Calculate the pH of a 0.050 M sodium benzoate solution
Sodium benzoate is the salt of a weak acid, benzoic acid, and a strong base, sodium hydroxide. That means the benzoate ion behaves as a weak base in water and raises the pH above 7. Use the calculator below to compute the exact pH using either pKa or Ka, then visualize how pH changes with concentration.
Interactive calculator
Expert guide: how to calculate the pH of a 0.050 M sodium benzoate solution
Calculating the pH of a sodium benzoate solution is a classic weak acid and weak base equilibrium problem. Even though sodium benzoate is a salt, the chemistry is not neutral. The sodium ion is effectively a spectator ion in water, but the benzoate ion, C6H5COO–, is the conjugate base of benzoic acid. Because benzoic acid is a weak acid, its conjugate base has measurable basicity and reacts with water to produce hydroxide ions. That hydroxide formation is what makes the solution basic.
In many general chemistry and analytical chemistry problems, the wording may say 0.050 m or 0.050 M. Strictly speaking, lowercase m refers to molality, while uppercase M refers to molarity. In dilute aqueous solutions such as this one, the numerical difference is often small enough that textbook exercises treat them similarly unless precision density data are supplied. For a standard classroom pH calculation, the accepted setup is to treat the concentration as 0.050 M sodium benzoate in water at 25 C.
Why sodium benzoate makes water basic
Sodium benzoate dissociates almost completely in water:
NaC7H5O2 → Na+ + C7H5O2–
The benzoate ion then hydrolyzes:
C7H5O2– + H2O ⇌ HC7H5O2 + OH–
This equilibrium produces hydroxide, so the pH rises above 7. The magnitude of the rise depends on three main factors:
- The concentration of sodium benzoate
- The acid dissociation constant of benzoic acid, usually reported as Ka or pKa
- The water ion-product constant, Kw, which changes with temperature
Core formula set
To solve the problem, you first convert the acid constant of benzoic acid into the base constant for benzoate:
- Find Ka from pKa using Ka = 10-pKa
- Calculate Kb from Kb = Kw / Ka
- Use the weak base equilibrium expression Kb = x2 / (C – x)
- Solve for x, where x = [OH–]
- Find pOH = -log[OH–]
- Find pH = 14.00 – pOH at 25 C
If benzoic acid has pKa = 4.20, then:
Ka = 10-4.20 = 6.31 x 10-5
Kb = 1.00 x 10-14 / 6.31 x 10-5 = 1.58 x 10-10
Step by step calculation for 0.050 M sodium benzoate
Let the initial concentration of benzoate be 0.050 M. Set up the ICE table for the hydrolysis reaction:
- Initial: [C7H5O2–] = 0.050, [HC7H5O2] = 0, [OH–] = 0
- Change: -x, +x, +x
- Equilibrium: 0.050 – x, x, x
Plug into the equilibrium expression:
Kb = x2 / (0.050 – x)
Since Kb is very small, many classes use the approximation 0.050 – x ≈ 0.050. Then:
x = √(KbC) = √[(1.58 x 10-10)(0.050)] = 2.81 x 10-6 M
Therefore:
pOH = -log(2.81 x 10-6) = 5.55
pH = 14.00 – 5.55 = 8.45
The more exact quadratic solution gives essentially the same answer to the displayed precision, so the expected pH is approximately 8.45 at 25 C.
What the result means chemically
A pH around 8.45 tells you the benzoate ion is only a weak base, not a strong one. The solution is basic, but not dramatically so. This is exactly what you would predict for the conjugate base of a weak acid with pKa near 4.20. Compare that with salts of much weaker acids, which can give much more basic solutions, or salts of stronger acids, whose conjugate bases are too weak to alter pH meaningfully.
Sodium benzoate is also a good example of why salt hydrolysis matters in real formulations. It is widely used as a preservative in foods, beverages, and pharmaceuticals. In actual products, pH control strongly affects benzoic acid and benzoate speciation, preservative performance, and stability. Although the pure chemistry exercise assumes only sodium benzoate in water, the same equilibrium logic is used in formulation science, environmental chemistry, and quality control.
