The dissociation constant of acetic acid is x×10⁻⁵. When 25 mL of 0.2 M CH₃COONa solution is mixed with 25 mL of 0.02 M CH₃COOH solution, the pH of the resultant solution is found to be equal to 5. The value of x is ____________

Solution

Chemistry ma’am · 4 steps

It is a buffer, so let Henderson's equation do the work; first get the salt-to-acid mole ratio.

1. Count millimoles. Salt: 25 × 0.2 = 5 mmol of CH₃COONa. Acid: 25 × 0.02 = 0.5 mmol of CH₃COOH. Both share the same 50 mL, so the volume cancels in the ratio.

   pH = pKₐ + log[CH₃COO⁻]/[CH₃COOH] = pKₐ + log 5/0.5

2. Find pKa. log 10 = 1, so 5 = pKₐ + 1, which gives pKₐ = 4.
3. Convert to Ka. Kₐ = 10⁻⁴. Written in the asked form, 10⁻⁴ = 10 × 10⁻⁵.
4. So. x = 10.

The common slip: 0.1. Putting the acid on top of the log ratio gives pKₐ = 6, so Kₐ = 0.1 × 10⁻⁵ and x = 0.1, not 10.

The idea to keep. For an acidic buffer, the salt (conjugate base) goes on top in the log term. When both solutions mix into one volume, use the mole ratio directly.

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