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.
To calculate the pH of water at 46⁰C, we need to use the equation for the ionization of water:
H2O ⇌ H+ + OH-
The equilibrium constant for this reaction is known as the ion product of water (Kw), which varies with temperature. Given that Kw = 1.219 x 10-14 at the given temperature, we can use this information to find the concentration of the H+ ions and OH- ions in water.
At equilibrium, the concentration of H+ ions is equal to the concentration of OH- ions. Let's assume that the concentration of both H+ and OH- is x (M).
Using the equation for the ion product of water, Kw = [H+][OH-], we can substitute the given value:
1.219 x 10-14 = x * x
Taking the square root of both sides:
√(1.219 x 10-14) = x
x ≈ 1.105 x 10-7 (M)
So, the concentration of H+ and OH- ions in water at equilibrium is approximately 1.105 x 10-7 M.
Now, to calculate the pH, we use the equation:
pH = -log[H+]
Taking the negative logarithm of the H+ concentration:
pH = -log(1.105 x 10-7)
By calculating this using a calculator or software, we find that the pH of water at 46⁰C with a Kw of 1.219 x 10-14 is approximately 6.96.
To calculate the pH of water at 46°C, we can use the concept of the autoionization of water and the equation for the ion product of water (Kw).
The equation for the ion product of water is Kw = [H+][OH-]
At 25°C, the value of Kw is 1.0 x 10^-14. However, at different temperatures, the value of Kw changes.
To find the pH of water at 46°C, we need to determine the concentration of H+ ions and OH- ions in water at that temperature.
Step 1: Find the concentration of H+ ions
We can assume that the concentration of OH- ions is equal to the concentration of H+ ions in pure water (as it is neutral). So, let's denote the concentration of H+ ions as x (M).
The concentration of H+ ions multiplied by the concentration of OH- ions should be equal to Kw:
x * x = 1.219 x 10^-14
Step 2: Solve for x
To find the value of x, we take the square root of both sides of the equation:
x = √(1.219 x 10^-14)
x ≈ 1.105 x 10^-7 M
Step 3: Calculate the pH
The pH is the negative logarithm of the H+ ion concentration. So,
pH = -log[H+]
pH ≈ -log(1.105 x 10^-7)
pH ≈ -(-6.957)
pH ≈ 6.957
Therefore, the pH of water at 46°C, with a Kw of 1.219 x 10^-14, is approximately 6.957.