a) NaOH 0,02 M
b) Mg(oH)2 0,05 M
c) Ca(oH)2 0,02 M
d)NH3 0,1 M Kb –
e)NH3 0,25M Kb = 1.8 X 10^-5
f)NH2oH 0,5 m Kb = 1,8×10^-5
Hitung pH dari larutan berikut
NaOH 0,02 M = [OH] = 0,02 x 1 = 0,02 / 2 x 10^-2
pOH = 2 – log 2
pH = 14 + log 2
Mg(OH)2 0,05 M = [OH] = 0.05 x 2 = 0,1 / 1 x 10^-1
pOh = 1 – log 1
pH = 13 + log 1
Ca(OH)2 0,02 M = [OH] = 0,02 x 2 = 0,04 / 4 x 10^-2
pOH = 2 – log 4
pH = 12 + log 4