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?

QUESTION POSTED AT 23/05/2020 - 04:32 PM

Answered by admin AT 23/05/2020 - 04:32 PM

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
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