Buffers - Chemistry | Scholardog

1. Fundamental Concepts

Buffers resist pH changes upon addition of small amounts of strong acid or strong base. Dilution does not change pH significantly but reduces buffer capacity.
Composed of a weak acid + its conjugate base or a weak base + its conjugate acid.
Buffers work by neutralizing small amounts of added H⁺ or OH⁻ using the conjugate pair.

Buffers have a limited capacity; large amounts of strong acid/base destroy the buffer.
Henderson–Hasselbalch equation:
$\mathrm{pH = p}K_a + \log\frac{[\text{conjugate base}]}{[\text{weak acid}]}$
When [conjugate base] = [weak acid], pH = pKa (maximum buffer capacity).
Effective buffer range: pKa ± 1.

Which mixture is a buffer?

A. HCl + NaCl

B. HC₂H₃O₂ + NaC₂H₃O₂

C. NaOH + NaCl

Explanation:

Buffers require a weak acid or weak base plus its conjugate salt.

HCl and NaOH are strong, so A and C are not buffers.
HC₂H₃O₂ is a weak acid; C₂H₃O₂⁻ (from NaC₂H₃O₂) is its conjugate base.

Answer: B

Calculate the pH of a buffer with 0.10 M CH₃COOH and 0.10 M CH₃COO⁻.

pKa(CH₃COOH) = 4.745.

Solution:

$\mathrm{pH = p}K_a + \log\frac{[\text{conj base}]}{[\text{weak acid}]} = 4.745 + \log\frac{0.10}{0.10} = 4.745$

Answer: pH = 4.745

A buffer contains 0.20 M HF and 0.10 M F⁻.

Ka(HF) = 6.6×10⁻⁴.

After adding enough HCl to increase [H⁺] by 0.010 M (assuming negligible volume change)

Step-by-Step Explanation:

Find pKa:
$\mathrm{p}K_a = -\log(6.6\times10^{-4}) \approx 3.18$
Added H⁺ reacts with conjugate base F⁻:
- [F⁻] decreases by 0.010 M → 0.09 M
- [HF] increases by 0.010 M → 0.21 M
Apply HH equation:
$\mathrm{pH} = 3.18 + \log\frac{0.09}{0.21} \approx 3.18 - 0.37 = 2.81$
Answer: pH ≈ 2.81

Confirm the pair is weak acid/conj base or weak base/conj acid.
Use the Henderson–Hasselbalch equation for pH calculations.
When strong acid/base is added:
- H⁺ + conjugate base → weak acid
- OH⁻ + weak acid → conjugate base + H₂O
Use moles or molarities for the ratio (volume cancels out).
Check if pH stays within pKa ± 1 (valid buffer range).