The Kw of water varies with temperature. Calculate the pH of water at 46⁰C with a Kw = 1.219 x 10-14. Show all calculations leading to an answer.
how do I Substitute this ?
I thought I did one of these for you a couple of days ago.
...........H2O ==> H^+ + OH^-
I.........liquid....0.....0
C.........liquid....x.....x
E.........liquid....x.....x
Substitute the E line into the Kw expression and solve for x = H^+, then convert to pH.
I don't quite understand the last step! Can you help me? I have a test next week!. Thanks!!
To calculate the pH of water at 46⁰C, we need to use the equation for the ion-product constant of water, Kw. The value of Kw changes with temperature, so be sure to use the correct value for the given temperature.
Given Kw = 1.219 x 10-14 at 46⁰C.
Kw = [H+][OH-]
Since water is neutral, the concentration of H+ ions is equal to the concentration of OH- ions.
Let's represent the concentration of H+ ions as x moles/L.
[H+] = x
[OH-] = x
Using the value of Kw, we can set up the following equation:
1.219 x 10-14 = x * x
To solve for x, take the square root of both sides:
√(1.219 x 10-14) = √(x * x)
1.102 x 10-7 = x
Now, we have the concentration of H+ ions.
To calculate the pH, we can use the equation:
pH = -log[H+]
pH = -log(1.102 x 10-7)
Using a calculator, we find:
pH ≈ 6.96
Therefore, the pH of water at 46⁰C, with a Kw = 1.219 x 10-14, is approximately 6.96.