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 \[\therefore \] \[{{[O{{H}^{-}}]}_{Total}}=\underbrace{1\times {{10}^{-7}}M}_{{{H}_{2}}O}+\underbrace{1\times {{10}^{-8}}M}_{NaOH}\] \[\therefore \]\[pOH=-\log [O{{H}^{-}}]=8-\log 11=8-104=6.96\] \[pH=14-pOH=14-6.96=7.04\]
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