Question

What is the Ka of a 0.0326 M solution of hydrofluoric acid with a pH of 3.94?

Answer 1

Kₐ = 4.06 × 10⁻⁷

Explanation:

Step 1. Calculate [H₃O⁺]

`\text{H}_(3)\text{O}^(+) = 10^{\text{-pH}}\text{ mol/L}`

pH = 3.94

`\text{H}_(3)\text{O}^(+) = 10^(-3.94)} \text{ mol/L}`

[H₃O⁺] = 1.15 × 10⁻⁴ mol·L⁻¹

Calculate
`K_{\text{a}}`

HF + H₂O ⇌ H₃O⁺ + F⁻

I/mol·L⁻¹: 0.0326 0 0

C/mol·L⁻¹: 0.0326-1.15 × 10⁻⁴ +1.15 × 10⁻⁴ +1.15 × 10⁻⁴

E/mol·L⁻¹: 0.0325 1.15 × 10⁻⁴ 1.15 × 10⁻⁴

So, at equilibrium,

[H₃O⁺] = [F⁻] = 1.15 × 10⁻⁴ mol·L⁻¹

[HF] = 0.326 – 1.15 × 10⁻⁴ mol·L⁻¹ = 0.0325 mol·L⁻¹

Kₐ = {[H₃O⁺][F⁻]}/[HF]

Kₐ = (1.15 × 10⁻⁴ × 1.15 × 10⁻⁴)/0.0325

Kₐ = 1.32 × 10⁻⁸/0.0325

Kₐ = 4.06 × 10⁻⁷

This is NOT a solution of HF (Kₐ = 7.2 × 10⁻⁴). It is more likely a solution of carbonic acid (H₂CO₃; Kₐ₁ = 4.27 × 10⁻⁷).