The balanced chemical equation for the reaction between benzoic acid and potassium hydroxide is:
HC7H5O2(aq) + KOH(aq) → KC7H5O2(aq) + H2O(l)
The number of moles of benzoic acid used in the titration is:
n(HC7H5O2) = M × V = 0.8748 mol/L × 150.0 mL / 1000 mL/L = 0.13122 mol
The volume of KOH solution required to reach equivalence point can be determined using the stoichiometry of the balanced equation. The stoichiometric ratio of benzoic acid to potassium hydroxide is 1:1. Therefore, the number of moles of KOH required to neutralize all of the benzoic acid is equal to n(HC7H5O2):
n(KOH) = 0.13122 mol
The volume of KOH solution required can be calculated as:
V(KOH) = n(KOH) / M(KOH) = 0.13122 mol / 0.3544 mol/L = 0.3701 L
At equivalence, the total volume of the solution is:
V(total) = V(HC7H5O2) + V(KOH) = 0.1500 L + 0.3701 L = 0.5201 L
The concentration of the salt formed, KC7H5O2, is:
M(KC7H5O2) = n(KC7H5O2) / V(total) = n(HC7H5O2) / V(total) = 0.13122 mol / 0.5201 L = 0.2523 M
The pH at equivalence can now be calculated using the acid dissociation constant (Ka) of benzoic acid:
Ka = [H+][C7H5O2-] / [HC7H5O2]
pKa = -log(Ka) = -log(6.3 × 10^-5) = 4.20
At equivalence, the concentration of benzoic acid and benzoate ions are equal, so [HC7H5O2] = [C7H5O2-]. Let x be the concentration of H+ ions at equivalence. Then:
Ka = x^2 / (0.2523 - x)
Solving this equation for x gives:
x = sqrt(Ka × (0.2523 - x)) = sqrt(6.3 × 10^-5 × 0.2523) = 0.002532
Therefore, the pH at equivalence is:
pH = -log[H+] = -log(0.002532) = 2.60
Rounding to two decimal places, the pH at equivalence is 2.60.