Calculate the pH of a 0.36 M CH3COONa Solution
Use this premium calculator to find the pH of a sodium acetate solution by applying weak-base hydrolysis, the relationship between Ka and Kb, and either the approximation method or the exact quadratic method.
Default value is 0.36 M.
Typical Ka for CH3COOH is 1.8 × 10-5.
At 25°C, Kw = 1.0 × 10-14.
Both methods should be very close for this dilute weak-base hydrolysis problem.
How to calculate the pH of a 0.36 M CH3COONa solution
Sodium acetate, written as CH3COONa, is the salt formed from a strong base, sodium hydroxide, and a weak acid, acetic acid. Because it comes from a weak acid, the acetate ion does not stay completely neutral in water. Instead, it reacts slightly with water to produce hydroxide ions. That hydrolysis reaction makes the solution basic, so the pH ends up above 7. If you are asked to calculate the pH of a 0.36 M CH3COONa solution, the chemistry is a classic weak-base equilibrium problem.
The key idea is that sodium ions are spectator ions in this context, while acetate ions are the species that matter for acid-base behavior. Once sodium acetate dissolves, it separates essentially completely into Na+ and CH3COO–. The acetate ion then reacts with water according to the equilibrium:
This equation shows why the solution becomes basic: hydroxide ions are produced. To quantify that basicity, we use the base dissociation constant Kb for acetate. In many textbook and exam problems, you are given Ka for acetic acid rather than Kb for acetate, so you convert with the standard relationship:
At 25°C, Kw is 1.0 × 10-14, and a commonly used value for acetic acid is Ka = 1.8 × 10-5. Substituting those constants gives:
Step-by-step setup
Now assign the initial concentration of acetate as 0.36 M, because sodium acetate fully dissociates in water:
- Initial [CH3COO–] = 0.36 M
- Initial [CH3COOH] = 0
- Initial [OH–] from hydrolysis = 0
Let x represent the amount of acetate that hydrolyzes. Then at equilibrium:
- [CH3COO–] = 0.36 – x
- [CH3COOH] = x
- [OH–] = x
Insert these values into the Kb expression:
Because Kb is very small, x is tiny compared with 0.36, so most chemistry courses allow the approximation 0.36 – x ≈ 0.36. That simplifies the algebra to:
Multiply both sides by 0.36:
Take the square root:
Now convert hydroxide concentration into pOH:
Finally, calculate pH:
So the pH of a 0.36 M CH3COONa solution is approximately 9.15 at 25°C, using Ka = 1.8 × 10-5.
Why sodium acetate gives a basic solution
This result is often surprising to students at first because salts are frequently imagined as neutral substances. In reality, the pH of a salt solution depends on the acid and base from which the salt was derived. Sodium acetate is produced from sodium hydroxide, a strong base, and acetic acid, a weak acid. Strong-base cations like Na+ usually do not affect pH, but the conjugate base of a weak acid does. Acetate is the conjugate base of acetic acid, so it has a measurable tendency to accept protons from water, generating OH–. That is why the solution is basic.
Approximation method versus exact method
For a concentration as large as 0.36 M and a Kb as small as 5.56 × 10-10, the approximation method is excellent. Still, it is useful to understand the exact method. If you solve the equilibrium expression exactly, you write:
Kb = x² / (0.36 – x), which rearranges to x² + Kb x – 0.36Kb = 0.
This is a quadratic equation in x. Solving it with the quadratic formula gives a value of x nearly identical to the square-root estimate. In practical terms, the difference is negligible for this problem, which confirms that the approximation is valid. A quick percent check shows that x / 0.36 is far below 5%, so the approximation is well within standard chemistry expectations.
| Parameter | Value Used | Meaning | Impact on Final pH |
|---|---|---|---|
| Salt concentration | 0.36 M | Initial acetate concentration after dissociation | Higher concentration slightly increases basicity |
| Ka of acetic acid | 1.8 × 10-5 | Acid strength of CH3COOH | Lower Ka means stronger conjugate base and higher pH |
| Kw at 25°C | 1.0 × 10-14 | Water autoionization constant | Used to convert Ka into Kb |
| Kb of acetate | 5.56 × 10-10 | Base strength of CH3COO– | Determines hydroxide production |
| [OH–] | 1.41 × 10-5 M | Equilibrium hydroxide concentration | Directly sets pOH and pH |
| Final pH | 9.15 | Basic solution | Final answer for the problem |
Common mistakes in this problem
- Treating sodium acetate as a strong base. Sodium acetate is not NaOH. It is a salt whose basicity comes from acetate hydrolysis, not complete OH– release.
