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When 200 ml of aqueous solution of HCl (pH = 2) is mixed with 300 ml of of an aqueous solution of NaOH (pH = 12), the pH of the resulting mixture is : |
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Answer» 10 For HCl solution `pH=-log[H^(+)]=2` `therefore [H^(+)]=10^(-2)M` Moles of `H^(+)` in 200 ml `=(10^(-2))/(1000)xx200` `=2xx10^(-3)` moles For NaOH solution, `pH=14-pOH=12` `therefore pOH = 2` `[OH^(-)]=10^(-2)M` Moles of `[OH^(-)]`in 300 ml solution `=(10^(-2))/(1000)xx300` `=3xx10^(-3)` moles `therefore` Moles of `OH^(-)` left `=3xx10^(-3)-2xx10^(-3)=10^(-3)` moles Total volume = 500 ml `therefore [OH^(-)]=(10^(-3))/(500)xx1000=2xx10^(-3)M` `pOH=-log[OH^(-)]=-log(2xx10^(-3))=2.699` `therefore pH=14-2.699=11.30` |
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