A) 8.0
B) 6.0
C) 6.96
D) 7.04
Correct Answer: D
Solution :
For \[{{10}^{-8}}M\]\[NaOH,\] \[[O{{H}^{-}}]={{10}^{-8}}+{{10}^{-7}}\](from water) \[=11\times {{10}^{-8}}\] \[\therefore \] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log \,\,11\times {{10}^{-8}}\] \[=8-1.04\] \[=6.96\] \[\therefore \] \[pH=14-6.96=7.04\]
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