Calculate the pH of a 15M Solution of CH3COONa
Use this premium sodium acetate pH calculator to estimate pH, pOH, hydroxide concentration, and hydrolysis behavior using the weak acid conjugate base approach.
Sodium Acetate pH Calculator
How to Calculate the pH of a 15M Solution of CH3COONa
When students first see sodium acetate, written as CH3COONa, they often assume the solution should be neutral because it is a salt. That assumption is only partly true for salts formed from a strong acid and a strong base. Sodium acetate is different. It is produced from a weak acid, acetic acid, and a strong base, sodium hydroxide. As a result, the acetate ion hydrolyzes in water and generates hydroxide ions, making the solution basic. If you want to calculate the pH of a 15M solution of CH3COONa, the key idea is to treat acetate as a weak base and solve its equilibrium with water.
In water, sodium acetate dissociates essentially completely into sodium ions and acetate ions. The sodium ion is a spectator ion for acid-base purposes, but the acetate ion reacts with water according to the equilibrium:
CH3COO- + H2O ⇌ CH3COOH + OH-
This reaction shows why the solution becomes basic. Each time acetate accepts a proton from water, hydroxide forms. The more hydroxide present, the lower the pOH and the higher the pH. Therefore, the whole problem becomes a standard weak-base equilibrium calculation.
Step 1: Start with the Known Acid Constant of Acetic Acid
Most textbook and laboratory calculations use the acid dissociation constant of acetic acid at 25 degrees Celsius as approximately 1.8 × 10-5. Because acetate is the conjugate base of acetic acid, you can find the base dissociation constant using:
Kb = Kw / Ka
With Kw = 1.0 × 10-14 and Ka = 1.8 × 10-5, the acetate ion has:
Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10
This tells us acetate is a weak base, which means it does not convert all of itself into hydroxide. Instead, only a small fraction hydrolyzes.
Step 2: Set Up the Weak Base Equilibrium
Let the initial acetate concentration be 15.0 M. In an idealized introductory chemistry calculation, we assume this is the starting concentration of CH3COO-. Let x represent the hydroxide concentration produced by hydrolysis. Then the equilibrium table is:
- Initial: [CH3COO-] = 15.0, [CH3COOH] = 0, [OH-] = 0
- Change: [CH3COO-] decreases by x, [CH3COOH] increases by x, [OH-] increases by x
- Equilibrium: [CH3COO-] = 15.0 – x, [CH3COOH] = x, [OH-] = x
The equilibrium expression is:
Kb = [CH3COOH][OH-] / [CH3COO-] = x² / (15.0 – x)
Step 3: Use the Common Approximation
Because Kb is very small and the concentration is large, x is tiny compared with 15.0. That allows the standard approximation:
15.0 – x ≈ 15.0
So the equilibrium equation simplifies to:
x² / 15.0 = 5.56 × 10^-10
x² = 8.34 × 10^-9
x = 9.13 × 10^-5 M
Since x equals the hydroxide concentration, we now have:
[OH-] = 9.13 × 10^-5 M
Step 4: Convert Hydroxide Concentration to pOH and pH
Next compute the pOH:
pOH = -log(9.13 × 10^-5) = 4.04
At 25 degrees Celsius, pH + pOH = 14.00, so:
pH = 14.00 – 4.04 = 9.96
Therefore, the idealized calculated pH of a 15M solution of CH3COONa is approximately 9.96.
Why the Solution Is Basic Instead of Neutral
This is one of the most important conceptual points in acid-base chemistry. The sodium ion comes from a strong base and does not significantly affect pH. The acetate ion, however, is the conjugate base of a weak acid, so it has measurable proton-accepting ability. That proton acceptance pulls equilibrium toward hydroxide production. In practical terms, whenever a salt contains the conjugate base of a weak acid, its aqueous solution typically has a pH above 7.
A useful comparison is sodium chloride. Chloride is the conjugate base of a strong acid, hydrochloric acid, so chloride is too weak a base to hydrolyze appreciably. Sodium acetate is different because acetic acid is weak, so acetate has enough basic character to alter solution pH.
