What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid (HCOOH) requires 29.80 mL of 0.3567 mol L^-1 NaOH(aq) ? K_a =1.8 × 10^-4
For formic acid.

Question

What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid (HCOOH) requires 29.80 mL of 0.3567 mol L^-1 NaOH(aq) ? K_a =1.8 × 10^-4

For formic acid.

Quizplus · Accepted Answer

At the equivalence point, moles of NaOH = moles of HCOOH
Moles of NaOH = 0.0298 L x 0.3567 mol/L = 0.01061 mol
0.01061 mol of HCOOH was present in 25 mL, so the initial concentration was 0.01061 mol/0.025 L = 0.4244 M
Using the equation for the dissociation of formic acid, we can set up the Ka expression:
Ka = [H+][HCOO-]/[HCOOH] = 1.8x10^-4
Since we are at the equivalence point, [HCOOH] = [HCOO-], so
Ka = [H+]^2/[HCOOH] = 1.8x10^-4
[H+] = sqrt(1.8x10^-4 x 0.4244) = 0.00821 M
pH = -log[H+] = 8.52

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