What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid requires 29.80 mL of 0.0567 M NaOH? K_a = 1.8 × 10^-4for formic acid. A)2.46 B)5.88 C)8.12 D)11.54

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Quizplus · Accepted Answer

At equivalence point, moles of NaOH = moles of formic acid initially present.
moles of formic acid = (25 mL / 1000 mL) x (0.1 M) = 0.0025 mol
moles of NaOH = (29.80 mL / 1000 mL) x (0.0567 M) = 0.00169 mol
As NaOH is a strong base, it will dissociate completely in water, so the concentration of OH- ions will be 0.00169 mol / 0.0548 L = 0.0309 M.
Now, we can use the expression for Kb of formate ion to find the concentration of H+ ions at equivalence point:
Kb = [HCOO-][OH-] / [HCOOH]
1.8 x 10^-4 = x^2 / (0.0025 - x)
Assuming x is small compared to 0.0025, we can simplify:
1.8 x 10^-4 = x^2 / 0.0025
x = 0.00824 M
pH = -log[H+] = -log(0.00824) = 2.085
However, this is the pH before any excess NaOH is added. In this case, NaOH is a strong base, so it will react completely with any remaining formic acid to form its conjugate base (and water). Therefore, the pH at equivalence point will be slightly greater than 7 (neutral), but closer to the basic side since we have more OH- ions than H+ ions. The best choice is C, 8.12, as it is closest to a basic pH.

What Is the Approximate PH at the Equivalence Point of a Weak