Exhibit 15-2 The following question(s) pertain to the sequence of events below:
Part A: A solution initially only containing 500.0 ml of 1.00 M Formic Acid.
Part B: 800.0 ml of 0.500 M NaOH is added to the solution from Part A.
Part C: An additional 250.0 ml of the 0.500 M NaOH is added to the resulting solution from Part B.
Refer to Exhibit 15-2, Part B. What is the resulting pH?

Question

Quizplus · Accepted Answer

The correct answer is B because the pH of the resulting solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the dissociation constant of the acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid. In this case, the pKa of formic acid is 3.75, and the concentration of the conjugate base (formate ion) is 0.4 M (since 800.0 ml of 0.500 M NaOH was added to 500.0 ml of 1.00 M formic acid, resulting in a total volume of 1300.0 ml and a concentration of 0.4 M formate ion). The concentration of the acid (formic acid) is 0.1 M (since 500.0 ml of 1.00 M formic acid was added to 800.0 ml of 0.500 M NaOH, resulting in a total volume of 1300.0 ml and a concentration of 0.1 M formic acid). Plugging these values into the Henderson-Hasselbalch equation, we get: pH = 3.75 + log(0.4/0.1) = 4.35.

Exhibit 15-2 the Following Question(s) Pertain to the Sequence of Events