Ph larutan 50ml ch3cooh 0,2m yang direaksikan dengan 25ml koh 0,2m ka ch3cooh 10-5

CH3COOH 0,2 M , 50mL
KOH 0,2M , 25mL
Ka = 10^-5

CH3COOH+KOH –> KCH3COO+H2O
m : 10 5 – –
r : 5 5 5 5
s : 5 – 5 5

ka x n asam
[H+] = ——————-
n garam
10^-5 x 5
= ————-
5
[H+] = 10^-5

pH = – log H+
= – log 10^-5
pH = 5

Maaf kalo salah 🙂

Nch3cooh= VxM = 50×0,2 = 10 mol
nKOH = 25×0,2 = 5 mol
[H+] = Ka x n = 10^-5 x 5 mol
= 5 x 10^-5
pH= -log[H+]
= -log5 x 10^-5
= 5 – log5
= 5 – 0,7
= 4,3

Ps : maaf kalau tdk tepat,hehe