Sodium benzoate
is a product of sodium hydroxide and benzoic acid
(
). What is the pH of 0.12 M Sodium benzoate? Please enter 3 significant figures (Sample problem 18.12 and follow-up problems)
The Correct Answer and Explanation is:
Answer: 8.64
Explanation:
Sodium benzoate (C₆H₅COONa) is the salt formed from a strong base (sodium hydroxide, NaOH) and a weak acid (benzoic acid, C₆H₅COOH). When dissolved in water, it dissociates completely into its constituent ions:
C₆H₅COONa(aq) → Na⁺(aq) + C₆H₅COO⁻(aq)
The sodium ion (Na⁺) is the conjugate acid of a strong base and acts as a spectator ion, meaning it does not react with water or affect the pH. The benzoate ion (C₆H₅COO⁻), however, is the conjugate base of a weak acid. It will react with water in a hydrolysis reaction, accepting a proton from water to form benzoic acid and hydroxide ions (OH⁻).
C₆H₅COO⁻(aq) + H₂O(l) ⇌ C₆H₅COOH(aq) + OH⁻(aq)
The production of OH⁻ ions makes the solution basic. The equilibrium for this reaction is described by the base dissociation constant, Kₑ. The value of Kₑ can be1.59 × 10⁻¹⁰
3. Equilibrium Calculation
We can now set up an ICE (Initial, Change, Equilibrium) table to determine the equilibrium concentration of OH⁻. Let ‘x’ be the concentration of OH⁻ produced.
| C₆H₅COO⁻ | C₆H₅COOH | OH⁻ | |
| Initial (M) | 0.12 | 0 | 0 |
| Change (M) | -x | +x | +x |
| Equilibrium (M) | 0.12 – x | x | x |
The Kₑ expression is:
Kₑ = [C₆H₅COOH][OH⁻] / [C₆H₅COO⁻]
1.59 × 10⁻¹⁰ = (x)(x) / (0.12 – x)
Since Kₑ is very small, we can assume that x is negligible compared to 0.12 M (i.e., 0.12 – x ≈ 0.12).
1.59 × 10⁻¹⁰ ≈ x² / 0.12
x² = (1.59 × 10⁻¹⁰)(0.12) = 1.908 × 10⁻¹¹
x = √(1.908 × 10⁻¹¹) = 4.37 × 10⁻⁶ M
So, [OH⁻] = 4.37 × 10⁻⁶ M.
4. Calculate pOH and pH
Now we can calculate the pOH and then the pH:
pOH = -log[OH⁻] = -log(4.37 × 10⁻⁶) = 5.36
pH = 14.00 – pOH = 14.00 – 5.36 = 8.64
The final pH of the 0.12 M sodium calculated from the acid dissociation constant (Kₐ) of its conjugate acid, benzoic acid, using the ion product of water (Kₑ = 1.0 × 10⁻¹⁴).
Kₑ = Kₑ / Kₐ = (1.0 × 10⁻¹⁴) / (6.3 × 10⁻⁵) = 1.59 × 10⁻¹⁰
To find the hydroxide ion concentration, [OH⁻], an ICE table is used, where ‘x’ represents the change in concentration at equilibrium.
| C₆H₅COO⁻ | C₆H₅COOH | OH⁻ | |
| Initial | 0.12 M | 0 | ~0 |
| Change | -x | +x | +x |
| Equilibrium | 0.12 – x | x | x |
The Kₑ expression is:
Kₑ = [C₆H₅COOH][OH⁻] / [C₆H₅COO⁻] = (x)(x) / (0.12 – x)
Since Kₑ is benzoate solution, rounded to three significant figures, is 8.64. very small, we can assume that x is negligible compared to the initial concentration of 0.12 M. Thus, (0.12 – x) ≈ 0.12.
1.59 × 10⁻¹⁰ = x² / 0.12
x² = (1.59 × 10⁻¹⁰)(0.12) = 1.908 × 10⁻¹¹
x = √(1.908 × 10⁻¹¹) = 4.37 × 10⁻⁶ M
So, [OH⁻] = 4.37 × 10⁻⁶ M.
Next, the pOH is calculated:
pOH = -log[OH⁻] = -log(4.37 × 10⁻⁶) = 5.36
Finally, the pH is found using the relationship pH + pOH = 14.00:
pH = 14.00 – pOH = 14.00 – 5.36 = 8.64
The pH of the 0.12thumb_upthumb_down
