3.30
13.00
10.70
11.30
Correct answer is A
PH+POH=14 p\(^{H}\) = Log [H\(^+\)]
p[H\(^+\) ] = 2.0 X 10\(^{-11}\)moldm\(^{-3}\)
p\(^{H}\) = Log [2.0 X 10\(^{-11}\)]
p\(^{H}\) = 10.70
10.70+ p\(^{OH}\) =14 p\(^{OH}\) = 14 -10.70
= 3.30
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