3g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500mL. To 20 mL of this solution 1/2 mL of 5M NaOH is added. The pH of this solution is [Given pKa of acetic acid = 7.75, molar mass of acetic acid = 60 g/mol, log 3 = 0.4771] Neglect any changes in volume.

Given data:It is given to identify the pH of resulting solution.

Explanation:

$$\begin{gathered} Acetic acid molecular mass =2\times 12+4\times 1+16\times 2=60 \mathrm{g} \\ \mathrm{Number} \mathrm{of} \mathrm{mole} \mathrm{in} \mathrm{acetic} \mathrm{acid} =\frac{3}{60}=\frac{1}{20}=0.05\text{ mole } \\ \mathrm{Acetic} \mathrm{acid} \mathrm{millimoles} \mathrm{in} 20 \mathrm{ml}=\frac{0.05\times 20\times 1000}{500}=2\text{ millimoles } \\ \mathrm{Millimoles} \mathrm{of} \mathrm{HCl} \mathrm{in} 20\mathrm{ml}=1 \mathrm{millimoles} \\ \mathrm{Millimoles} \mathrm{of} \mathrm{NaOH}=2.5 \mathrm{millimoles} \\ \mathrm{NaOH} \mathrm{millimoles} \mathrm{remaining} =2.51=1.5 \mathrm{millimoles} \\ {\mathrm{CH}}_3\mathrm{COOH}+\mathrm{NaOH}\text{ (Remaining) }\to {\mathrm{CH}}_3\mathrm{COONa}+\text{ water } \\ \mathrm{Initital} 2 3/2 0 0 \\ \mathrm{Remained} 0.5 0 3/2 \\ \begin{array}{l}\mathrm{pH}=\mathrm{pka}+\log \frac{\left[\mathrm{Salt}\right]}{\left[\text{ Acid }\right]}\\ \mathrm{pH}=7.75+\log \frac{[3/2]}{[1/2]}\\ \mathrm{pH}=7.75+\log \left(3\right)=7.75+0.48=8.23\end{array} \end{gathered}$$

Hence, correct option is: (D) 8.22 to 8.24