A) 12
B) 13
C) 1
D) 12.96
Correct Answer: B
Solution :
\[Ba{{(OH)}_{2}}\ \to \ \underset{.05M}{\mathop{B{{a}^{+2}}}}\,+\underset{2\times 0.5M}{\mathop{2O{{H}^{-}}}}\,\] \[pOH=\log \frac{1}{{{[OH]}^{-}}}=\log \frac{1}{.1}=1\] \[pH+pOH=14\]; \[pH+1=14\]; \[pH=14-1=13\]You need to login to perform this action.
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