A 100.0 mL sample of 0.20 mol L^-1 HF is titrated with 0.10 mol L^-1 KOH. Determine the pH of the solution after the addition of 300.0 mL of KOH. The K_a of HF is 3.5 × 10^-4. A) 12.40 B) 9.33 C) 5.06 D) 8.94 E) 12.00

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After adding 300.0 mL of 0.10 mol L^-1 KOH to 100.0 mL of 0.20 mol L^-1 HF, the solution is in excess of KOH. The moles of HF initially are 0.02 mol, and the moles of KOH added are 0.03 mol. The excess moles of KOH are 0.01 mol in a total volume of 400.0 mL, resulting in a concentration of 0.025 mol L^-1 KOH. The pH is determined by the concentration of OH^- ions, which is 0.025 mol L^-1, giving a pOH of 1.60 and a pH of 12.40.

A 1000 ML Sample of 0