What is the pH of 6.5 × 10^-5 M KOH(aq)at 25°C?

Question

Quizplus · Accepted Answer

The pH of a solution can be calculated using the formula: pH = -log[H+], where [H+] is the hydrogen ion concentration.
For a strong base like KOH, the hydroxide ion concentration can be calculated using the formula [OH-] = Molarity of KOH.
Therefore, [OH-] = 6.5 × 10^-5 M
Using the Kw expression at 25°C:
Kw = [H+][OH-] = 1.0 × 10^-14
[H+] = Kw/[OH-] = 1.0 × 10^-14/6.5 × 10^-5 = 1.54 × 10^-10
pH = -log[H+] = -log(1.54 × 10^-10) = 9.81
Thus, the pH of 6.5 × 10^-5 M KOH(aq) at 25°C is 9.81.

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