contains ${\displaystyle (0.02mol-0.005mol)}$ 0.015 mol of acetic acid, and 0.005 mol of it's conjugate base.
$\overline{\begin{array}{ccccc}& O{H}^{-}& C{H}_{3}COO{H}^{-}& \Rightarrow C{H}_{3}CO{O}^{-}+& H_{2}O\\ Initial(mol)& 0.005& 0.020& 0& /\\ Change(mol)& -0.005& -0.005& +0.005& /\\ Equilibrium(mol)& 0& 0.015& 0.005& /\end{array}}$
Step 2
3) Now, let us calculate the concentration of ${\displaystyle C{H}_{3}COOH}$ and ${\displaystyle C{H}_{3}CO{O}^{-}}$
$$
$[C{H}_{3}COOH]=\frac{Numberofmolesofaceticacid}{Totalvolumeofasolution}=\frac{0.015mol}{0.150L}=0.100M$
$[C{H}_{3}CO{O}^{-}]=\frac{Numberofmolesofconjugatebase}{Totalvolumeofasolution}=\frac{0.005mol}{0.150L}=0.033M$
4) Let us calculate the pH using Henderson-Hasselbalch equation