The Ka value of nitrous acid, HNO2 , is 4.0 × 10¯4 . What is the equilibrium constant for the reaction; NO2¯ + H2O <-> HNO2 + OH¯ ?
To determine the equilibrium constant for the given reaction, you need to use the Ka value of nitrous acid, HNO2, and the relationship between Ka and Kb.
The equilibrium constant, denoted as Kb, is related to Ka by the equation: Kw = Ka x Kb, where Kw is the ionization constant of water.
The value of Kw at 25°C is 1.0 x 10^-14.
Since the reaction involves both HNO2 and OH-, we can write the following equation:
HNO2 + OH- ↔ NO2- + H2O
Now, let's represent the equilibrium constant using the concentrations of the species involved:
Kb = [NO2-] x [H2O] / [HNO2] x [OH-]
We know that the concentration of OH- is 1.0 x 10^-7 M (assuming the concentration of water is constant).
Now, let's substitute the values and solve for Kb:
Kw = Ka x Kb
1.0 x 10^-14 = (4.0 x 10^-4) x Kb
Divide both sides by (4.0 x 10^-4):
Kb = (1.0 x 10^-14) / (4.0 x 10^-4)
Simplifying the equation:
Kb = 2.5 x 10^-11
Therefore, the equilibrium constant for the given reaction NO2- + H2O ↔ HNO2 + OH- is 2.5 x 10^-11.