What is the equilibrium constant for the reaction NO2 + H2O ⇌ HNO2 + OH⁻ if the Ka value of nitrous acid (HNO2) is 4.0 x 10⁻⁴?

To find the equilibrium constant for the reaction NO₂ + H₂O ⇌ HNO₂ + OH^–, we start with the ionization reaction of nitrous acid:

HNO₂ ⇌ H⁺ + NO₂^–

The equilibrium constant expression for this reaction is:

Ka = [H⁺][NO₂^–]/[HNO₂]

We are given that Ka = 4.0 x 10⁻⁴.

For the reaction we are considering, we can relate it to the ionization reaction of HNO₂. The reaction can be rewritten as:

NO₂ + H₂O ⇌ HNO₂ + OH^–

Applying the principles of equilibrium, the equilibrium constant (K) for our target reaction will be related to the Ka of HNO₂ by the following relationship:

K = Ka / Kw

Where Kw is the ion-product constant of water, which at 25°C is 1.0 x 10⁻¹⁴.

Substituting the values:

K = (4.0 x 10⁻⁴) / (1.0 x 10⁻¹⁴)

K = 4.0 x 10⁹

Thus, the equilibrium constant for the reaction NO₂ + H₂O ⇌ HNO₂ + OH^– is 4.0 x 10⁹.