Question

3 g of acetic acid Ls added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution $\frac{1}{2}$ mL of 5 M NaOH is added. The pH of the solution is ________ .
[Given : pKa of acetic acid = 4.75, molar mass of acetic acid = 60 g / mol, log3 = 0.4771]
Neglect any changes in volume.
Given 6.5

Answer 1

Answer: 5.22

Molar mass of acetic acid $=2\times 12+4\times 1+16\times 2=60g$
No of moles of acetic acid $=\frac{3}{60}=\frac{1}{20}=0.05$ mole
Millimoles of acetic acid in $20ml=\frac{0.05\times 20\times 1000}{500}=2$ millimoles
Millimoles of $HCl$ in $20ml=1$ millimoles
Millimoles of $NaOH=2.5$ millimoles
Millimoles of $NaOH$ remaining $=2.5-1=1.5$ millimoles
Millimoles of $NaOH=2.5$ millimoles
Millimoles of $NaOH$ remaining $=2.5-1=1.5$ millimoles
$$\begin{gathered} pH=pka+\log \frac{[Salt]}{[Acid]} \\ pH=4.74+\log \frac{[3/2]}{[1/2]} \\ pH=4.74+\log (3)=4.74+0.48=5.22 \end{gathered}$$