- Using Ka directly instead of Kb. The hydrolyzing species is CH3COO–, so the equilibrium constant must be Kb, not Ka.
- Forgetting full dissociation of the salt. The starting acetate concentration is the same as the sodium acetate concentration because CH3COONa dissociates essentially completely.
- Confusing pOH and pH. Once you get [OH–], you first find pOH, then subtract from 14 at 25°C to get pH.
- Using the Henderson-Hasselbalch equation incorrectly. That formula is for buffer systems with both weak acid and conjugate base present in substantial amounts. A pure sodium acetate solution is not the same setup.
What happens if the concentration changes?
The pH of sodium acetate depends on concentration because the hydroxide concentration produced by hydrolysis scales with the square root of Kb times concentration. That means the pH does increase as concentration increases, but not linearly. Doubling the concentration does not double the pH shift. Instead, the change is more gradual because of the logarithmic pH scale and the square-root relationship inside the equilibrium expression.
| CH3COONa Concentration (M) | Approximate [OH–] (M) | Approximate pOH | Approximate pH at 25°C |
|---|---|---|---|
| 0.010 | 2.36 × 10-6 | 5.63 | 8.37 |
| 0.050 | 5.27 × 10-6 | 5.28 | 8.72 |
| 0.100 | 7.46 × 10-6 | 5.13 | 8.87 |
| 0.360 | 1.41 × 10-5 | 4.85 | 9.15 |
| 0.500 | 1.67 × 10-5 | 4.78 | 9.22 |
| 1.000 | 2.36 × 10-5 | 4.63 | 9.37 |
This table is especially helpful when comparing textbook scenarios. It shows that even fairly concentrated sodium acetate solutions are only mildly basic, not strongly alkaline. That is because acetate is a weak base, so it hydrolyzes only slightly.
Physical chemistry perspective
From a deeper equilibrium perspective, the system is controlled by the conjugate acid-base pair CH3COOH/CH3COO– and the self-ionization of water. The ratio between Ka and Kb reflects a fundamental thermodynamic relationship at a fixed temperature. If temperature changes, Kw changes as well, so pH values and the Ka-to-Kb conversion also shift. In general chemistry, however, unless another temperature is stated, 25°C is assumed.
The acetate ion is resonance-stabilized, which reduces its basic strength compared with strong bases such as hydroxide or alkoxides. That resonance stabilization helps explain why sodium acetate solutions are only moderately basic despite sometimes being fairly concentrated. The molecule is basic enough to hydrolyze water measurably, but not enough to produce large OH– concentrations.
When this calculation matters in real applications
Sodium acetate appears in laboratory buffers, chemical manufacturing, food processing, textile operations, and educational chemistry demonstrations. Understanding its pH behavior matters when preparing acetate buffers, predicting reagent compatibility, and interpreting titration curves. In analytical chemistry, acetate systems are common because acetic acid and sodium acetate form one of the most widely used buffer pairs in introductory and applied laboratories.
If you were preparing an acetate solution in a lab, you would also think about ionic strength, temperature, and activity effects if high precision were required. However, for most coursework, exam, and practical preparation problems, the ideal approximation used here is the accepted approach.
Fast exam strategy
- Identify the salt as coming from a weak acid and a strong base.
- Conclude the solution is basic.
- Write the hydrolysis reaction for acetate.
- Convert Ka to Kb using Kb = Kw / Ka.
- Use x = √(KbC) for [OH–].
- Find pOH, then pH.
- Check that x is much smaller than the starting concentration.
Final answer summary
For a 0.36 M sodium acetate solution at 25°C, using Ka for acetic acid equal to 1.8 × 10-5 and Kw equal to 1.0 × 10-14, the acetate ion behaves as a weak base with Kb = 5.56 × 10-10. Solving the hydrolysis equilibrium gives [OH–] ≈ 1.41 × 10-5 M, pOH ≈ 4.85, and therefore pH ≈ 9.15. That makes the solution mildly basic, exactly as expected for a salt of a weak acid and a strong base.