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mol CH₃COOH = (50)(1)
= 50 mmol
mol CH₃COONa = (50)(1)
= 50 mmol
Setelah pencampuran akuades:
[CH₃COOH] = 50/100
= 0,5 M
[CH₃COONa] = 50/100
= 0,5 M
Maka
[H⁺] = Ka. [CH₃COOH]/[CH₃COONa]
= (1,8.10⁻⁵)(0,5/0,5)
= 1,8.10⁻⁵ M
pH = -log [H⁺]
= 5 - log 1,8
SOLVED :)