20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is ___________ × 10⁻² (Nearest integer)
Given : pKa(CH₃ COOH)=4.76
log 2=0.30
log 3=0.48

Solution

Chemistry ma’am · 4 steps

The base is the limiting reagent, so you are left with a mix of acetic acid and acetate: a buffer.

1. Count millimoles. NaOH: 20 × 0.1 = 2 mmol. Acetic acid: 50 × 0.1 = 5 mmol. The base runs out first.
2. After neutralisation. CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O uses up all 2 mmol of OH⁻. That leaves 3 mmol of CH₃COOH and forms 2 mmol of CH₃COO⁻. Both are in the same 70 mL, so the volume cancels in the ratio.

   pH = pKₐ + log[CH₃COO⁻]/[CH₃COOH] = 4.76 + log 2/3

3. Put in logs. log 2/3 = 0.30 - 0.48 = -0.18, so pH = 4.76 - 0.18 = 4.58.
4. So. pH = 4.58 = 458 × 10⁻², so the answer is 458.

The common slip: 494. Putting the ratio upside down, acid over salt, gives log 3/2 = +0.18 and a pH of 4.94, so 494 instead of 458.

The idea to keep. When a strong base partly neutralises a weak acid, work in millimoles first. The leftover acid and the salt formed make a buffer, and you can use their mole ratio directly in the Henderson equation.

Answer from the NTA final answer key. Solution by PrepWiser.