A) 8
B) more than 8
C) between 6 and 7
D) slightly more than 7
Correct Answer: C
Solution :
From\[{{H}_{2}}O,[{{H}^{+}}]=1\times {{10}^{-7}}M\] From \[HCl,[{{H}^{+}}]=1\times {{10}^{-8}}M\] Total \[[{{H}^{+}}]=(1\times {{10}^{-7}}+1\times {{10}^{-8}})M\] \[=(1\times {{10}^{-7}}+0.1\times {{10}^{-7}})M\] \[=1.1\times {{10}^{-7}}M\] \[pH=-\log (1.1\times {{10}^{-7}})\] \[=6.9586\]
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