pH of a solution of 10 ml. 1 N sodium acetate and 50 ml 2 N acetic acid $\left({\mathrm{K}}_{\mathrm{a}} = 1.8 \times {10}^{-5}\right)$ is approximately

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}} + \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]} ; \left[\mathrm{Salt}\right] =\frac{10}{1000}\times 1={10}^{-2} ;\left[\mathrm{Acid}\right] =\frac{50}{1000}\times 2 ={10}^{-1} \\ \mathrm{pH} =-\log \left(1.8\times {10}^{-5}\right) + \log \frac{\left[{10}^{-2}\right]}{\left[{10}^{-1}\right]} \\ =-\log 1.8 + 5+\log {10}^{-1} \\ =-0.2553+5-1=3.7447 \mathrm{or} = 4 \end{gathered}$$