Calculate the $P^H$ of the following mixtures given  $P{k}_a=P{k}_b=4.74$

$\left(1\right) 50ml0.1MNaOH+50ml0.1MC{H}_{3}COOH$

$\left(2\right) 50ml0.1MNaOH+50ml0.05MC{H}_{3}COOH$

$\left(3\right) 50ml0.05MNaOH+50ml0.1MC{H}_{3}COOH$

$\left(1\right)$ $\begin{array}{l}NaOH+C{H}_{3}COOH\to C{H}_{3}COONa+H_{2}O\\ 50ml,0.1M50ml,0.1M0\\ 5mEq5mEq0\\ 005mEq\\ \left[C{H}_{3}COONa\right]=\left[Salt\right]=\frac{5}{100}=0.05M\\ forsalthydrolysis{P}^H=\frac{1}{2}(P_{kw}+P_{ka}+\log C)\\ forunivalentsalt=\frac{1}{2}(14+4.74+\log \left(0.05\right))\\ =\frac{1}{2}(14+4.74-2+0.7)=8.72\end{array}$

$\left(2\right) \left[NaOH\right]=\left[O{H}^{-}\right]=\frac{2.5}{100}=2.5\times {10}^{-2}M$

For base, $\begin{array}{l}{P}^{OH}=-\log \left[O{H}^{-}\right]=-\log 2.5\times {10}^{-2}\\ =-\log 2.5+2\\ =-4+2=1.60\end{array}$

$\left(3\right)$ For buffer solution $\begin{array}{l}{P}^H=P_{ka}+\log \frac{\left[S\right]}{\left[A\right]}\\ =4.74+\log \frac{\left[2.5\right]}{\left[2.5\right]}\\ =4.74\end{array}$