pH confusion/dumb question

[jrobb220]

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So I thought I had pH stuff down pretty well, but after going through the destroyer again I came across a problem I didn't quite get.

#156) at 37 degrees C the Kw of H2O = 5.0X10^-14. What is the pH at this temperature?

Alright so I understand how they went through and got the pH,

Kw = [H3O][OH]

5.0E-14 = x^2

X= ~2.2E-7 so [H30] = 2.2E-7 -log(2.2E-7) = pH = ~ 6.65


HOWEVER...

Doesn't the [OH] also = 2.2E-7 so -log(2.2E-7) = pOH = ~ 6.65

14 - pOH = pH = 7.35

So wouldn't those just average out to a pH of 7!?

The answer is just 6.65?

 

[bustedgrill]

I think that in the Destroyer he was just trying to show you that the Kw is not a constant and changes with the temp. I wouldn't go too much more in depth.

 

[dentrilla]

So I thought I had pH stuff down pretty well, but after going through the destroyer again I came across a problem I didn't quite get.

#156) at 37 degrees C the Kw of H2O = 5.0X10^-14. What is the pH at this temperature?

Alright so I understand how they went through and got the pH,

Kw = [H3O][OH]

5.0E-14 = x^2

X= ~2.2E-7 so [H30] = 2.2E-7 -log(2.2E-7) = pH = ~ 6.65


HOWEVER...

Doesn't the [OH] also = 2.2E-7 so -log(2.2E-7) = pOH = ~ 6.65

14 - pOH = pH = 7.35

So wouldn't those just average out to a pH of 7!?

The answer is just 6.65?

I'm pretty sure that the formula pH + pOH = 14 is based on the fact that [H30][OH] = 10^-14 , and since the latter equation is for water @ 25degree celcius then that must mean "pH + pOH = 14" is only true at that temp. I'm not too sure but thats the only thing i can think off