Answer :
To find the Ka of the monoprotic acid solution, the percent ionization is used to calculate the concentrations of the ionic and non-ionic forms. Those concentrations are then plugged into the equilibrium expression, resulting in the calculation of the Ka value to be approximately 3.35 × 10⁻³.
To calculate the acid dissociation constant (Ka) for a monoprotic acid in aqueous solution where 4.54% of the acid is ionized, we use the given concentration of the acid solution (1.55 M) and the degree of ionization. For a monoprotic acid, which we can denote as HA, the dissociation can be represented as:
HA ⇌ H⁺ + A⁻
Since 4.54% of the acid is ionized, we can calculate the concentrations of H⁺ and A⁻ at equilibrium. First, we determine the ionized concentration:
[H⁺] = [A⁻] = 1.55 M × 0.0454 = 0.0704 M
Also, the concentration of non-ionized acid would be:
[HA] remaining = 1.55 M × (1-0.0454) = 1.479 M
Then, we apply these values to the equilibrium expression for the dissociation of a weak acid:
Ka = [H⁺][A⁻] / [HA]
Substituting in the calculated concentrations:
Ka = 0.0704 M × 0.0704 M / 1.479 M
Ka = 0.00495776 / 1.479
Ka ≈ 3.35 × 10⁻³
