Sebanyak 250 ml larutan Ba(OH)2 0,005 M mempunyai pH...

250 mL Ba(OH)2 0,005 M

pH ... ?

Ba(OH)2 adalah basa kuat, jadi
[OH^-] = b × Mb
[OH^-] = 2 × 0,005
[OH^-] = 0,01
[OH^-] = 10^-2

pOH = - log [OH^-]
pOH = - log 10^-2
pOH = 2

pH = 14 - pOH
pH = 14 - 2
pH = 12