Quick comparison table for similar concentrations
| Sodium benzoate concentration (M) | Estimated [OH-] at 25 C (M) | Estimated pOH | Estimated pH |
|---|---|---|---|
| 0.010 | 1.26 x 10^-6 | 5.90 | 8.10 |
| 0.050 | 2.81 x 10^-6 | 5.55 | 8.45 |
| 0.100 | 3.98 x 10^-6 | 5.40 | 8.60 |
| 0.500 | 8.90 x 10^-6 | 5.05 | 8.95 |
These values show a useful trend: increasing the concentration of a weak base increases pH, but not in a one to one linear way. Because the hydroxide concentration depends on the square root of concentration in the weak base approximation, even a tenfold concentration increase gives only a moderate pH shift.
Comparison of exact vs approximation methods
| Method | Equation used | Computed [OH-] for 0.050 M | Computed pH | Practical note |
|---|---|---|---|---|
| Square root approximation | x ≈ √(KbC) | 2.81 x 10^-6 M | 8.45 | Fast and usually accepted in coursework |
| Exact quadratic solution | x = [-Kb + √(Kb² + 4KbC)] / 2 | 2.81 x 10^-6 M | 8.45 | Best for calculators and precise reporting |
Common mistakes to avoid
- Using Ka directly to compute pH without converting to Kb first
- Assuming the salt solution is neutral just because it contains sodium
- Forgetting that sodium is a spectator ion while benzoate controls the pH
- Mixing up pKa and Ka values
- Using pH = 14 – pOH without adjusting for temperature when Kw is not the 25 C value
- Entering 6.3 x 10^-5 as 6.3 instead of 0.000063 in a calculator
How temperature influences the answer
Temperature changes the ionization of water, so neutral pH is not always exactly 7.00. At 25 C, Kw is commonly taken as 1.0 x 10^-14, which makes pH + pOH = 14.00. At higher temperatures, Kw increases, and at lower temperatures, it decreases. For many classroom problems, 25 C is assumed unless the problem says otherwise. That is why the standard answer for this sodium benzoate calculation is reported near pH 8.45. If you choose another temperature, the difference is usually modest but still measurable.
When to use Henderson-Hasselbalch and when not to
Students often ask whether the Henderson-Hasselbalch equation can be used here. The answer is generally no for a pure sodium benzoate solution, because Henderson-Hasselbalch is designed for buffer systems containing appreciable amounts of both the weak acid and its conjugate base. In a solution that begins with only sodium benzoate, you should use salt hydrolysis and the weak base equilibrium expression. If benzoic acid were also present, then the system would become a benzoic acid benzoate buffer and Henderson-Hasselbalch would be the natural shortcut.
Real world relevance of benzoate chemistry
Benzoate chemistry is not just an exam topic. It appears in food preservation, cosmetic formulations, pharmaceutical liquids, and water quality discussions. The effectiveness of benzoate preservatives depends strongly on pH because the undissociated acid and the benzoate ion differ in antimicrobial behavior. Understanding equilibrium relationships helps chemists predict solubility, speciation, and product performance. It also explains why labels and technical sheets often specify ideal pH ranges for products containing benzoate salts.
Authoritative references for pH and equilibrium context
- USGS: pH and water overview
- NIST: chemical thermodynamics resources
- EPA: alkalinity and acid-base conditions in water
Final takeaway
To calculate the pH of a 0.050 M sodium benzoate solution, treat benzoate as a weak base, convert the benzoic acid constant into Kb, solve for hydroxide concentration, and then convert to pOH and pH. Using pKa = 4.20 for benzoic acid at 25 C gives an answer of about pH 8.45. That result is chemically reasonable, mathematically consistent, and directly tied to the hydrolysis of the benzoate ion in water.