The ion-product of pure water, Kw, is
2.51 * 10(-14) at the body temp of 37 degrees Celcius. What are the molar H2O and OH concentrations in the pure water temperature?
I expect you meant H3O^+ and not H2O.
(H3O^+)(OH^-) = Kw
(H3O^+)=(OH^-) = sqrt Kw.
Yes you are correct for the change to H3O^
But do I need the plug the 37 degrees in somewhere or is the response above the final answer?
I saw only this one extra question whereas your post said "additional" questions. If there are more than this please just post everything again at the top of the page.
You don't do anything with the 37C temperature. That's the temperature of the water at body temperature and Kw changes with T. At this T it is 2.51E-14 (at room T it is 1E-14). So (H3O^+) = (OH^-) = sqrt 2.51E-14.
So I would write (H3O^+) = 1.58E-7
and (OH^-) = 1.58E-7
By the way it makes it hard to follow the thread when you change screen names.
I didn't put units on my answers and I should have done so. They are
(H3O^+) = 1.58E-7 M
(OH^-) = 1.58E-7 M.
And just to make things a little more interesting let me point out that this is a NEUTRAL solution with a pH = 6.80. We always talk about pH = 7.0 as being neutral and solutions less than 7.0 as being acid BUT all of that is based on the ion product for H2O being 1E-14. :-)
To determine the molar H2O and OH- concentrations in pure water at 37 degrees Celsius, we first need to understand that the ion-product of water (Kw) is equal to the product of the concentrations of H+ ions and OH- ions in water.
The ion-product (Kw) of water at any temperature can be calculated using the equation:
Kw = [H+][OH-]
Given that the ion-product (Kw) at 37 degrees Celsius is 2.51 * 10^(-14), we can assign this value to Kw:
2.51 * 10^(-14) = [H+][OH-]
Since pure water is neutral, the concentration of H+ ions is equal to the concentration of OH- ions. Therefore, let's assume that the concentration of H+ ions is x. This means that the concentration of OH- ions is also x.
Now we can substitute these values into the equation:
2.51 * 10^(-14) = x * x
To solve for x, we take the square root of both sides of the equation:
√(2.51 * 10^(-14)) = √(x^2)
Simplifying, we get:
1.585 * 10^(-7) = x
So, the concentration of both H2O and OH- ions in pure water at 37 degrees Celsius is 1.585 * 10^(-7) Molar.