Finding the pH of a buffer can be a rather complicated task. The best way to learn how to solve questions like these is to take a look at an example.
We have 20cm³ of 2 moldm⁻³ CH₃COOH. We add 5cm³ of 0.5 moldm⁻³ NaOH to it. What is the final pH?
Step 1: Find the remaining no. of moles of CH₃COOH
CH₃COOH + NaOH → CH₃COO⁻Na⁺ + H₂O
Original amt of CH₃COOH = \( \frac{20}{1000} (2) \) = 0.04 mol
Original amt of NaOH = \( \frac{5}{1000} (0.5) \) = 0.0025 mol
Remaining CH₃COOH = 0.04 - 0.0025 = 0.0375 mol
Step 2: Find the new concentration of CH₃COOH
New conc. = \(\frac{0.0375}{0.025}\) = 1.5 moldm⁻³
Step 3: Use the Ka constant to find [H⁺] and then find the pH.
Ka of CH₃COOH = 1.8 x 10⁻⁵
Ka = 1.8 x 10⁻⁵ = \( \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \)
As one NaOH reacts with one CH₃COOH, one CH₃COO⁻ is formed for every NaOH that reacts. Thus, 0.0025 mol of NaOH produces 0.0025mol of CH₃COO⁻.
[CH₃COO⁻] = \(\frac{0.0025}{0.025}\) = 0.1 moldm⁻³
[H⁺] = \( \frac{[CH_3COOH]}{[CH_3COO^-]} = \frac{1.5 (1.8 \times 10^{-5})}{0.1} \) = 2.7 x 10⁻⁴
pH = -log₁₀ 2.7 x 10⁻⁴ = 3.57