A) 8
B) between 7 and 8
C) between 6 and 7
D) between 8 and 9
Correct Answer: C
Solution :
Key Idea: In a dilute solution, the \[{{H}^{+}}\] furnished by water cannot be neglected Given \[HCl={{10}^{-8}}M\] \[\underset{{{10}^{-8}}}{\mathop{HCl}}\,\xrightarrow{{}}\underset{{{10}^{-8}}}{\mathop{{{H}^{+}}}}\,+\underset{{{10}^{-8}}}{\mathop{C{{l}^{-}}}}\,\] \[{{H}_{2}}O\underset{{{10}^{-7}}}{\mathop{{{H}^{+}}}}\,\,\,+\underset{{{10}^{-7}}}{\mathop{O{{H}^{-}}}}\,\] \[\therefore \] Total \[[{{H}^{+}}]={{10}^{-7}}+{{10}^{-8}}\] \[={{10}^{-7}}(1+{{10}^{-1}})\] \[pH=-\log \,\,[{{H}^{+}}]\] \[=-\log \,\,{{10}^{-7}}\,(1+{{10}^{-1}})\] = 6.98 \[\therefore \] The pH is between 6 and 7.
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