A) 3
B) \[10-11\]
C) 1
D) 11
Correct Answer: D
Solution :
In \[N/1000\,\,KOH,\] \[[O{{H}^{-}}]={{10}^{-3}}M\] \[pOH=-\log [O{{H}^{-}}]=-\log {{10}^{-3}}=3\] \[pH=14-pOH=14-3=11\]You need to login to perform this action.
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