- 2067 Mangsir · 1+4 marks
Buffer; pH and degree of ionization of acetic acid
Short answer
A buffer solution resists pH change on adding small amounts of strong acid/base; it contains a weak acid and its salt. For 200 mL 0.2 M CH3COOH with 1.64 g CH3COONa, [salt]=0.10 M, [acid]=0.2 M, pH=4.74+log(0.1/0.2)=4.44; degree of ionization α=[H⁺]/C=3.6×10⁻⁵/0.2=1.8×10⁻⁴.
What is a buffer? 1.64 g of anhydrous sodium acetate is added to 200 ml of 0.2M acetic acid. What is the pH of buffer? Calculate the degree of ionization of the acid in the solution. (Ka of acid = 1.8×10⁻⁵)
Buffer
A buffer solution resists a change in its pH when small amounts of a strong acid or base are added. It contains a weak acid and its salt (or a weak base and its salt).
pH of the buffer
Moles CH₃COONa = 1.64/82 = 0.020 mol
[salt] = 0.020/0.200 = 0.10 M ; [acid] = 0.2 M
pKa = −log(1.8×10⁻⁵) = 4.745
pH = pKa + log([salt]/[acid]) = 4.745 + log(0.10/0.20)
= 4.745 − 0.301 = 4.44
Degree of ionisation of the acid
[H⁺] = Ka [acid]/[salt] = 1.8×10⁻⁵ × 0.2/0.1 = 3.6×10⁻⁵ M
α = [H⁺]/C = 3.6×10⁻⁵/0.2 = 1.8×10⁻⁴
Answer: pH = 4.44; degree of ionisation α = 1.8×10⁻⁴ (0.018 %). (Without the salt, α would be √(Ka/C) = 9.5×10⁻³; the common ion greatly suppresses ionisation.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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