JEE Main 2022 · 28 June · evening · Single correct · +4 / −1
A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4. The ratio of [CH₃CH₂COO - ]/[CH₃CH₂COOH] required to make buffer is ___________.
Given : Kₐ(CH₃CH₂COOH) = 1.3 × 10⁻⁵
- (a) 0.03
- (b) 0.13
- (c) 0.23
- (d) 0.33
Answer and solution
Answer (b) 0.13
Solution
Chemistry ma’am · 4 steps
Skip the logs here: the salt-to-acid ratio is just Kₐ divided by [H⁺].
- Acid equilibrium. For CH₃CH₂COOH ⇌ CH₃CH₂COO⁻ + H⁺, the Kₐ expression links the ratio you want directly to [H⁺]. This is the Henderson equation before taking logs.
Kₐ = [CH₃CH₂COO⁻][H⁺]/[CH₃CH₂COOH] ⇒ [CH₃CH₂COO⁻]/[CH₃CH₂COOH] = Kₐ/[H⁺]
- Find H+. pH 4 means [H⁺] = 10⁻⁴ mol L⁻¹.
- Put in numbers. The ratio is (1.3 × 10⁻⁵)/(1 × 10⁻⁴) = 0.13. Since pKₐ = 4.89 is above the target pH of 4, you need more acid than salt, so a ratio below 1 makes sense.
- So. The required ratio [CH₃CH₂COO⁻]/[CH₃CH₂COOH] is 0.13, option (b).
The idea to keep. In an acid buffer, salt/acid = Kₐ/[H⁺]. If the target pH is below pKₐ, the ratio must be less than 1; that quick check catches an inverted answer.
Answer from the NTA final answer key. Solution by PrepWiser.