Respuesta :

Answer is: Ka for the monoprotic acid is 3.03·10⁻⁴.

Chemical reaction: HA(aq) 
⇄ A⁻(aq) + H⁺(aq).
c(monoprotic acid) = 0.0165 M.
pH = 2.68.
[A
⁻] = [H⁺] = 10∧(-2.68).
[A
⁻] = [H⁺] = 0.00209 M; equilibrium concentration.
[HA] = 0.0165 M - 0.00209 M.
[HA] = 0.0144 M.
Ka = [A
⁻]·[H⁺] / [HA]. 
Ka = (0.00209 M)² / 0.0144 M.
Ka = 0.000303 M = 3.03·10
⁻⁴ M.
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