Calculate the pH of a 0.35 M CH3COONa Solution
Use this interactive sodium acetate calculator to determine the pH, pOH, Kb, hydroxide concentration, and hydrolysis extent for a 0.35 M CH3COONa solution at 25 C or other common temperatures. The tool applies weak-base hydrolysis from the acetate ion, not a strong-base shortcut.
Sodium Acetate pH Calculator
CH3COONa fully dissociates in water to produce Na+ and CH3COO-. The acetate ion acts as a weak base: CH3COO- + H2O ⇌ CH3COOH + OH-.
Calculated pH
Calculated pOH
How to Calculate the pH of a 0.35 M CH3COONa Solution
To calculate the pH of a 0.35 M CH3COONa solution, you need to recognize what sodium acetate does in water. Sodium acetate is a salt formed from a strong base, sodium hydroxide, and a weak acid, acetic acid. Because the acid portion is weak, its conjugate base, acetate ion, can react with water and produce hydroxide ions. That hydrolysis reaction makes the solution basic.
The key point is that CH3COONa does not behave like a strong base on its own. Instead, it dissociates completely into sodium ions and acetate ions, and then the acetate ion establishes a weak-base equilibrium with water. This is why the correct route is to calculate Kb from the known Ka of acetic acid and then solve for the hydroxide concentration.
Step 1: Write the dissociation and hydrolysis reactions
In water, sodium acetate dissociates essentially completely:
CH3COONa → Na+ + CH3COO-
The sodium ion is a spectator ion under these conditions, so the important chemistry comes from acetate:
CH3COO- + H2O ⇌ CH3COOH + OH-
This equilibrium produces hydroxide ions, which means the final pH must be greater than 7 at 25 C.
Step 2: Convert Ka to Kb
For a conjugate acid-base pair, the relation between the acid and base dissociation constants is:
Ka × Kb = Kw
At 25 C, the ionic product of water is usually taken as:
Kw = 1.0 × 10^-14
If the acetic acid dissociation constant is:
Ka = 1.8 × 10^-5
Then the base dissociation constant of acetate is:
Kb = Kw / Ka = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10
Step 3: Set up the ICE table
Because the sodium acetate concentration is 0.35 M, the initial acetate concentration is also 0.35 M after full dissociation.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| CH3COO- | 0.35 | -x | 0.35 – x |
| CH3COOH | 0 | +x | x |
| OH- | 0 | +x | x |
The equilibrium expression is:
Kb = [CH3COOH][OH-] / [CH3COO-]
Substituting the ICE values gives:
5.56 × 10^-10 = x² / (0.35 – x)
Step 4: Solve for x = [OH-]
Because Kb is very small, x will be much smaller than 0.35. That means most textbook solutions use the approximation:
0.35 – x ≈ 0.35
So:
x² = (5.56 × 10^-10)(0.35) = 1.946 × 10^-10
x = √(1.946 × 10^-10) ≈ 1.40 × 10^-5 M
Therefore:
[OH-] ≈ 1.40 × 10^-5 M
If you solve the quadratic exactly, you get essentially the same answer because the approximation is excellent here. The percent hydrolysis is tiny, which confirms that the weak-base approximation is justified.
Step 5: Find pOH and pH
Now calculate pOH:
pOH = -log(1.40 × 10^-5) ≈ 4.855
At 25 C:
pH = 14.000 – 4.855 = 9.145
Why Sodium Acetate Makes Water Basic
Students often ask why sodium acetate, which looks like a neutral salt, produces a basic pH. The reason is rooted in conjugate acid-base theory. Acetic acid is weak, meaning it does not completely donate protons in water. Its conjugate base, acetate, therefore still has enough basic character to accept a proton from water. When that happens, hydroxide ions are formed. Those hydroxide ions shift the solution to the basic side.
By contrast, a salt like sodium chloride comes from a strong acid and a strong base. Chloride ion is such a weak base that it does not appreciably hydrolyze in water, so NaCl solutions are close to neutral. This makes sodium acetate an excellent example of how salt hydrolysis controls pH.
Comparison of common dissolved species and pH behavior
| Compound in water | Parent acid | Parent base | Expected pH trend | Reason |
|---|---|---|---|---|
| CH3COONa | Weak acid | Strong base | Basic | Acetate hydrolyzes to form OH- |
| NaCl | Strong acid | Strong base | Near neutral | No meaningful hydrolysis |
| NH4Cl | Strong acid | Weak base | Acidic | NH4+ hydrolyzes to form H3O+ |
How Accurate Is the Approximation?
