A) 8.0
B) 7.0
C) 6.98
D) 14.0
Correct Answer: C
Solution :
\[{{H}_{2}}O\]⇌\[[{{H}^{+}}]\,\,[O{{H}^{-}}]\] HCl ⇌\[[{{H}^{+}}]\,\,[C{{l}^{-}}]\] Total \[[{{H}^{+}}]={{[{{H}^{+}}]}_{{{H}_{2}}O}}+{{[{{H}^{+}}]}_{HCl}}\] \[={{10}^{-7}}+{{10}^{-8}}\] \[={{10}^{-7}}[1+{{10}^{-1}}]\] \[[{{H}^{+}}]={{10}^{-7}}\times \frac{11}{10}\]\[\] \[pH=-\log \,[{{H}^{+}}]=-\log \left( {{10}^{-7}}+\frac{11}{10} \right)\];\[pH=6.958\]
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