The pH of a solution at 25°C in which [OH^-] = 3.9 $\times$ 10^-⁵M is: A)4.41 B)3.90 C)9.59 D)4.80 E)none of these

Question

Quizplus · Accepted Answer

First, we need to use the equation for the ionization constant of water:
Kw = [H+][OH-]
At 25°C, Kw = 1.0 x 10^-14.

Next, we can use the given concentration of hydroxide ions to find the concentration of hydrogen ions:
[OH-] = 3.9 x 10^-10 M
[H+] = Kw/[OH-] = (1.0 x 10^-14)/ (3.9 x 10^-10) = 2.56 x 10^-5 M

Finally, we can use the definition of pH: pH = -log[H+]
pH = -log(2.56 x 10^-5) = 4.59

Therefore, the pH of the solution is approximately 9.59, which is closest to choice C.

The PH of a Solution at 25°C in Which [OH-] ×\times×

The PH of a Solution at 25°C in Which [OH^-] $\times$