Question

For 0.2 M nitrous acid, HNO₂, in aqueous solution at equilibrium, which is the correct expression for Ka if x represents the concentration of H₃O+?

Answer 1

For 0.2 M nitrous acid (HNO₂), the correct expression for the acid dissociation constant (Ka) using x as the hydronium ion ([H₃O⁺]) concentration at equilibrium is Ka = (x²) / 0.2, assuming x is much smaller than the initial concentration.

Step-by-step explanation:

The correct expression for the acid dissociation constant (Ka) of nitrous acid (HNO₂) when x represents the concentration of hydronium ions ([H₃O⁺]) is based on the equilibrium reaction:

HNO₂ (aq) ⇌ H₃O⁺ (aq) + NO₂⁻ (aq)

At equilibrium, the concentration changes for the reactants and products can be represented using an ICE (Initial, Change, Equilibrium) table. The initial concentration of HNO₂ is 0.2 M and the initial concentrations of H₃O⁺ and NO₂⁻ are approximately zero. As the reaction proceeds to equilibrium, the concentration of HNO₂ will decrease by x, while the concentrations of H₃O⁺ and NO₂⁻ will increase by x.

The Ka expression is therefore:

Ka = ([H₃O⁺][NO₂⁻]) / [HNO₂]

Because [H₃O⁺] and [NO₂⁻] are both equal to x at equilibrium, and the equilibrium concentration of HNO₂ is 0.2 M - x, we can simplify the expression to:

Ka = (x²) / (0.2 - x)

Assuming x is much smaller than the initial concentration (which is a valid assumption for weak acids), this simplifies further to:

Ka = (x²) / 0.2