InterviewSolution
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InterviewSolution
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Ionic product of water at `310 K` is `2.7xx10^(-14)`. What is the `pH` of netural water at this temperature? |
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Answer» Correct Answer - `pH = 6.78` Ionic product, `K_(w)=[H^(+)][OH^(-)]` Let `[H^(+)]=x` ltBrgt Since `[H^(+)]=[OH^(-)], K_(w)=x^(2)` `rArrK_(w)"at310K is "2.7xx10^(-14)` `therefore 2.7xx10^(-14)=x^(2)` `rArrx=1.64xx10^(-7)` `rArr[H^(+)]=1.64xx10^(-7)` `rArrpH=-log[H^(+)]` `=-log[1.64xx10^(-7)]` = 6.78 Hence, the pH of neutral water is 6.78. |
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