Question

What is the Ka of a 2.5 x 10^-2 M solution of nitrous acid (HNO2) with a pH of 4.27?

Answer 1

Answer:

Kₐ = 1.2 × 10⁻⁷

Explanation:

Step 1. Calculate [H₃O⁺]

$\text{pH = 4.27}\\\text{H _{3} O ^{+} } = 10^{\text{-pH}}\text{ mol/L} = 10^{-4.27} \text{ mol/L} =5.37 \times 10^{-5} \text{ mol/L}$

2. Calculate Kₐ

                        HNO₂ + H₂O ⇌   H₃O⁺    +      NO₂⁻

I/mol·L⁻¹:          0.025                     0                  0

C/mol·L⁻¹: 0.025-5.4 × 10⁻⁵   +5.4 × 10⁻⁵   +5.4 × 10⁻⁵

E/mol·L⁻¹:         0.025              5.4 × 10⁻⁵     5.4 × 10⁻⁵

So, at equilibrium,

[H₃O⁺] = [NO₂⁻] = 5.4 × 10⁻⁵ mol·L⁻¹

[HF] = 0.025 mol·L⁻¹

Kₐ = {[H₃O⁺][NO₂⁻]}/[HNO₂

Kₐ = (5.4 × 10⁻⁵ × 5.4 × 10⁻⁵)/0.025

Kₐ = 2.9 × 10⁻⁹/0.025

Kₐ = 1.2 × 10⁻⁷

The acid is not nitrous acid, because the Kₐ of HNO₂ is 7.2 × 10⁻⁴