Chemistry
joedlouiselina
34

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?

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(1) Answers
Maryanne545

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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