Calculate the pH of a 36 M CH3COONa Solution
Use this interactive calculator to determine the pH of a sodium acetate solution from acetate hydrolysis. The default setup solves the classic problem for a 36 M CH3COONa solution using the accepted acetic acid dissociation constant near 25 degrees Celsius.
Default answer for 36 M CH3COONa at 25 degrees Celsius is approximately pH 10.15.
How to calculate the pH of a 36 M CH3COONa solution
To calculate the pH of a 36 M CH3COONa solution, you need to recognize what sodium acetate is doing in water. CH3COONa is sodium acetate, a salt formed from acetic acid, which is a weak acid, and sodium hydroxide, which is a strong base. Because sodium comes from a strong base and does not appreciably hydrolyze, the chemistry that matters is the acetate ion, CH3COO–. That ion behaves as a weak base in water by accepting a proton from water and generating hydroxide ions.
The hydrolysis reaction is:
Since hydroxide ions are produced, the resulting solution is basic, which means its pH will be greater than 7. The job is to determine how much OH– forms at equilibrium and then convert that value into pOH and pH. In most textbook problems, this is done using the base dissociation constant Kb for acetate, which is related to the acid dissociation constant Ka of acetic acid by the familiar equation Kb = Kw / Ka.
Step 1: Identify the relevant equilibrium constants
For acetic acid at about 25 degrees Celsius, a common value is Ka = 1.8 × 10-5. For water, Kw = 1.0 × 10-14. Therefore:
This very small Kb tells you acetate is only a weak base. Even so, because the starting concentration in this problem is extremely high at 36 M, enough acetate is present to generate a measurable concentration of hydroxide ions.
Step 2: Set up the ICE table
For the reaction CH3COO– + H2O ⇌ CH3COOH + OH–, assume an initial acetate concentration of 36 M.
- Initial: [CH3COO–] = 36, [CH3COOH] = 0, [OH–] = 0
- Change: [-x, +x, +x]
- Equilibrium: [CH3COO–] = 36 – x, [CH3COOH] = x, [OH–] = x
Insert these equilibrium concentrations into the expression for Kb:
Since Kb is very small compared with the concentration 36 M, the approximation 36 – x ≈ 36 is excellent. Then:
That value is the hydroxide ion concentration:
Step 3: Convert hydroxide concentration to pOH and pH
Once [OH–] is known, calculate pOH:
Then use:
So the pH of a 36 M CH3COONa solution is approximately 10.15 under standard textbook assumptions.
Why sodium acetate makes the solution basic
Students often wonder why sodium acetate does not have a neutral pH even though it is a salt. The answer depends on the parent acid and base used to form the salt. Sodium acetate comes from acetic acid and sodium hydroxide. Acetic acid is weak, so its conjugate base, acetate, has a measurable tendency to pull protons from water. Sodium hydroxide is strong, and sodium ions do not alter pH in any meaningful way in a simple equilibrium treatment. As a result, the acetate ion dominates the acid-base behavior.
This principle is general. Salts from a strong acid and strong base tend to be neutral. Salts from a weak acid and strong base tend to be basic. Salts from a strong acid and weak base tend to be acidic. Recognizing this pattern helps you decide very quickly whether to use Ka or Kb in a pH calculation.
Exact vs approximate calculation
The approximation x = √(KbC) is widely used because it is simple and usually accurate when x is small compared with the initial concentration. For this problem, the approximation is especially good because the hydroxide concentration is around 1.4 × 10-4 M while the initial acetate concentration is 36 M. The ratio x/C is tiny, so subtracting x from 36 changes the denominator by a negligible amount.
If you want the exact expression, rearrange:
into a quadratic:
Solving gives:
Numerically, the exact result differs from the approximation by an amount too small to matter for most educational purposes. The calculator above allows both methods so you can compare them directly.
Important realism note about a 36 M solution
From a pure calculation standpoint, chemistry textbooks sometimes present very large concentrations to practice equilibrium methods. In reality, a 36 M sodium acetate solution is extraordinarily concentrated. At such high concentrations, ideal-solution assumptions become weak, activity effects become important, and actual solubility and solution behavior may limit whether such a solution can be prepared or treated as ideal. That means the answer pH 10.15 is the standard equilibrium result under simplified assumptions, not necessarily an experimental value under all real-world conditions.
