20 mL of 0.1 M acetic acid is mixed with 50 mL of potassium...

For an acidic buffer, $\mathrm{pH}=\mathrm{p} K_\alpha+\log \frac{[\text{ Salt }]}{[\text{ Acid }]}\ldots(i)$ Let the concentration of potassium acetate solution $=x \mathrm{M}$ 20 mL of 0.1 M acetic acid $=20 \times 0.1 \mathrm{millimol}=2$ millimol 50 mL of $x \mathrm{M}$ potassium acetate $=x \times 50 \mathrm{millimol}$$\begin{aligned} & =50 x \text{ milimol } \\ \mathrm{pH} & =4.8 (Given)\end{aligned}$$$Substitute all value in eq. (i),$$\begin{aligned} 4.8 & =-\left(\log 1.8+10^{-5}\right)+\log \frac{50 x}{2} \\ \log 25 x & =0.06 \\ x & =0.04 \mathrm{M} \end{aligned}$$