Calculate the pH of a 0.29 M CH3COONa Solution
Instantly determine the pH, pOH, hydroxide concentration, and base hydrolysis behavior of sodium acetate in water using a clean, interactive calculator and expert chemistry guide.
Sodium Acetate pH Calculator
How to calculate the pH of a 0.29 M CH3COONa solution
To calculate the pH of a 0.29 M CH3COONa solution, you need to recognize what sodium acetate is doing in water. CH3COONa, commonly called sodium acetate, is a salt formed from acetic acid, which is a weak acid, and sodium hydroxide, which is a strong base. Because it contains the conjugate base of a weak acid, the acetate ion hydrolyzes in water and generates hydroxide ions. That means the solution is basic, so the pH will be greater than 7.
This is one of the most common equilibrium problems in introductory chemistry, analytical chemistry, and acid-base theory. Even though the solution contains a salt rather than an obvious base such as NaOH, the acetate ion still behaves as a weak base. In practical terms, the sodium ion is a spectator ion, while the acetate ion controls the acid-base chemistry. Once you frame the problem that way, the math becomes straightforward.
Step 1: Write the hydrolysis reaction
When sodium acetate dissolves, it dissociates almost completely:
CH3COONa → Na+ + CH3COO-
The sodium ion does not affect pH significantly in this context. The acetate ion reacts with water:
CH3COO- + H2O ⇌ CH3COOH + OH-
This reaction shows why the solution becomes basic. Hydroxide ions are produced, increasing the pH above neutral.
Step 2: Convert Ka of acetic acid into Kb for acetate
The conjugate acid-base relationship connects the acid dissociation constant and the base dissociation constant:
Ka × Kb = Kw
At 25 degrees C, the ion product of water is usually taken as:
Kw = 1.0 × 10-14
For acetic acid, a common textbook value is:
Ka = 1.8 × 10-5
So the base dissociation constant for acetate is:
Kb = Kw / Ka = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
Step 3: Set up the equilibrium expression
Let the initial concentration of acetate be 0.29 M. If x is the amount that reacts with water, then the equilibrium concentrations are:
- [CH3COO-] = 0.29 – x
- [CH3COOH] = x
- [OH-] = x
The equilibrium expression for the base reaction is:
Kb = [CH3COOH][OH-] / [CH3COO-]
Substitute the equilibrium concentrations:
5.56 × 10-10 = x² / (0.29 – x)
Step 4: Use the weak base approximation
Because Kb is very small, x will be tiny compared with 0.29, so you can approximate:
0.29 – x ≈ 0.29
This simplifies the equation to:
x² = (5.56 × 10-10)(0.29)
x² = 1.6124 × 10-10
x = 1.27 × 10-5
Since x = [OH-], the hydroxide concentration is:
[OH-] = 1.27 × 10-5 M
Step 5: Calculate pOH and pH
Now compute pOH:
pOH = -log(1.27 × 10-5) ≈ 4.90
Then:
pH = 14.00 – 4.90 = 9.10
So the final answer is:
The pH of a 0.29 M CH3COONa solution is approximately 9.10 at 25 degrees C.
Why the solution is basic
Students often wonder why a salt can make water basic. The reason is tied to the parent acid and base that formed the salt. Sodium acetate comes from NaOH and CH3COOH. NaOH is a strong base, so Na+ has essentially no acidity in water. CH3COOH is a weak acid, so its conjugate base CH3COO- is strong enough to react with water measurably. That hydrolysis consumes water and generates OH-, pushing the pH above 7.
This principle is very general. Salts from strong acids and strong bases are usually neutral. Salts from strong acids and weak bases are acidic. Salts from weak acids and strong bases are basic. Sodium acetate belongs to the last group, which is why your answer lands in the mildly basic range rather than near neutral.
Exact versus approximate solution
For most classroom problems, the square-root approximation is accurate enough. Still, an exact solution can be obtained by solving the quadratic equation:
x² + Kb x – KbC = 0
where C = 0.29 M. Solving for the positive root gives nearly the same answer because the extent of hydrolysis is extremely small compared with the starting concentration. The percent ionization is far below 1%, so the approximation is well justified.
| Parameter | Value used | Meaning |
|---|---|---|
| Salt concentration | 0.29 M | Initial acetate concentration after complete dissociation |
| Ka for acetic acid | 1.8 × 10-5 | Standard textbook value near 25 degrees C |
| Kw | 1.0 × 10-14 | Water ion product at 25 degrees C |
| Kb for acetate | 5.56 × 10-10 | Basicity of CH3COO- in water |
| [OH-] | 1.27 × 10-5 M | Hydroxide produced by hydrolysis |
| pOH | 4.90 | Negative log of hydroxide concentration |
| pH | 9.10 | Final basic solution pH |
What happens if concentration changes?
