JEE Main 2023 — Ionic Equilibrium Question with Solution

First, we need to find the moles of NaOH and acetic acid (CH₃COOH) in the solution:

Moles of NaOH = Volume × Molarity = 20 mL × 0.1 M = 2 mmol

Moles of acetic acid = Volume × Molarity = 50 mL × 0.1 M = 5 mmol

Since NaOH is a strong base, it will react with acetic acid to form acetate ions (CH₃COO⁻) and water:

CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O

2 mmol of NaOH will react with 2 mmol of acetic acid, resulting in 2 mmol of acetate ions and leaving 3 mmol of acetic acid unreacted.

Next, we need to find the concentrations of acetic acid and acetate ions in the resulting 70 mL solution:

Concentration of acetic acid = Moles / Total volume = 3 mmol / 70 mL = 0.04286 M

Concentration of acetate ions = Moles / Total volume = 2 mmol / 70 mL = 0.02857 M

Now we can use the Henderson-Hasselbalch equation to find the pH of the resulting solution:

pH = pKa + log ([A⁻] / [HA])

Given the pKa of acetic acid is 4.76, we can substitute the values:

pH = 4.76 + log (0.02857 / 0.04286)

Using the given log values, we can approximate the log value:

log (0.02857 / 0.04286) ≈ log (2/3) ≈ log 2 - log 3 ≈ 0.30 - 0.48 = -0.18

Now substitute this value back into the Henderson-Hasselbalch equation:

pH = 4.76 - 0.18 = 4.58

To express the pH as the nearest integer multiplied by 10⁻², multiply the pH value by 100:

4.58 × 100 = 458

So, the pH of the resulting solution is approximately 458 × 10⁻².