A) 10
B) 7.01
C) 6.99
D) 4
Correct Answer: B
Solution :
When the solution is very dilute, the concentration of \[O{{H}^{-}}\] produced from water cannot be neglected. Hence, \[[O{{H}^{-}}]={{10}^{-10}}+{{10}^{-7}}\] (obtained from water) \[={{10}^{-7}}(0.001+1)\] \[=1.001\times {{10}^{-7}}\] \[\therefore \] \[pOH=-\log \,[O{{H}^{-}}]\] \[=-\log \,(1.001\times {{10}^{-7}})\] \[=7-0.01=6.99\] \[\therefore \] \[pH=14-pOH\] \[=14-6.99=7.01\]
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