Question

Chemistry
A solution was prepared by dissolving 0.62 g of KOH(s) to form 500 mL of solution.

The pH of the potassium hydroxide solution is

Answer 1

are you sure?

Answer 2

The pH of a solution of KOH can be calculated using the formula:

pH = -log10([OH-])

First, we need to find the concentration of hydroxide ions ([OH-]) in the solution. We can do this by first finding the number of moles of KOH and then dividing by the volume of the solution in liters.

Molar mass of KOH = 39.10 g/mol + 16.00 g/mol + 1.01 g/mol = 56.11 g/mol

Number of moles of KOH = mass of KOH / molar mass of KOH = 0.62 g / 56.11 g/mol = 0.011 mol

Volume of the solution in liters = 500 mL / 1000 mL/L = 0.5 L

Concentration of hydroxide ions = number of moles of KOH / volume of the solution = 0.011 mol / 0.5 L = 0.022 M

Now, we can calculate the pH using the formula:

pH = -log10(0.022) ≈ 1.66

Therefore, the pH of the potassium hydroxide solution is approximately 1.66.

Answer 3

I apologize, there was a mistake in my previous response. Since KOH is a strong base, it dissociates completely in water, producing an equal concentration of hydroxide ions (OH-). Therefore, the concentration of hydroxide ions in the solution is equal to the concentration of KOH.

Concentration of KOH = 0.62 g / 56.11 g/mol / 0.5 L = 0.022 M

pH = -log10(0.022) ≈ 1.66

Therefore, the correct pH of the potassium hydroxide solution is approximately 1.66.