Answer:
Step-by-step explanation:
To determine the pH of pure water at 40.0°C, we need to use the equation for the ionization constant of water (Kw) and the relationship between pH and the concentration of hydrogen ions (H+).
The equation for Kw is:
Kw = [H+][OH-]
In pure water, the concentration of hydrogen ions (H+) and hydroxide ions (OH-) are equal, so we can rewrite the equation as:
Kw = [H+][H+]
Taking the square root of both sides of the equation, we have:
√(Kw) = [H+]
Given that Kw at 40.0°C is 2.92 x 10^(-14), we can substitute this value into the equation:
[H+] = √(2.92 x 10^(-14))
Calculating the square root, we find:
[H+] ≈ 1.71 x 10^(-7)
To find the pH, we use the formula:
pH = -log[H+]
Substituting the value of [H+], we have:
pH = -log(1.71 x 10^(-7))
pH ≈ 6.767
Therefore, the pH of pure water at 40.0°C is approximately 6.767.
The correct answer is B) 6.767.