Answer:
When HCl is dissolved in water, \[{{H}^{+}}\]
ion contribution of water will also be taken since it is very dilute solution.
\[{{[{{H}^{+}}]}_{Total}}={{[{{H}^{+}}]}_{HCl}}+{{[{{H}^{+}}]}_{Water}}\]
\[={{10}^{-8}}+{{10}^{-7}}\]
\[=11\times
{{10}^{-8}}M\]
\[pH=-\log
[{{H}^{+}}]\]
\[=-\log [11\times
{{10}^{-8}}]=6.9586\]
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