tentukan ph dari HCI 0,008 m

H+ = _/a.Ma
= _/1.(8.10^-3)
= _/8.10^-3
=2_/3 . 10^-1.5
Ph= - [H+]
= - log 2_/3 . 10^-1.5
= 1.5- 2_/3
Maaf kalo salah

[H+] = a Ma
= 1 x 8.10-3
= 8.10-3 M
pH = -log [H+]
= -log 8.10-3
= 3 - log 8