Jawabanmu

2014-02-11T19:13:26+07:00
b. Ba(OH)2 (aq) → Ba 2+(aq) + 2 OH – (aq)
[OH –] = 2 x [Ba(OH)2] = 2 x 0.01 = 2.10 –2 M
pOH = - log [OH–] = - log 2.10 –2

pOH
= 2 - log 2
Jadi pH = 14 - pOH = 14 - (2 - log 2) = 12 + log 2

2 2 2
2014-02-11T19:36:32+07:00
Ba(OH)2 ~ Ba2+ ~ 2OH-
0.1 M  ----- 0.1M ---- 2*0.1 M
0.1 M  ----- 0.1M ---- 0.2 M

atau [OH-] = b x Mb
[OH-] = 2x 0.1M
[OH-] = 0.2M

pOH= -log [OH-]
pOH= -log[2 x 10^-1 M]
pOH= 1- log 2

pH = pKw - pOH
pH = 14-(1 - log2)
pH = 13 + log2 
pH = 13,301