The pH of sodium acetate depends on concentration. As concentration increases, the hydroxide concentration also increases, but not in a simple linear way because the equilibrium involves a square-root relationship under the weak-base approximation. Specifically:
[OH-] ≈ √(KbC)
That means if the concentration increases by a factor of 4, the hydroxide concentration only increases by a factor of 2. Therefore the pH rises gradually rather than dramatically.
| CH3COONa concentration (M) | Approximate [OH-] (M) | Approximate pH at 25 degrees C |
|---|---|---|
| 0.01 | 2.36 × 10-6 | 8.37 |
| 0.05 | 5.27 × 10-6 | 8.72 |
| 0.10 | 7.45 × 10-6 | 8.87 |
| 0.29 | 1.27 × 10-5 | 9.10 |
| 0.50 | 1.67 × 10-5 | 9.22 |
| 1.00 | 2.36 × 10-5 | 9.37 |
Common mistakes when solving this problem
- Treating sodium acetate like a neutral salt. It is not neutral because acetate is the conjugate base of a weak acid.
- Using Ka directly in the ICE table. The reacting species is CH3COO-, so you need Kb, not Ka, unless you solve through an equivalent alternate approach.
- Forgetting that Na+ is a spectator ion. Sodium does not significantly influence pH here.
- Using pH = -log(0.29). That would only apply to a strong acid with hydrogen ion concentration of 0.29 M, which is not this system.
- Ignoring temperature dependence. If the temperature changes, Kw and sometimes tabulated Ka values may shift, changing the exact numerical answer.
Does the symbol 0.29 m change the answer?
Strictly speaking, lowercase m often means molality, while uppercase M means molarity. In many educational problems, especially simplified online queries, people use the lowercase letter loosely even when they mean molarity. If the problem truly means 0.29 molal sodium acetate, the most rigorous treatment would account for solution density and activity effects. However, for a typical general chemistry calculation in a dilute aqueous solution, 0.29 m and 0.29 M are often treated as approximately equivalent for a quick pH estimate. That is why this calculator allows the notation but uses the standard aqueous approximation unless more advanced physical data are supplied.
Why this calculation matters in real chemistry
Sodium acetate appears in buffer preparation, laboratory calibration work, food chemistry, and industrial formulations. Understanding its pH behavior helps in several practical contexts:
- Designing acetate buffer systems with acetic acid
- Predicting whether a salt solution is acidic, basic, or neutral
- Estimating pH before experimental verification with a pH meter
- Learning conjugate acid-base relationships in equilibrium chemistry
- Comparing theoretical pH with measured pH when ionic strength becomes important
In more advanced chemistry, chemists may move beyond concentration-based approximations and use activities, especially at higher ionic strengths. But for standard coursework and routine calculations, the hydrolysis method shown here is the accepted and reliable approach.
Authoritative chemistry references
For deeper reading on acid-base equilibria, weak electrolytes, and aqueous chemistry, these authoritative sources are excellent starting points:
- LibreTexts Chemistry for structured acid-base equilibrium explanations.
- National Institute of Standards and Technology (NIST) for trusted scientific reference data.
- U.S. Environmental Protection Agency (EPA) for water chemistry context and pH fundamentals.
- University of California, Berkeley Chemistry for academic chemistry resources.
Final answer summary
If you are asked to calculate the pH of a 0.29 M CH3COONa solution at 25 degrees C, use the acetate hydrolysis reaction, convert acetic acid Ka into acetate Kb, solve for hydroxide concentration, and then convert to pOH and pH. With Ka = 1.8 × 10-5 and Kw = 1.0 × 10-14, the result is:
pH ≈ 9.10
This makes sense chemically because sodium acetate is the salt of a weak acid and a strong base, so its aqueous solution is basic. The calculator above automates these steps and also shows how the pH changes as concentration varies, making it useful for both homework checking and concept review.