$50 \mathrm{mL}$ of $0.02\mathrm{MNaOH}$ solution is mixed with $50 \mathrm{mL}$ of $0.06\mathrm{M}$ acetic acid solution, the $\mathrm{pH}$ of the resulting solution is_________
$(\mathrm{pKa}$ of acetic acid is $4.76, \log 5=0.70)$. 

When a strong base is added to a weak acid solution, it results in the formation of a salt. Here, acid is present in a limiting reagent and base is present in excess amounts. So, by using the pH formula:

$\begin{array}{ccc}pH& =& pKa+\log \frac{\left[salt\right]}{\left[acid\right]}\\ pH& =& 4.46+ \log \frac{\left[0.02\times 50\right]}{\left[(0.06-0.02)\times 50\right]} \\ pH& =& 4.76+\log (0.5)\\ pH& =& 4.46\end{array}$