A) 8.0
B) 6.0
C) 6.98
D) 7.04
Correct Answer: D
Solution :
\[\because \] \[[NaOH]={{10}^{-8}}M\] \[\therefore \] \[[O{{H}^{-}}]=1\times {{10}^{-8}}M\] Here, \[[O{{H}^{-}}]\] of water in this dilute solution should be added \[\therefore \] \[{{[O{{H}^{-}}]}_{Total}}=\underbrace{1\times {{10}^{-7}}M}_{{{H}_{2}}O}+\underbrace{1\times {{10}^{-8}}M}_{NaOH}\] \[=11\times {{10}^{-8}}\] \[\therefore \]\[pOH=-\log [O{{H}^{-}}]=8-\log 11=8-104=6.96\] \[pH=14-pOH=14-6.96=7.04\]
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