A) 6.96
B) 8
C) 6
D) 7
Correct Answer: A
Solution :
\[{{[{{H}^{+}}]}_{{{H}_{2}}O}}=1\times {{10}^{-7}}\,M\] \[{{[{{H}^{+}}]}_{HCl}}=1\times {{10}^{-8}}\,\text{M}\] Total \[[{{H}^{+}}]\]concentration \[={{[{{H}^{+}}]}_{{{H}_{2}}O}}+{{[{{H}^{+}}]}_{HCl}}\] \[=1\times {{10}^{-7}}+1\times {{10}^{-8}}\] \[=11\times {{10}^{-8}}\] \[\therefore \] \[pH=-\log [{{H}^{+}}]\] \[=-\log \,11\times {{10}^{-8}}\] \[=8-\,\log 11=8-1.04\] \[=6.96\]
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