A) 10
B) 7.01
C) 6.99
D) 4
Correct Answer: B
Solution :
Key Idea: In a very dilute solution, the concentration of\[O{{H}^{-}}\]produced from water cannot be neglected \[\left( \begin{align} & pOH=-\log [O{{H}^{-}}] \\ & \,\,\,\,pH=14-pOH \\ \end{align} \right)\] \[[O{{H}^{-}}]=\underset{(from\,\,NaOH)}{\mathop{{{10}^{-10}}}}\,+\underset{(from\,water)}{\mathop{{{10}^{-7}}}}\,\] \[={{10}^{-7}}(0.001+1)\] \[=1.001\times {{10}^{-7}}\] \[pOH=-\log [O{{H}^{-}}]\] \[=-\log (1.001\times {{10}^{-7}})\] \[=7-0.01\] \[=6.99\] \[pH=14-pOH\] \[=14-6.99\] \[=7.01\]
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