A) 7.00
B) 4.00
C) 9.00
D) 1.00
Correct Answer: B
Solution :
\[[{{H}_{3}}{{O}^{+}}]=[{{H}^{+}}]={{10}^{-10}}\] \[pH+pOH=14\] \[pH=-\log [{{H}^{+}}]\] \[pH=-\log \,[{{10}^{-10}}]\] pH = 10 pOH + 10 = 14 pOH = 14 - 10 = 4You need to login to perform this action.
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