Methanoic acid HCOOH is a weak acid present in the sting of some ants - VCE - SSCE Chemistry - Question 4 - 2003 - Paper 1

The expression for the acid dissociation constant (Kₐ) of methanoic acid (HCOOH) can be written as:

$K_a = \frac{[HCOO^-][H^+]}{[HCOOH]}$

Let the concentration of H⁺ produced by the dissociation of methanoic acid be represented by 'x'. Therefore, we have:

• Initial concentration of HCOOH = 0.10 M
• At equilibrium:
  • [HCOO⁻] = x
  • [H⁺] = x
  • [HCOOH] = 0.10 - x

Using the expression for Kₐ:

$K_a = \frac{x \times x}{0.10 - x}$

Given that Kₐ = 1.8 × 10⁻⁵, and assuming 'x' is very small compared to 0.10, we can simplify:

$1.8 \times 10^{-5} \approx \frac{x^2}{0.10}$

This results in:

$x^2 = (1.8 \times 10^{-5}) \times 0.10 = 1.8 \times 10^{-6}$

Taking the square root to find 'x':

$x = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \, M$

Thus:

• H⁺ ≈ 1.34 × 10⁻³ M
• HCOO⁻ ≈ 1.34 × 10⁻³ M