Answer is: Ka for the monoprotic acid is 6.8·10⁻⁴.
Chemical reaction:
HA(aq) ⇄ A⁻(aq) + H⁺(aq).
c(monoprotic acid) = 0.0128 M.
pH = 2.58.
[A⁻]
= [H⁺] = 10∧(-2.58).
[A⁻] = [H⁺] = 0.00263 M; equilibrium concentration.
[HA] = 0.0128 M - 0.00263 M.
[HA] = 0.01017 M.
Ka = [A⁻]·[H⁺] / [HA].
Ka =
(0.00263 M)² / 0.01017 M.
Ka = 0.00068 M = 6.8·10⁻⁴ M.
