Calculate pH of a 0.1 M CH3COONa Solution
Use this premium sodium acetate pH calculator to determine pH, pOH, hydroxide concentration, hydrolysis constant, and degree of hydrolysis for aqueous CH3COONa solutions.
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How to calculate pH of a 0.1 M CH3COONa solution
To calculate the pH of a 0.1 M CH3COONa solution, you need to recognize that sodium acetate is the salt of a weak acid, acetic acid, and a strong base, sodium hydroxide. In water, sodium acetate dissociates almost completely into Na+ and CH3COO–. The sodium ion is essentially a spectator ion, but the acetate ion reacts with water through hydrolysis to produce a basic solution. That is why a sodium acetate solution has a pH above 7.
The key equilibrium is:
CH3COO– + H2O ⇌ CH3COOH + OH–
Because hydroxide ions are formed, the solution becomes basic. The calculation is not done with the acid dissociation constant of acetic acid directly, but with the base hydrolysis constant of acetate. However, the two constants are connected:
Kb = Kw / Ka
At 25°C, water has Kw = 1.0 × 10-14, and acetic acid has Ka ≈ 1.8 × 10-5. Therefore:
Kb = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
For a 0.1 M solution of CH3COONa, let the hydroxide concentration generated by hydrolysis be x. Then:
- Initial acetate concentration = 0.1 M
- Change in acetate concentration = -x
- Equilibrium acetate concentration = 0.1 – x
- Equilibrium acetic acid concentration = x
- Equilibrium hydroxide concentration = x
The equilibrium expression becomes:
Kb = x2 / (0.1 – x)
Since Kb is very small, chemists often use the approximation 0.1 – x ≈ 0.1. This gives:
x = √(Kb × C) = √(5.56 × 10-10 × 0.1) = 7.46 × 10-6 M
Now calculate pOH:
pOH = -log(7.46 × 10-6) = 5.13
Finally:
pH = 14.00 – 5.13 = 8.87
Why sodium acetate gives a basic pH
Students often wonder why a salt can change pH at all. The answer depends on the strength of the parent acid and base that formed the salt. Sodium acetate comes from acetic acid, which is weak, and sodium hydroxide, which is strong. A strong base contributes a cation, Na+, that does not significantly react with water. But the conjugate base of a weak acid, acetate, is strong enough to hydrolyze water slightly and generate OH–. The result is a basic solution.
This principle can be generalized:
- Strong acid + strong base salt: usually neutral
- Weak acid + strong base salt: basic
- Strong acid + weak base salt: acidic
- Weak acid + weak base salt: pH depends on both Ka and Kb
Step by step derivation for CH3COONa
1. Write the dissociation and hydrolysis reactions
Sodium acetate dissolves as:
CH3COONa → Na+ + CH3COO–
The hydrolysis reaction that controls pH is:
CH3COO– + H2O ⇌ CH3COOH + OH–
2. Convert Ka to Kb
Since acetate is the conjugate base of acetic acid, use:
Kb = Kw / Ka
With Ka = 1.8 × 10-5 and Kw = 1.0 × 10-14:
Kb = 5.56 × 10-10
3. Build an ICE table
- Initial: [CH3COO–] = 0.1, [CH3COOH] = 0, [OH–] = 0
- Change: -x, +x, +x
- Equilibrium: 0.1 – x, x, x
4. Solve for x
Substitute into the Kb expression:
5.56 × 10-10 = x2 / (0.1 – x)
Using the approximation gives x = 7.46 × 10-6 M. Using the exact quadratic equation gives essentially the same answer, which validates the approximation because x is tiny compared with 0.1.
5. Convert to pOH and pH
- pOH = -log[OH–]
- pH = 14 – pOH at 25°C
That gives pH ≈ 8.87.
