Answer :
To calculate the pH and pOH of a solution containing 38.65 g of Potassium Hydroxide (KOH) dissolved in 750 mL of water, we first need to determine the moles of KOH present in the solution.
1. Calculate the molar mass of KOH:
- The molar mass of KOH is approximately 56.11 g/mol (potassium = 39.10 g/mol, oxygen = 16.00 g/mol, hydrogen = 1.01 g/mol).
2. Determine the number of moles of KOH:
- Divide the mass of KOH by its molar mass:
38.65 g / 56.11 g/mol ≈ 0.688 moles of KOH
3. Calculate the concentration of KOH in the solution:
- Convert the volume of the solution to liters:
750 mL = 0.75 L
- Calculate the molarity (M) of KOH:
Molarity = moles of solute / liters of solution
Molarity = 0.688 moles / 0.75 L ≈ 0.917 M
4. Determine the OH- concentration:
- Since KOH dissociates completely in water, the concentration of hydroxide ions (OH-) is equal to the molarity of KOH:
[OH-] = 0.917 M
5. Calculate the pOH:
- pOH = -log[OH-]
- pOH = -log(0.917)
- pOH ≈ 0.037
6. Calculate the pH:
- pH + pOH = 14 (at 25°C)
- pH = 14 - pOH
- pH = 14 - 0.037 ≈ 13.963
Therefore, the pH of the solution prepared by dissolving 38.65 g of Potassium Hydroxide in 750 mL of water is approximately 13.963, and the pOH is approximately 0.037.
