Answer:
\[pH=5\]
\[\therefore \]\[[{{H}^{+}}]={{10}^{-5}}M\]
Let 1 L of this solution is
diluted to 100 L.
\[{{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}\]
\[{{10}^{-5}}\times 1={{M}_{2}}\times
100\]
\[{{M}_{2}}={{10}^{-7}}M\]
\[{{H}^{+}}\]ion contribution of
water will also be considered because solution of acid is very dilute.
\[{{[{{H}^{+}}]}_{Total}}={{[{{H}^{+}}]}_{Acid}}+{{[{{H}^{+}}]}_{Water}}\]
\[={{10}^{-7}}+{{10}^{-7}}\]
\[=2\times {{10}^{-7}}M\]
\[pH=-{{\log }_{10}}[{{H}^{+}}]\]
\[=-{{\log }_{10}}[2\times
{{10}^{-7}}]=6.6989\]
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