JEE PYQ: Ionic Equilibrium Question 17
Question 17 - 2020 (07 Jan Shift 2)
Type: Numerical
3 g of acetic acid is 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, $\log 3 = 0.4771$]
Neglect any changes in volume.
Show Answer
Answer: 5.22
Solution
No. of moles of $\text{CH}_3\text{COOH} = \frac{3}{60} = 0.05$ mol $= 50$ mmol
No. of millimoles of HCl $= 250 \times 0.1 = 25$ mmol
500 mL solution: 50 mmol $\text{CH}_3\text{COOH}$ + 25 mmol HCl
20 mL solution contains: $\frac{50}{500} \times 20 = 2$ mmol $\text{CH}_3\text{COOH}$, $\frac{25}{500} \times 20 = 1$ mmol HCl
$\frac{1}{2}$ mL of 5 M NaOH $= \frac{5}{2} \times \frac{1}{1000} \times 1000 = 2.5$ mmol NaOH
HCl + NaOH $\rightarrow$ NaCl + $\text{H}_2\text{O}$: 1 mmol HCl consumed, remaining NaOH $= 2.5 - 1 = 1.5$ mmol
$\text{CH}_3\text{COOH}$ + NaOH $\rightarrow$ $\text{CH}_3\text{COONa}$ + $\text{H}_2\text{O}$: 1.5 mmol consumed
Remaining $\text{CH}_3\text{COOH} = 2 - 1.5 = 0.5$ mmol $\text{CH}_3\text{COONa} = 1.5$ mmol
pH $= \text{pKa} + \log \frac{[\text{Salt}]}{[\text{Acid}]} = 4.75 + \log \frac{1.5}{0.5} = 4.75 + \log 3 = 4.75 + 0.4771 = 5.22$
BETA
