CHEMISTRY COMMUNITY

The question asks, " The value of Kw for water at body temperature (37 degrees Celcius) is 2.1x10^-14. what is the molarity of H3O+ ions and the pH of neutral water at 37 degrees Celcius? what is the molarity of OH- in neutral water at 37 degrees Celcius?
I'm having a hard time understanding the question. Please help.
Also I thought we only had to do questions with room temperature (25 degrees Celcius). Don't we have to change the way we do this types of problems to satisfy this? thanks

I think that as long as the temp stays the same throughout the problem then it should not cause any changes in the way that you do the problem.

Kw = [H3O+][OH-]. So if Kw=2.1 x 10^-14, this is equal to [H3O+][OH-]. This comes from the equation 2H2O (l) --> H3O+ (aq) + OH- (aq), we take x^2 = 2.1 x 10^-14, so x=1.4 x 10^-7. You then take the -log(1.4 x 10^-7) to get a pH of 6.8. Once you have the pH you can calculate the pOH knowing that pH + pOH = 14.

Why is Kw=x^2? Doesn't that imply that H3O and OH are the same value? And if so, then how are we supposed to know that?

elliekauffman1I wrote:Why is Kw=x^2? Doesn't that imply that H3O and OH are the same value? And if so, then how are we supposed to know that?

I also need help with this question

[H3O+] = [OH-] when a solution is neutral, so if you set [H3O+] = x, then [OH-] = x. Therefore, if [H3O+][OH-] = 2.1 x 10^-14, then x^2 = 2.1 x 10^-14.

Hope this helps!

So does the Equilibrium constant still equal 10^-14 even if the temperature does not equal 25 degrees?

The equilibrium constant K changes with temperature so Kw is 10^-14 only at 25 degrees Celsius.