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Calculate the [ H + ] and pH of a 0.0040 M butanoic acid solution. The K a of butanoic acid is 1.52 × 10 − 5 . Use the method of successive approximations in your calculations.

Respuesta :

Answer:

Ka= [H+][But-]/H[Hbut]

Let k moles of H+ be formed

1.52*10^-5=k^2/0.004-k

K^2=0.004*1.52*10^-5-1.52*10^-5k

K^2- 1.52*10^-5k-6.08*10^-8=0

Solving this quadratic equation.

K=7.6*10^-6

PH=-log(7.6*10^-6)=5.2

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