Answer:
[tex]\boxed{\text{(D) 2 HCHO}_{2}}[/tex]
Explanation:
HCOOH + H₂O ⇌ H₃O⁺ + HCOO⁻
HCHO₂ is a weak acid. It dissociates only to a few percent, so there will be more HCHO₂ than H₃O⁺ present.
After H₂O, the most abundant species will be undissociated HCHO₂, so the answer will be either (B) or (D).
We can use an ICE table to organize the calculation of the pH.
HCOOH +H₂O ⇌ H₃O⁺ + HCOO⁻
I/mol·L⁻¹: 0.5 0 0
C/mol·L⁻¹: -x +x +x
E/mol·L⁻¹: 0.5 - x x x
[tex]K_{\text{a}} = \dfrac{\text{[H}_{3}\text{O}^{+}]\text{HCOO}^{-}]} {\text{[HCOOH]}} = 2 \times 10^{-4}\\\\\dfrac{x^{2}}{0.5-x} = 2 \times 10^{-4}[/tex]
Check for negligibility of x
[tex]\dfrac{ 0.5 }{2 \times 10^{-4}} = 2500 > 400.[/tex]
∴ x ≪ 0.5
[tex]\dfrac{x^{2}}{0.5} = 2 \times 10^{-4}[/tex]
x² = 0.5 × 2 × 10⁴ = 1 × 10⁻⁴
x = √(1 × 10⁻⁴) = 1 × 10⁻²
[H₃O⁺] = x mol·L⁻¹ = 1 × 10⁻² mol·L⁻¹
pH = -log[H₃O⁺] = -log(1 × 10⁻²) = 2
The correct answer is [tex]\boxed{\textbf{(D) 2 HCHO}_{2}}[/tex].