This is an important distinction for advanced chemistry. Introductory calculations usually use concentrations directly in equilibrium expressions, but rigorous thermodynamics uses activities. At very high ionic strengths, activity coefficients can shift the effective acid-base behavior. For homework and exam contexts, though, the accepted method is the one shown here.
Worked summary of the full calculation
- Write the base hydrolysis reaction for acetate: CH3COO– + H2O ⇌ CH3COOH + OH–.
- Find Kb from Ka: Kb = Kw / Ka = 1.0 × 10-14 / 1.8 × 10-5 = 5.56 × 10-10.
- Use the initial concentration C = 36 M.
- Apply x = √(KbC) to estimate [OH–].
- Compute [OH–] = √(5.56 × 10-10 × 36) ≈ 1.41 × 10-4 M.
- Calculate pOH = -log(1.41 × 10-4) ≈ 3.85.
- Calculate pH = 14.00 – 3.85 = 10.15.
Comparison table: pH of sodium acetate at different concentrations
The concentration of the salt strongly influences the hydroxide ion concentration. The table below uses Ka = 1.8 × 10-5 and Kw = 1.0 × 10-14, applying the standard weak-base approximation for acetate.
| CH3COONa Concentration (M) | Kb of Acetate | Estimated [OH-] (M) | pOH | Estimated pH |
|---|---|---|---|---|
| 0.01 | 5.56 × 10-10 | 2.36 × 10-6 | 5.63 | 8.37 |
| 0.10 | 5.56 × 10-10 | 7.45 × 10-6 | 5.13 | 8.87 |
| 1.00 | 5.56 × 10-10 | 2.36 × 10-5 | 4.63 | 9.37 |
| 10.0 | 5.56 × 10-10 | 7.45 × 10-5 | 4.13 | 9.87 |
| 36.0 | 5.56 × 10-10 | 1.41 × 10-4 | 3.85 | 10.15 |
Comparison table: common acid-base salt behavior
A lot of pH mistakes happen because learners use the wrong conceptual model for salts. This table helps you quickly classify solution behavior.
| Salt Type | Example | Parent Species | Expected pH Trend | Main Reason |
|---|---|---|---|---|
| Strong acid + strong base | NaCl | HCl + NaOH | Near neutral | Neither ion hydrolyzes significantly |
| Weak acid + strong base | CH3COONa | CH3COOH + NaOH | Basic | Conjugate base hydrolyzes to form OH- |
| Strong acid + weak base | NH4Cl | HCl + NH3 | Acidic | Conjugate acid hydrolyzes to form H3O+ |
| Weak acid + weak base | NH4CH3COO | NH3 + CH3COOH | Depends on Ka and Kb | Both ions hydrolyze and compete |
Common mistakes when solving this problem
- Using Ka directly to find pH. For sodium acetate, the reacting species is acetate, so you need Kb or the Ka to Kb conversion.
- Assuming the solution is neutral because it is a salt. Not all salts are neutral. Sodium acetate is basic.
- Forgetting to calculate pOH first. Since hydrolysis makes OH–, you usually find pOH before converting to pH.
- Ignoring the distinction between concentration and activity at very high concentration. For standard problems, concentration-based equilibrium is acceptable, but advanced work may require activities.
- Incorrect logarithm handling. Keep scientific notation clear when taking negative logarithms.
Authoritative chemistry references
If you want to verify acid-base constants, equilibrium concepts, and water chemistry from reliable educational and government sources, these references are useful:
- Chemistry LibreTexts for acid-base equilibrium explanations and worked examples.
- U.S. Environmental Protection Agency for foundational pH and water chemistry context.
- University of Wisconsin chemistry resources for acid-base equilibrium instruction.
Final answer
Under the standard equilibrium model used in general chemistry, the pH of a 36 M CH3COONa solution is approximately 10.15. This comes from acetate hydrolysis, where acetate acts as a weak base and generates hydroxide ions in water.