Bantu soal kimia ini ya.. Pliss..

CH3COOH + NaOH ----> NaCH3COO + H2O
mula : 0,15. 0,1
reaksi : 0,1. 0,1. 0,1
sisa : 0,05. - 0,1

Diket pH buffer = 5, maka [H+] = 10^-5
Rumus buffer asam :
[H+] = Ka x (na/ng)
10^-5 = Ka x (0,05/0,1)
10^-5 = Ka x (5x 10^-1)
Ka = (10^-5)/ (5 x 10^-1)
Ka = 2 x 10^-5