Answer:
The neutral water has \[[{{H}^{+}}]=1\times
{{10}^{-7}}M\]
By adding\[1.0\times
{{10}^{-8}}M\,HCl\], a concentration of \[1.0\times
{{10}^{-8}}\]\[M\,\,{{H}^{+}}\]ions has increased in solution.
Thus,
total \[[{{H}^{+}}]=(1\times {{10}^{-7}}+1\times {{10}^{-8}})M\]
\[=(1\times
{{10}^{-7}}+0.1\times {{10}^{-7}})M=1.1\times {{10}^{-7}}M\]
\[pH=-\log
(1.1\times {{10}^{-7}})=-[log1.1+log{{10}^{-7}}]\]
\[=-[0.0414-7.0]=6.9586\]
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