A 25.0 mL sample of 0.10 M sodium benzoate is titrated with 0.10 M HCl(aq).What is the pH after the addition of 32.0 mL of HCl(aq)? (K_b of C₆H₅CO₂^- = 1.6 × 10^-10)

Question

Quizplus · Accepted Answer

1. Write the balanced equation for the reaction:
C6H5COO-Na+ (sodium benzoate) + HCl (hydrochloric acid)  →  C6H5COOH (benzoic acid) + NaCl (sodium chloride)

2. Write the equation for the dissociation of sodium benzoate:
C6H5COO-Na+ (sodium benzoate)  ↔  C6H5COO- (benzoate ion)  + Na+ (sodium ion)

3. Calculate the moles of sodium benzoate in the original solution:
moles of sodium benzoate = (0.10 M) x (25.0 mL/1000 mL) = 0.0025 mol

4. Since the initial solution of sodium benzoate is a salt and does not contain any significant amount of free H+ or OH- ions, we can assume that the initial pH of the solution is determined by the autoionization of water:
2H2O ↔ H3O+ + OH-
Kw = [H3O+][OH-] = 1.0 x 10^-14 mol^2/L^2
pKw = -logKw = 14.0
pH = 1/2(pKw - log[H3O+]) = 1/2(14.0 - 0) = 7.00

5. Calculate the moles of HCl added to the solution:
moles of HCl = (0.10 M) x (32.0 mL/1000 mL) = 0.0032 mol

6. Since the number of moles of HCl added is greater than the number of moles of sodium benzoate, the H+ ions from HCl will fully react with the benzoate ions to form benzoic acid:
moles of benzoate ion reacted = 0.0025 mol
moles of H+ added = 0.0032 mol
Since the reaction is 1:1, the excess moles of H+ will remain in the solution.

7. Calculate the final concentration of H+ ions in the solution:
moles of H+ remaining = 0.0032 mol - 0.0025 mol = 0.0007 mol
final volume of solution = 25.0 mL + 32.0 mL = 57.0 mL
final concentration of H+ ions = 0.0007 mol/0.057 L = 0.0123 M

8. Calculate the final pH of the solution:
pH = -log[H+] = -log(0.0123) = 1.91

Therefore, the correct answer is (C) 1.91.

A 250 ML Sample of 0