The Kw of water varies with temperature. Calculate the pH of water at 46⁰C with a Kw = 1.219 x 10-14.
........H2O ==> H^+ + OH^-
I.......liquid..0......0
C.......liquid..x......x
E.......liquid..x......x
Substitute the E line into Kw for H2O and solve for x = (H^+) then convert to pH.
I don't have any idea how to substitute the E line and then how to convert to ph!! Can you walk me through it.
To calculate the pH of water at a specific temperature, we need to use the ion-product constant of water, Kw, and the relationship between Kw and the concentrations of the hydrogen ion (H+) and hydroxide ion (OH-) in water.
Kw represents the product of the concentration of H+ ions and OH- ions in water at a given temperature. Mathematically, Kw = [H+][OH-].
To calculate the concentration of H+ ions, we need to assume that at 46°C, the concentration of H+ ions and OH- ions are equal because water is neutral. Therefore, [H+] = [OH-].
Given that Kw = 1.219 x 10^-14, we can substitute the [H+] concentration with x to get the following equation:
(1.219 x 10^-14) = x * x
Using basic algebra, we can determine that the concentration of both H+ and OH- ions is the square root of Kw:
x = √(1.219 x 10^-14)
x ≈ 1.104 x 10^-7
This represents the concentration of both H+ and OH- ions at 46°C.
Now, to calculate the pH, we need to take the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log[H+]
pH = -log(1.104 x 10^-7)
pH ≈ 6.96
Therefore, the pH of water at 46°C with a Kw of 1.219 x 10^-14 is approximately 6.96.