Options:
A.
B.
C.
D.
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Earlier discussion
3 older comments kept for context.
Uche Emmanuel
This is also wrong
Because they said hydrogen ion concentration not hydroxide ion concentration...having found the pH you subtract from 14 to give 3.30
Favour Onwuchekwa
poH = – Log[\( OH^{-1} \)]
[\( OH^{-1} \)]=2.0×\( 10^{-11}moldm^{-3} \)
poH=– Log[\( 2.0×10^{-11} \)]
poH= 10.70
Option [C] is correct.
Oluwaseun Adeyemi
p\(^{OH}\) = -Log [OH\(^-\)]
[OH\(^-\) ] = 2.0 X 10\(^{-11}\)moldm\(^{-3}\)
p\(^{OH}\) = -Log [2.0 X 10\(^{-11}\)]
p\(^{OH}\) = 10.70