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 0.3567 M NaOH? 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 requires 29.80 mL of 0.3567 M NaOH? K_a =1.8 × 10^-4

For formic acid.

Quizplus · Accepted Answer

First, we need to calculate the amount of moles of NaOH used:
0.3567 M NaOH x 0.02980 L NaOH = 0.01062366 mol NaOH
As the stoichiometric ratio between formic acid (HA) and NaOH (OH-) for complete neutralization is 1:1, we can say that there are 0.01062366 moles of formic acid in the initial solution. 
Therefore, the initial concentration of formic acid (HA) is:
0.01062366 mol / 0.025 L (volume of formic acid used) = 0.4249464 M
Now, we can use the equation for the dissociation of formic acid:
HA + H2O ↔ H3O+ + A-
Ka = [H3O+] [A-] / [HA]
At the equivalence point, we have added enough NaOH to completely convert all HA to A-. Therefore, the concentration of A- is equal to the initial concentration of HA, which we already calculated to be 0.4249464 M. Since the initial concentration of HA and A- are equal, we can assume that their concentrations at equilibrium are also equal. Thus, we can simplify the Ka equation to:
Ka = [H3O+]^2 / [HA]
Rearranging:
[H3O+] = sqrt(Ka x [HA]) = sqrt(1.8e-4 x 0.4249464) = 0.008523
Taking the negative log of [H3O+], we get the pH:
pH = -log(0.008523) = 8.52
Therefore, the correct answer is C) 8.52.

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