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The pH of a `10^(-8)` molar solution of HCl in water isA. 8B. between 6 and 7C. between 7 and 8D. None of these |
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Answer» Correct Answer - C `pH=-log[H^(+)]=-log10^(-8)=8` It is not possible for acid, so it is `[H^(+)]`, the `[H^(+)]` of water is also added. Total `[H^(+)]` in solution `=[H^(+)]"of HCl"+[H^(+)]` of water `=(1xx10^(-8)+1xx10^(-7))M` `=(1+10)xx10^(-8)=11xx10^(-8)M` `therefore pH=-log[H^(+)]` `=-log 11xx10^(-8)` `=-log 11xx8 log 10` `=-1.0414+8=6.9586` |
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