DAT Question: Please Help Me Out

[sochun90]

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This is the question: At 37 Celsius degree, the Kw for H2O is 5*10^-14. What is the pH of water at this temperature?

I figured since Kw is [H30+][OH-], [H30]= square root of Kw. Then I get pH, and it was the answer.

However(!), [OH-] would also be the square root of Kw, right? Then pOH would the same value as pH that I got above. Then(!), pOH=14-pH, which would, then, give me the different value for pH from the one I got above.

Could anyone give me a solid reason why the second approach does not work?

Thanks.

 

[lutopie0730]

I got curious about it and here is what i found


If the pH falls as temperature increases, does this mean that water becomes more acidic at higher temperatures? NO!

A solution is acidic if there is an excess of hydrogen ions over hydroxide ions. In the case of pure water, there are always the same number of hydrogen ions and hydroxide ions. That means that the water remains neutral - even if its pH changes.

The problem is that we are all so familiar with 7 being the pH of pure water, that anything else feels really strange. Remember that you calculate the neutral value of pH from Kw. If that changes, then the neutral value for pH changes as well.

At 100°C, the pH of pure water is 6.14. That is the neutral point on the pH scale at this higher temperature. A solution with a pH of 7 at this temperature is slightly alkaline because its pH is a bit higher than the neutral value of 6.14.

Similarly, you can argue that a solution with a pH of 7 at 0°C is slightly acidic, because its pH is a bit lower than the neutral value of 7.47 at this temperature.



So i guess as Kw changes due to its temperature dependancy,

at 36 degree celcius, pKw= 13.3

pOH=pKw-pH=13.3-6.65=6.65