For weak acid and weak base calculations, the approximation that the change in concentration is small compared with the initial concentration is commonly used. In this case, the hydroxide concentration is only about 1.40 × 10^-5 M, while the initial acetate concentration is 0.35 M. The fraction hydrolyzed is:
(1.40 × 10^-5 / 0.35) × 100 ≈ 0.004%
That is far below the common 5% threshold used in equilibrium chemistry, so the approximation is exceptionally good.
Computed pH for sodium acetate at different concentrations
The table below uses Ka = 1.8 × 10^-5 and Kw = 1.0 × 10^-14 at 25 C. Values are textbook equilibrium estimates and illustrate how concentration affects pH.
| CH3COONa concentration (M) | Kb of acetate | Approximate [OH-] (M) | Approximate pOH | Approximate pH |
|---|---|---|---|---|
| 0.010 | 5.56 × 10^-10 | 2.36 × 10^-6 | 5.627 | 8.373 |
| 0.050 | 5.56 × 10^-10 | 5.27 × 10^-6 | 5.278 | 8.722 |
| 0.100 | 5.56 × 10^-10 | 7.45 × 10^-6 | 5.128 | 8.872 |
| 0.350 | 5.56 × 10^-10 | 1.40 × 10^-5 | 4.855 | 9.145 |
| 1.000 | 5.56 × 10^-10 | 2.36 × 10^-5 | 4.627 | 9.373 |
These numbers show a useful trend: as sodium acetate concentration rises, the hydroxide concentration also increases, and the pH becomes more basic. However, because the hydrolysis is governed by a small Kb, the pH does not skyrocket the way it would for a strong base of the same formal concentration.
Common Mistakes When Solving This Problem
- Treating sodium acetate as a strong base. It is not. The base behavior comes from hydrolysis of acetate, not full release of OH- like NaOH.
- Using Ka directly without converting to Kb. Since acetate is the base in solution, the equilibrium is governed by Kb.
- Forgetting that the sodium ion is a spectator. Na+ does not control the pH here.
- Using Henderson-Hasselbalch when only a salt is present. That equation is primarily for buffer systems containing appreciable acid and conjugate base together.
- Ignoring temperature. The relation pH + pOH = pKw changes slightly as Kw changes with temperature.
Exact Quadratic vs Approximate Method
For educational purposes, it is worth comparing the exact and approximate routes. Starting from:
Kb = x² / (C – x)
Rearrange into a quadratic:
x² + Kbx – KbC = 0
Then solve:
x = [-Kb + √(Kb² + 4KbC)] / 2
With C = 0.35 M and Kb = 5.56 × 10^-10, the exact root differs negligibly from the square-root estimate. This is why general chemistry textbooks frequently use the approximation for salts of weak acids or weak bases when the equilibrium constant is small and the concentration is moderate.
Practical Importance of Sodium Acetate pH
Sodium acetate appears in chemical laboratories, biological sample preparation, and industrial formulations. Knowing the pH of a sodium acetate solution matters because pH can affect solubility, buffering behavior, reaction pathways, and stability of sensitive compounds. In practice, sodium acetate is also often paired with acetic acid to make acetate buffer systems, which are widely used around pH 4 to 6. However, when sodium acetate is dissolved by itself in water, the resulting solution is alkaline because only the conjugate base remains available to hydrolyze.
Where the constants come from
- Ka of acetic acid: commonly reported near 1.8 × 10^-5 at 25 C in general chemistry references.
- Kw of water: commonly reported as 1.0 × 10^-14 at 25 C, but it varies with temperature.
- pH concept: formal environmental and chemistry references such as the U.S. EPA and NIST provide foundational definitions and data support.
Fast Summary for Exams
- Write acetate hydrolysis: CH3COO- + H2O ⇌ CH3COOH + OH-
- Find Kb = Kw/Ka = (1.0 × 10^-14)/(1.8 × 10^-5) = 5.56 × 10^-10
- Use x = √(KbC) = √[(5.56 × 10^-10)(0.35)] = 1.40 × 10^-5 M
- Compute pOH = 4.855
- Compute pH = 14.000 – 4.855 = 9.145
If you only need the final exam-style result, the pH of a 0.35 M CH3COONa solution is about 9.15 at 25 C. If your instructor uses a slightly different Ka value for acetic acid, such as 1.75 × 10^-5, your final pH may shift by a few thousandths, but it will still be very close to 9.14.