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 400.0 mL of KOH. The K_a of HF is 3.5 × 10^-4. A) 13.08 B) 12.60 C) 13.85 D) 12.30 E) 12.78

Question

Quizplus · Accepted Answer

Step 1: Calculate the amount of moles of HF initially present in the solution:

0.20 mol/L x 0.100 L = 0.020 mol HF

Step 2: Calculate the amount of moles of KOH added to the solution:

0.10 mol/L x 0.400 L = 0.040 mol KOH

Step 3: Determine which reactant is limiting:

HF + KOH → KF + H2O

The stoichiometry of the reaction is 1:1, therefore KOH is the limiting reactant because it is added in more excess than HF.

Step 4: Calculate the amount of moles of HF that reacted with the KOH:

0.020 mol HF - 0.040 mol KOH = -0.020 mol HF (negative sign indicates that HF is completely consumed)

Step 5: Calculate the concentration of HF after the addition of KOH:

100.0 mL + 400.0 mL = 0.500 L

[HF] = (0.020 mol / 0.500 L) = 0.040 mol/L

Step 6: Calculate the pKa of HF using the given Ka value:

Ka = 3.5 × 10^-4
pKa = -log(Ka) = -log(3.5 × 10^-4) = 3.46

Step 7: Calculate the pH of the solution using the Henderson-Hasselbalch equation:

pH = pKa + log([A-] / [HA])

Since KOH is a strong base and fully dissociates, the solution becomes basic and A- is equal to the concentration of potassium ion:

[A-] = [K+] = (0.040 mol KOH / 0.500 L) = 0.080 mol/L

[HA] = [HF] = 0.040 mol/L

pH = 3.46 + log(0.080/0.04) = 12.60

Therefore, the pH of the solution after the addition of 400.0 mL of KOH is 12.60. The best choice is B.

A 1000 ML Sample of 0