Jawabanmu

2014-02-03T21:31:20+07:00
OH^- = a. ma (rumus basa kuat)
       = 2. 0.01
       = 0.02 M
pOH = - log OH^-
     = - log 2 x 10^-2
      = 2- Log 2
pH = 14- (2-log2)
      = 14-2+log 2
       = 12+log2
2014-02-03T22:29:21+07:00
[H⁺] = M . a
[H⁺] = 0,01 . 2
[H⁺] = 2 . 10⁻²

pH = - log [H⁺]
pH = - log 2 . 10⁻²
pH = 2 - log 2