Approximation Versus Quadratic Solution
In many classroom settings, the square root approximation is accepted because it is fast and accurate for weak-base systems where the ionization is small relative to the starting concentration. For sodium acetate at 15.0 M, the approximation works extremely well. Still, it is worth understanding the exact form.
Starting from:
Kb = x² / (C – x)
Rearranging gives the quadratic:
x² + Kb x – Kb C = 0
Solving this equation yields an hydroxide concentration that is nearly identical to the square root approximation because x is much smaller than C. The calculator above lets you use either method.
Important Real World Note About a 15M Solution
A 15M sodium acetate solution is extremely concentrated. In real chemical systems, activity effects become significant at high ionic strength, and ideal formulas become less exact. The simple calculation above assumes ideal behavior, meaning it uses concentrations directly as if they were activities. In advanced analytical chemistry, physical chemistry, or process chemistry, you would often need to correct for non-ideal behavior, especially at concentrations this high.
Even so, for textbook and many exam contexts, the ideal equilibrium method is the expected approach. If the problem simply asks for the pH of a 15M solution of CH3COONa, the standard answer remains about 9.96.
Reference Data for Acetic Acid and Water
| Property | Typical Value at 25 degrees Celsius | Why It Matters |
|---|---|---|
| Ka of acetic acid | 1.8 × 10^-5 | Used to derive Kb for acetate |
| pKa of acetic acid | 4.76 | Common logarithmic form used in buffer work |
| Kw of water | 1.0 × 10^-14 | Relates acid and base strengths and gives pH + pOH = 14 |
| Kb of acetate | 5.56 × 10^-10 | Directly controls hydroxide formation in sodium acetate solution |
pH Trend of Sodium Acetate at Different Concentrations
The table below uses the same ideal weak-base approximation with Ka = 1.8 × 10-5 and Kw = 1.0 × 10-14. It shows how pH rises as sodium acetate concentration increases. These values are useful for comparison and help explain why a 15M solution reaches the upper end of the typical sodium acetate pH range under ideal assumptions.
| CH3COONa Concentration | Calculated [OH-] | Calculated pOH | Calculated pH |
|---|---|---|---|
| 0.01 M | 2.36 × 10^-6 M | 5.63 | 8.37 |
| 0.10 M | 7.45 × 10^-6 M | 5.13 | 8.87 |
| 1.00 M | 2.36 × 10^-5 M | 4.63 | 9.37 |
| 5.00 M | 5.27 × 10^-5 M | 4.28 | 9.72 |
| 15.0 M | 9.13 × 10^-5 M | 4.04 | 9.96 |
Common Mistakes When Solving This Problem
- Treating sodium acetate as neutral. This ignores acetate hydrolysis and leads to an incorrect pH of 7.
- Using Ka directly instead of converting to Kb. Since acetate is a base, you must use the conjugate base constant.
- Forgetting to calculate pOH first. The equilibrium gives hydroxide, so pOH is the direct logarithmic step.
- Confusing molarity and molality. Many textbook examples say 15M, while some problems use 15m. Under concentrated conditions, these are not identical.
- Ignoring non-ideal behavior in advanced work. At 15M, activity corrections can matter in research or process calculations.
Quick Method Summary
- Write the hydrolysis reaction: CH3COO- + H2O ⇌ CH3COOH + OH-
- Calculate Kb from Kw/Ka
- Use [OH-] ≈ √(Kb × C) for the standard approximation
- Find pOH from the hydroxide concentration
- Compute pH = 14 – pOH at 25 degrees Celsius
Authoritative References for Further Study
LibreTexts Chemistry is a useful open educational reference, but for official and academic sources, see:
- NIST Chemistry WebBook for thermodynamic and chemical reference data.
- U.S. Environmental Protection Agency for water chemistry and pH background resources.
- Brigham Young University Chemistry for educational chemistry materials from a university domain.
Final Answer
If you calculate the pH of a 15M solution of CH3COONa using standard ideal equilibrium assumptions at 25 degrees Celsius, with Ka for acetic acid equal to 1.8 × 10-5, you obtain:
pH ≈ 9.96
That result comes from acetate hydrolysis, which produces hydroxide ions in water. The calculator on this page automates the full process and also lets you compare approximation and quadratic methods for better understanding.