Larutan asam kuat dengan ph 13 sebanyak 20 ml dicampur dengan 30 ml larutan CH3COOH 0.1M, ph campuran yang terjadi adalah ?

Diket: M CH3COOH =0,1 mol
V CH3COOH = 20 mL di campur dg 30 mL
(Ka CH3COOH = 1,8×10^-5
pH=13
Di tanya: pH campuran...??
Di jawab :
[H+] = - log[H+] =pH
= - log [H+] =13
= - log 10^-13
[H+]= 10^-13
M CH3COOH = M×V
= 0,1×(20+30)
= 5mmol
[H+] = ka×na/ng
10^-13= 1,8×10^-5× 5/ng
10^-13=9×10^-5/ng
ng = 9×10^-8
pH= - log [H+]
= - log 9×10^-8
= 8-log 9
Semoga benar