19 Calculate the pH of a 0.36 M CH3COONa Solution
Use this premium hydrolysis calculator to find the pH of a sodium acetate solution. Enter concentration, acetic acid Ka, and water ion product to calculate hydroxide concentration, pOH, and final pH with a visual chart.
Calculator
Default example: 0.36 M sodium acetate.
Used only if Custom Ka is selected.
Default at 25°C: 1.0 × 10^-14.
For 0.36 M CH3COONa at 25°C with Ka = 1.8 × 10^-5, the pH is expected to be mildly basic.
Solution Profile
How to calculate the pH of a 0.36 M CH3COONa solution
To calculate the pH of a 0.36 M CH3COONa solution, you need to recognize that sodium acetate is a salt formed from a strong base, sodium hydroxide, and a weak acid, acetic acid. That combination is important because salts of weak acids and strong bases hydrolyze in water to produce hydroxide ions. As a result, the solution becomes basic, and its pH rises above 7.
In water, sodium acetate dissociates almost completely into sodium ions and acetate ions. The sodium ion is essentially a spectator ion in this context, but the acetate ion reacts with water according to the equilibrium:
CH3COO- + H2O ⇌ CH3COOH + OH-
This hydrolysis is controlled by the base dissociation constant of acetate, Kb. Since acetic acid is the conjugate acid of acetate, Kb is related to the acid dissociation constant Ka through the water ion product Kw:
Kb = Kw / Ka
At 25°C, a widely used value for acetic acid is Ka = 1.8 × 10^-5, and Kw = 1.0 × 10^-14. Therefore:
Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10
Now use the hydrolysis approximation for a weak base salt:
[OH-] ≈ √(Kb × C)
where C = 0.36 M. Substituting:
[OH-] ≈ √((5.56 × 10^-10)(0.36)) = √(2.00 × 10^-10) ≈ 1.41 × 10^-5 M
Then calculate pOH:
pOH = -log(1.41 × 10^-5) ≈ 4.85
Finally:
pH = 14.00 – 4.85 = 9.15
Why sodium acetate gives a basic solution
Many students initially expect a salt solution to be neutral because salts often come from acid-base neutralization. However, neutrality depends on the nature of the parent acid and base. Sodium acetate does not come from a strong acid and strong base pair. Instead, it comes from weak acetic acid and strong sodium hydroxide. That means the acetate ion retains a measurable tendency to attract a proton from water, generating OH-.
Put differently, the conjugate base of a weak acid is strong enough to react with water. Acetate is not a strong base in the everyday sense, but it is sufficiently basic to shift the equilibrium toward hydroxide formation. This is why sodium acetate solutions commonly have pH values in the mildly basic range, usually around 8.8 to 9.3 depending on concentration and temperature.
Step-by-step chemistry setup
- Write the salt dissociation: CH3COONa → CH3COO- + Na+
- Ignore Na+ for pH because it does not hydrolyze appreciably.
- Write acetate hydrolysis: CH3COO- + H2O ⇌ CH3COOH + OH-
- Calculate Kb = Kw / Ka.
- Use either the approximation x = √(KbC) or solve the quadratic exactly.
- Find pOH = -log[OH-].
- Convert to pH = 14 – pOH at 25°C.
Exact calculation versus approximation
For most classroom and exam questions, the approximation method is acceptable because Kb is small and the fraction hydrolyzed is tiny compared with the original salt concentration. In this case:
x / C = (1.41 × 10^-5) / 0.36 ≈ 3.9 × 10^-5
That is far below 5%, so the approximation is excellent. If you solve the quadratic exactly, the answer changes only negligibly. This is a good example of when a weak-equilibrium shortcut is fully justified.
| Quantity | Value used | Meaning |
|---|---|---|
| Salt concentration, C | 0.36 M | Initial acetate concentration from sodium acetate |
| Ka of acetic acid | 1.8 × 10^-5 | Acid strength of CH3COOH at 25°C |
| Kw of water | 1.0 × 10^-14 | Water autoionization constant at 25°C |
| Kb of acetate | 5.56 × 10^-10 | Base strength of CH3COO- |
| [OH-] | 1.41 × 10^-5 M | Hydroxide concentration from hydrolysis |
| pOH | 4.85 | Negative log of hydroxide concentration |
| pH | 9.15 | Final basicity of the solution |
Comparison table: how concentration changes the pH of sodium acetate
The pH of a sodium acetate solution rises as concentration increases, but the increase is not linear because pH is logarithmic. Using Ka = 1.8 × 10^-5 and Kw = 1.0 × 10^-14 at 25°C, the following calculated values illustrate the trend:
| CH3COONa concentration (M) | Estimated [OH-] (M) | pOH | pH |
|---|---|---|---|
| 0.01 | 2.36 × 10^-6 | 5.63 | 8.37 |
| 0.05 | 5.27 × 10^-6 | 5.28 | 8.72 |
| 0.10 | 7.45 × 10^-6 | 5.13 | 8.87 |
| 0.36 | 1.41 × 10^-5 | 4.85 | 9.15 |
| 0.50 | 1.67 × 10^-5 | 4.78 | 9.22 |
| 1.00 | 2.36 × 10^-5 | 4.63 | 9.37 |
Important assumptions in this pH calculation
- The solution is sufficiently dilute that activity effects are ignored and concentration is used directly.
