What is the ka of a 2.5 x 10^-2 m solution of nitrous acid (hno2) with a ph of 4.27?

Kₐ = 1.2 × 10⁻⁷

Explanation:

Step 1. Calculate [H₃O⁺]

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

i believe the answer is 5,000

it would only be the stuff in module eight so you need to study all the lessons in that you also could check the flvs

hope this : )