The Kw of water varies with temperature. Calculate the pH of water at 46℃ with a Kw=1.219 x 10 to the negative 14th.

Help please! I have a test next week!

(H^+)(OH^-) = Kw.

You know Kw. You know from the ionization of H2O that (H^+) = (OH^-) from H2O ==> H^+ + OH^-
Solve for (H^+) and convert to pH. I suspect you've already done this and didn't get pH = 7 so you think something is wrong. It isn't. The pH will not be 7. Why? because Kw is not 1E-14.

To calculate the pH of water at 46℃ using the given value of Kw (1.219 x 10^-14), we need to use the formula for the autoionization of water:

Kw = [H+][OH-]

At 25℃ (room temperature), the value of Kw is 1.0 x 10^-14. However, as the temperature increases, the value of Kw also increases.

To determine the concentration of [H+] (which is in turn used to calculate the pH), we'll first need to find the concentration of [OH-] ions using the Kw equation. Since water is neutral, the concentration of [H+] ions is equal to the concentration of [OH-] ions.

At 46℃, we have:

Kw = [H+][OH-]
1.219 x 10^-14 = [H+][H+]
[H+]^2 = 1.219 x 10^-14
[H+] ≈ √(1.219 x 10^-14)

Let's do the calculation:

[H+] ≈ √(1.219 x 10^-14)
[H+] ≈ 1.105 x 10^-7

Now that we have the concentration of [H+] ions, we can calculate the pH. The pH is defined as the negative logarithm (base 10) of the concentration of [H+] ions:

pH = -log[H+]
pH = -log(1.105 x 10^-7)
pH ≈ -(-6.96)
pH ≈ 6.96

Therefore, the approximate pH of water at 46℃ with a Kw of 1.219 x 10^-14 is around 6.96.