A) \[7\]
B) \[_{p}{{K}_{\omega }}\]
C) \[1\]
D) zero
Correct Answer: B
Solution :
\[pH+pOH=14\] \[pH+pOH{{=}_{p}}{{K}_{a}}\] \[[{{H}^{+}}]\times [O{{H}^{-}}]={{10}^{-14}}\] \[-\log [{{H}^{+}}]-\log [O{{H}^{-}}]=14\] \[_{p}{{K}_{\omega }}=14\]
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