A) 11.3
B) 2.7
C) 11
D) 3
Correct Answer: A
$ [O{H^{-}}]=2[Ba{{(OH)}_2}]=2\times .001=2\times {10^{-3}} $
$ pOH=-\log 2\times {10^{-3}},=2.7 $
$ pH=14-pOH=14-2.7=11.3 $
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