Kₐ for butyric acid (C₃ H₇ COOH) is 2 × 10⁻⁵. The pH of 0.2 M solution of butyric acid is __________ × 10⁻¹. (Nearest integer)
[Given log 2=0.30]

Solution

Chemistry ma’am · 4 steps

A weak acid barely ionises, so take the square root of Kₐ times C to get [H⁺].

1. Weak acid ionisation. Butyric acid ionises only slightly: C₃H₇COOH ⇌ C₃H₇COO⁻ + H⁺. If α is the fraction ionised, Kₐ = Cα² because 1 - α ≈ 1.

   [H⁺] = Cα = √(Kₐ C)

2. Put in numbers. [H⁺] = √(2 × 10⁻⁵ × 0.2) = √(4 × 10⁻⁶) = 2 × 10⁻³ M. Here α = 0.01, so ignoring it next to 1 is safe.
3. Take the log. pH = -log(2 × 10⁻³) = 3 - log 2 = 3 - 0.30 = 2.7.
4. So. pH = 2.7 = 27 × 10⁻¹, so the answer is 27.

The common slip: 54. Forgetting the square root and taking [H⁺] = Kₐ C = 4 × 10⁻⁶ gives pH 5.4, so 54 instead of 27.

The idea to keep. For a weak acid that ionises less than about 5%, [H⁺] = √(Kₐ C), or pH = 1/2(pKₐ - log C). Check α is small before trusting it.

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