Question

How do you do this question?

Answer 1

`\boxed{\text{(D) 2 HCHO}_(2)}`

Step-by-step 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

`K_{\text{a}} = \frac{\text{[H}_(3)\text{O}^(+)]\text{HCOO}^(-)]} {\text{[HCOOH]}} = 2 * 10^(-4)\\\\(x^(2))/(0.5-x) = 2 * 10^(-4)`

Check for negligibility of x

`( 0.5 )/(2 * 10^(-4)) = 2500 > 400.`

∴ x ≪ 0.5

`(x^(2))/(0.5) = 2 * 10^(-4)`

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
`\boxed{\textbf{(D) 2 HCHO}_(2)}`.