1. Understand the Reaction
Benzoic acid (C₆H₅COOH) is a weak acid, and NaOH is a strong base. The reaction at the equivalence point forms sodium benzoate (C₆H₅COONa), the salt of the weak acid:
C₆H₅COOH (aq) + NaOH (aq) → C₆H₅COONa (aq) + H₂O (l)
2. Calculate the Moles of Benzoic Acid
* Moles of benzoic acid = (volume in L) × (molarity)
* Moles of benzoic acid = (0.050 L) × (0.43 mol/L) = 0.0215 mol
3. Determine the Moles of NaOH at Equivalence
At the equivalence point, the moles of NaOH will be equal to the moles of benzoic acid:
* Moles of NaOH = 0.0215 mol
4. Calculate the Volume of NaOH Used
* Volume of NaOH = (moles of NaOH) / (molarity of NaOH)
* Volume of NaOH = (0.0215 mol) / (0.50 mol/L) = 0.043 L = 43.0 mL
5. Determine the Concentration of the Sodium Benzoate
* The total volume at the equivalence point is 50.0 mL (benzoic acid) + 43.0 mL (NaOH) = 93.0 mL
* Concentration of sodium benzoate = (moles of sodium benzoate) / (total volume in L)
* Concentration of sodium benzoate = (0.0215 mol) / (0.093 L) = 0.231 mol/L
6. Calculate the pH of the Sodium Benzoate Solution
Sodium benzoate is the salt of a weak acid (benzoic acid) and a strong base (NaOH). It will undergo hydrolysis, producing hydroxide ions (OH⁻):
C₆H₅COO⁻ (aq) + H₂O (l) ⇌ C₆H₅COOH (aq) + OH⁻ (aq)
We can use the following steps to calculate the pH:
* Calculate the Kb:
* Kb = Kw / Ka (where Kw = 1.0 × 10⁻¹⁴)
* Kb = (1.0 × 10⁻¹⁴) / (6.5 × 10⁻⁵) = 1.54 × 10⁻¹⁰
* Set up an ICE table:
| | C₆H₅COO⁻ | H₂O | C₆H₅COOH | OH⁻ |
|-----|-----------|-------|-----------|-------|
| I | 0.231 | | 0 | 0 |
| C | -x | | +x | +x |
| E | 0.231-x | | x | x |
* Use the Kb expression:
* Kb = [C₆H₅COOH][OH⁻] / [C₆H₅COO⁻]
* 1.54 × 10⁻¹⁰ = (x)(x) / (0.231 - x)
* Since Kb is very small, we can approximate (0.231 - x) ≈ 0.231
* 1.54 × 10⁻¹⁰ = x² / 0.231
* x² = 3.55 × 10⁻¹¹
* x = [OH⁻] = 5.96 × 10⁻⁶ M
* Calculate pOH:
* pOH = -log[OH⁻] = -log(5.96 × 10⁻⁶) = 5.23
* Calculate pH:
* pH + pOH = 14
* pH = 14 - 5.23 = 8.77
Therefore, the pH at the equivalence point of the titration is approximately 8.77.
