Final answer:
After adding 300.0 mL of 0.10 M KOH to 100.0 mL of 0.20 M HF, we find there's an excess of 0.010 moles KOH, which makes the solution's pH approximately 12.40, none of the provided choices match this calculated pH.
Step-by-step explanation:
To determine the pH of the solution after the addition of 300.0 mL of 0.10 M KOH to a 100.0 mL sample of 0.20 M HF, we need to first find out how many moles of KOH and HF we have. Since molarity (M) is moles per liter (mol/L), we calculate:
- Moles of KOH = 0.10 mol/L × 0.300 L = 0.030 mol
- Moles of HF = 0.20 mol/L × 0.100 L = 0.020 mol
During the titration of HF with KOH, a stoichiometric reaction occurs, where one mole of HF reacts with one mole of KOH. Since there is more KOH than HF, all the HF will react and KOH will be left in excess. The excess moles of KOH are 0.030 mol - 0.020 mol = 0.010 mol.
Since the reaction now forms a solution with different ions present, we need to calculate the concentration of KOH in the total volume of the solution (100.0 mL of HF + 300.0 mL of KOH = 400.0 mL or 0.400 L).
Excess KOH concentration = 0.010 mol / 0.400 L = 0.025 M
This excess OH- will determine the pH of the solution. The pOH is calculated using the formula pOH = -log[OH-]. Therefore:
- pOH = -log[0.025] = 1.60
- pH = 14 - pOH = 14 - 1.60 = 12.40
The correct pH of the solution, after accounting for significant figures, will be approximately closer to 12.40, which isn't an option from the provided choices. Therefore, based on the information given and calculations performed, none of the answer choices (a-d) are correct for the pH after the addition of 300.0 mL of KOH.