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: 1.2 x 10^-7

Step-by-step explanation:

Answer 2

Kₐ = 1.2 × 10⁻⁷

Step-by-step 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 * 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