A) 11.3
B) 2.7
C) 11
D) 3
Correct Answer: A
Solution :
[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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