A) 4
B) 10
C) 6
D) between 6-7
Correct Answer: B
Solution :
\[1\times {{10}^{-4}}M\,NaOH=1\times {{10}^{-4}}M[O{{H}^{-}}]\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log (1\times {{10}^{-4}})=4\] \[pH=14-pOH\] \[=14-4=10\]You need to login to perform this action.
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