Answer:
The pH of the solution is 4.28
Explanation:
The dissolution reaction as below
CH₃COOH ⇔ CH₃COO⁻ + H⁺
[tex]K_{a} = \frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH} = 10^{-4.76}[/tex]
Assume the concentration of the ion, [H⁺] = a,
so [CH₃COO⁻] = a and [CH₃COOH] = 3a
Then use the formula of Ka, we get
Ka = a * a / 3a = 10^-4.76 ⇔ a = 3 x 10^-4.76 = 5.21 x 10^-5
Hence pH = -log(a) = - log(5.21 x 10^-5) = 4.28