Calculating pH, pOH, [OH-], and Kw at 37°C

[spiff711]

Problem:
The pH of pure water at 37°C is 6.80. Calculate Kw,
pOH, and [OH^-] at this temperature.

Work:
pH: 6.8
pOH = 14 -6.8 = 7.2
[OH]: 10^-7.2 = 6.3 x 10^-8 M
K_w= [H₃O] [OH] = 1.5 x 10^-7 x 6.3 x 10^-8 = 1x10^-14


Real answer:
pH:6.8
pOH:6.8
[OH]: 1.6x10^-7
K_w: 2.5 x 10^-14

What am I doing wrong? Any help would be greatly appreciated.

 

[symbolipoint]

Beware my response here may be fundamentally flawed: but the question should give you both pH and pOH if you want to find K_w for water at 37 celslius.

In your "Work" section you bluntly assumed 1x10^-14 for water, but this is NOT the value for water at 37 celsius.

 

[spiff711]

Thanks for the response.

Well K_w= [H₃O] x [OH]
So I got the value for pOH from the equation pk = 14 = pOH + pH.
I then found the concentrations for H₃O and OH by using the formulas for pH (pH = -log [H₃O]) and pOH ( pOH= -log[OH]). I then plugged those values into the orginal equation and got K_w = 1x10^-14. I realize that is the value for water at 25^o C but apart from the fact that K increases with temperature I don't know how else it ties in.

 

[Borek]

Where does the OH^- comes from? Think about stoichiometry of the reaction. How is [OH^-] related to [H⁺] if there are no other sources of both but water autodissociation?

 

[symbolipoint]

Now this is clear (Borek, #4). The [H+] concentration is equal to the [OH-] concentration. All the values needed to determine the dissociation constant are present.

 

[spiff711]

Edit- Never mind, I understand. The water dissociates into equal parts so the pH and pOH will both be 6.8. I think at least.

Thanks for the help!