Enough of a monoprotic acid is dissolved in water to produce a 0.0165 m solution. the ph of the resulting solution is 2.68. calculate the ka for the acid.

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.

Particle exchange energy though elastic collision , hope this .

It shows that the experiment is designed properly.