4. Calculate what the pH and the pOH would be if you dissolved 38.65 g of Potassium Hydroxide in
water to prepare 750. mL of solution.



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.