1 mol of Ba(OH)2 gives 2 mol of OH-
So,
[OH-] = 2*[Ba(OH)2]
= 2*0.075 M
= 0.150 M
use:
pOH = -log [OH-]
= -log (0.15)
= 0.8239
use:
PH = 14 - pOH
= 14 - 0.8239
= 13.1761
Answer: 13.18
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