What is the ka of a 0.0326 m solution of hydrofluoric acid with a ph of 3.94?

Kₐ = 4.06 × 10⁻⁷

Explanation:

Step 1. Calculate [H₃O⁺]

pH = 3.94

[H₃O⁺] = 1.15 × 10⁻⁴ mol·L⁻¹

Calculate

                             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).

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