Chemistry Q55 – JEE-MAIN 2026

20 mL of a solution of acetic acid required 28.4 mL of 0.1 M NaOH for its neutralization. A solution (X) was prepared by mixing 20 mL of the above acetic acid and 14.2 mL of 0.1 M NaOH solution. What is the pH of the solution (X)? (

p K_{a}

value of acetic acid is 4.75).

Clue 1 of 3: Stoichiometry of Neutralization

First, determine the initial concentration of the acetic acid using the neutralization data.

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Step 1: Determine the initial concentration of acetic acid✦ Active

At neutralization, moles of acetic acid equal moles of NaOH. First, calculate the moles of NaOH used for neutralization:

Moles of NaOH = V_{NaOH} \times M_{NaOH} = 28.4 mL \times 0.1 M = 2.84 mmol

Therefore, moles of CH₃COOH = $2.84 mmol$ . The concentration of acetic acid is:

M_{acid} = \frac{Moles of CH₃COOH}{V_{acid}} = \frac{2.84 mmol}{20 mL} = 0.142 M

Step 2: Calculate moles of reactants and products in solution (X)○ Expand

Solution (X) is prepared by mixing 20 mL of 0.142 M acetic acid and 14.2 mL of 0.1 M NaOH. Calculate initial moles:

Initial moles of CH₃COOH = 20 mL \times 0.142 M = 2.84 mmol

Initial moles of NaOH = 14.2 mL \times 0.1 M = 1.42 mmol

The reaction is CH₃COOH + NaOH → CH₃COONa + H₂O. NaOH is the limiting reactant. After the reaction:

Moles of CH₃COOH remaining = 2.84 mmol - 1.42 mmol = 1.42 mmol

Moles of CH₃COONa formed = 1.42 mmol

The solution contains a weak acid (CH₃COOH) and its conjugate base (CH₃COONa) in equal molar amounts, forming a buffer solution.

Step 3: Calculate the pH of the buffer solution○ Expand

Use the Henderson-Hasselbalch equation for the buffer solution:

p H = p K_{a} + \log (\frac{[CH₃COONa]}{[CH₃COOH]})

Since the moles of CH₃COOH and CH₃COONa are equal ( $1.42 mmol$ each) and they are in the same total volume, their concentrations are equal. Thus, the ratio $\frac{[CH₃COONa]}{[CH₃COOH]}$ is 1.

p H = p K_{a} + \log (1)

p H = p K_{a} + 0

Given $p K_{a} = 4.75$ :

p H = 4.75

💡 Teacher's Secret Hint

Remember that when the concentrations of the weak acid and its conjugate base are equal in a buffer, the pH is equal to the pKa.

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Chemistry Question 55 – JEE-MAIN 2026

Clue 1 of 3: Stoichiometry of Neutralization

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