sebanyak 0,2 mol H2SO4 dalam 1 liter larutan.berapa ph larutan H2SO4 tersebut

Dik: n=0,2, V=1 L
Dit: PH...?
jwb: [H+]=
[tex] \frac{n}{v} = \frac{02}{1} = 0.2 = 2 \times {10}^{ - 1} [/tex]
PH= -log[H+]
=
[tex] - log(2 \times {10}^{ - 1} ) \\ = 1 - log(2) [/tex]