Question

If the ka of a monoprotic weak acid is 5.8 × 10-6, what is the ph of a 0.45 m solution of this acid?

Answer 1

Kₐ = 5.8*10⁻⁶
C = 0.45 mol/L

[H⁺] = √(KₐC)

pH = -lg[H⁺] = -lg{√(KₐC)}

pH = -lg{√(5.8*10⁻⁶*0.45)} = 2.79