Sebanyak 50ml larutan ch3cooh 0,2M (ka=10^-5) dicampur dengan 50ml larutan Naoh 0,1M. hitunglah ph larutan tersebut

jumlah mol CH3COOH = v x M = 50 x 0,2 = 10mmol
jumlah mol NaOH = v x M = 50 x 0,1 = 5mmol
CH3COOH + NaOH ➡ NaCH3COO +H2O
m : 10 5 - -
r : 5 5 5 5
----------------------------------------------------------- -
s : 5 - 5 5
karena yg tersisa larutan asam lemah dg garamnya berarti menggunakan rumus penyangga.
[H+] = ka x a/g
= 10^-5 x 10/5
= 2.10^-5
PH = - log [H+]
= - log 2.10^-5
= 5 - log 2