- The temperature is 25°C, so Kw is treated as 1.0 × 10^-14.
- Sodium acetate fully dissociates in water.
- The approximation x << C is valid, which it clearly is here.
- No additional acids, bases, or buffer components are present.
If the solution were highly concentrated or the temperature changed substantially, a more advanced treatment using activities and temperature-adjusted equilibrium constants would be preferred. For most general chemistry settings, though, the standard method is entirely appropriate.
Common mistakes students make
1. Treating CH3COONa as if it were neutral
This is the most common error. Because acetate is the conjugate base of a weak acid, the solution is basic, not neutral.
2. Using Ka directly instead of Kb
The reacting species in solution is acetate, not acetic acid. So you either convert Ka to Kb or use the pKa relationship carefully. Using Ka directly in the hydrolysis expression gives the wrong direction of equilibrium.
3. Forgetting to calculate pOH first
Since the hydrolysis produces OH-, your first logarithmic result is pOH, not pH. You then convert to pH using 14.00 – pOH at 25°C.
4. Ignoring units and scientific notation
With values like 1.8 × 10^-5 and 1.0 × 10^-14, scientific notation matters. A small arithmetic slip can shift the pH noticeably.
Relation to buffer chemistry
Sodium acetate is also a major component in acetate buffer systems. If acetic acid were present along with sodium acetate, you would no longer solve the problem as a simple salt hydrolysis. Instead, you would apply the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
But in this problem, only CH3COONa is specified, so salt hydrolysis is the correct path. This distinction matters because many exam questions deliberately compare pure weak acid salts with true buffer mixtures.
Real data reference points and source-based context
Reliable equilibrium constants and water chemistry references are important when calculating pH. Authoritative educational and public sources consistently report acetic acid as a weak acid with Ka around 1.8 × 10^-5 near room temperature. This supports the standard classroom result of a mildly basic sodium acetate solution.
- LibreTexts Chemistry provides broad university-level explanations of weak acid and salt hydrolysis concepts.
- U.S. Environmental Protection Agency publishes water chemistry guidance relevant to pH and acid-base equilibria.
- National Institute of Standards and Technology offers high-quality scientific data resources and standards context.
For directly academic or government reading, these pages are especially relevant:
- EPA overview of pH in aqueous systems
- NIST Physical Measurement Laboratory
- University of Wisconsin chemistry resources
How the 0.36 M result fits into expected chemistry behavior
A pH of about 9.15 is chemically reasonable. Acetate is a weak base, so the solution should not be extremely alkaline. You would expect a pH somewhat above neutral but well below strongly basic solutions such as sodium hydroxide. That moderate basicity is exactly what the calculation gives. If your answer were below 7 or above 11, it would be a signal to check your setup.
Another useful check is the magnitude of hydroxide concentration. A value of approximately 1.41 × 10^-5 M means the solution has more OH- than pure water by over three orders of magnitude, which is enough to make the solution clearly basic while still remaining in the mild range.
Quick exam shortcut
If you are solving this under timed conditions, the fastest route is:
- Identify acetate as a conjugate base.
- Compute Kb = 10^-14 / 1.8 × 10^-5 = 5.56 × 10^-10.
- Use [OH-] = √(Kb × 0.36).
- Get [OH-] ≈ 1.41 × 10^-5.
- Find pOH ≈ 4.85.
- Conclude pH ≈ 9.15.
Final takeaway
To calculate the pH of a 0.36 M CH3COONa solution, treat acetate as a weak base produced by the salt of a weak acid and strong base. Convert acetic acid Ka to acetate Kb, estimate hydroxide concentration using the square-root hydrolysis formula, and then convert pOH to pH. Using standard 25°C values gives a final answer of pH ≈ 9.15. This is the expected result for a moderately concentrated sodium acetate solution and is fully consistent with weak acid conjugate base chemistry.