If 50 ml of 0.2 M KOH is added to 40 ml of 0.5 M HCOOH, the pH of the resulting solution is $\left({\mathrm{K}}_{\mathrm{a}} = 1.8 \times {10}^{-4}\right)$

$$\begin{gathered} \mathrm{pH}=-\log {\mathrm{K}}_{\mathrm{a}} + \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]} \\ \left[\mathrm{Salt}\right] = \frac{0.2 \times 50}{1000} = 0.01 ; \left[\mathrm{Acid}\right] = \frac{0.5 \times 40}{1000} = 0.02 \\ \mathrm{pH}=-\log \left(1.8 \times {10}^{-4}\right) + \log \frac{0.01}{0.02} \\ \mathrm{pH}=4-\log \left(1.8\right) + \log 0.5 \\ \mathrm{pH} = 4-\log \left(1.8\right) - 0.301 \\ \mathrm{pH} = 3.4 \end{gathered}$$