Comparison table: pH of sodium acetate at different concentrations
The pH of CH3COONa changes with concentration because the hydroxide ion concentration from hydrolysis changes with the initial acetate concentration. The values below assume 25°C, Ka of acetic acid = 1.8 × 10-5, and the standard hydrolysis approximation.
| CH3COONa Concentration (M) | Kb of Acetate | Approx. [OH-] (M) | pOH | pH |
|---|---|---|---|---|
| 0.001 | 5.56 × 10^-10 | 7.46 × 10^-7 | 6.13 | 7.87 |
| 0.010 | 5.56 × 10^-10 | 2.36 × 10^-6 | 5.63 | 8.37 |
| 0.100 | 5.56 × 10^-10 | 7.46 × 10^-6 | 5.13 | 8.87 |
| 0.500 | 5.56 × 10^-10 | 1.67 × 10^-5 | 4.78 | 9.22 |
| 1.000 | 5.56 × 10^-10 | 2.36 × 10^-5 | 4.63 | 9.37 |
Exact method versus approximation
In introductory chemistry, the approximation method is preferred because it is fast and usually accurate for weak hydrolysis systems. Still, an exact quadratic solution is more rigorous and especially useful when concentration is very low or when the equilibrium constant is not extremely small compared with the concentration. For a 0.1 M sodium acetate solution, both methods agree to several decimal places.
| Method | Equation Used | [OH-] for 0.1 M | Calculated pH | Practical Comment |
|---|---|---|---|---|
| Approximation | x ≈ √(KbC) | 7.46 × 10^-6 M | 8.87 | Fast and standard for classroom use |
| Exact quadratic | x = (-Kb + √(Kb² + 4KbC)) / 2 | 7.45 × 10^-6 M | 8.87 | Best for rigorous calculation and software tools |
Common mistakes when calculating pH of CH3COONa
- Using Ka directly to find pH. The reacting species is acetate, which is a base in water, so convert Ka to Kb first.
- Assuming the solution is neutral because it is a salt. Not all salts are neutral. Sodium acetate is basic.
- Forgetting to calculate pOH first. Hydrolysis gives OH–, so pOH is found before pH.
- Ignoring temperature. At temperatures other than 25°C, Kw changes, and neutral pH is not always exactly 7.00.
- Mixing up sodium acetate with acetic acid. Acetic acid is acidic; sodium acetate is basic in water.
Practical applications of sodium acetate pH calculations
Understanding the pH of sodium acetate solutions matters in more than exam questions. Sodium acetate appears in analytical chemistry, biological buffer preparation, textile processing, food systems, and laboratory protocols where acetate buffers are used to stabilize pH. Since sodium acetate is one component of an acetate buffer, knowing its hydrolysis behavior is foundational to predicting the final pH after mixing with acetic acid.
In biochemistry and molecular biology, acetate systems may appear in precipitation and washing procedures. In analytical chemistry, acetate solutions are used when a mildly basic environment is desirable without the much stronger effect of sodium hydroxide. In educational labs, sodium acetate is one of the best examples for illustrating salt hydrolysis because the chemistry is clean, the constants are well tabulated, and the result is easy to verify experimentally.
Authority sources for constants and reference data
If you want to verify equilibrium constants or learn more about sodium acetate and acetic acid chemistry, these sources are useful:
- NIST Chemistry WebBook for trusted thermodynamic and chemical reference data.
- PubChem at the National Institutes of Health for compound properties and identifiers.
- University of California Davis chemistry educational resources for acid-base and equilibrium instruction.
Quick formula summary
- Find acetate base constant: Kb = Kw / Ka
- Use concentration C = 0.1 M
- Estimate hydroxide: [OH-] ≈ √(KbC)
- Compute pOH = -log[OH-]
- Compute pH = 14 – pOH at 25°C
For a standard 0.1 M CH3COONa solution, the accepted answer is pH ≈ 8.87. This value can vary slightly if a different Ka for acetic acid is used, if temperature is changed, or if ionic strength corrections are applied in more advanced treatments. For general chemistry, however, 8.87 is the right target value and the one you should expect in most solved examples.
Final answer
When asked to calculate the pH of a 0.1 M CH3COONa solution, treat acetate as a weak base in water. Convert the Ka of acetic acid to Kb, solve for hydroxide concentration by hydrolysis, then convert pOH to pH. Using Ka = 1.8 × 10-5 at 25°C, the pH comes out to approximately 8.87. That means sodium acetate produces a mildly basic aqueous solution.