Enough of a monoprotic acid is dissolved in water to produce a 0.0178 M solution. The resulting pH is 2.53. Calculate the Ka for the acid.?

Denote the monoprotic acid as HA.

HA(aq) + H₂O(l) ⇌ A⁻(aq) + H₃O⁺(aq) ... Kₐ

Initial concentrations :
[HA]ₒ = 0.0178 M
[A⁻]ₒ = [H₃O⁺]ₒ = 0 M

At equilibrium :
pH = 2.53, thus [H₃O⁺] = 10⁻²·⁵³ M
[HA] = (0.0178 - 10⁻²·⁵³) M
[A⁻] = [H₃O⁺] = 10⁻²·⁵³ M

Kₐ = [A⁻] [H₃O⁺] / [HA] = (10⁻²·⁵³)² / (0.0178 - 10⁻²·⁵³) = 5.87 × 10⁻⁴ (M)