Calculate the pH of a 0.23 M CH3COONa Solution
Use this premium calculator to determine the pH, pOH, hydroxide concentration, and acetate hydrolysis behavior for a sodium acetate solution. The default setup is tuned for a 0.23 M CH3COONa solution at 25 degrees Celsius using the accepted acetic acid dissociation constant.
Sodium Acetate pH Calculator
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How to calculate the pH of a 0.23 M CH3COONa solution
To calculate the pH of a 0.23 M CH3COONa solution, you need to recognize what sodium acetate does in water. Sodium acetate, written as CH3COONa, dissociates essentially completely into sodium ions and acetate ions:
CH3COONa → Na+ + CH3COO-
The sodium ion is a spectator ion because it comes from the strong base sodium hydroxide. The acetate ion, however, is the conjugate base of acetic acid. That means it can react with water and generate hydroxide ions:
CH3COO- + H2O ⇌ CH3COOH + OH-
Because hydroxide is produced, the solution becomes basic, so the pH is greater than 7 at 25 degrees Celsius. The core of the problem is therefore a weak base hydrolysis calculation, not a simple neutral salt calculation.
Step 1: Identify the relevant equilibrium constant
Most chemistry students are given the acid dissociation constant, Ka, for acetic acid rather than the base dissociation constant, Kb, for acetate. At 25 degrees Celsius, acetic acid has a Ka of about 1.8 × 10-5. Since acetate is the conjugate base of acetic acid, its Kb is found from:
Kb = Kw / Ka
With Kw = 1.0 × 10-14 at 25 degrees Celsius:
Kb = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
This number tells you acetate is a weak base, which means it only partially reacts with water.
Step 2: Set up the ICE table
Let the initial acetate concentration be 0.23 M. If x is the amount that hydrolyzes, then:
- Initial: [CH3COO-] = 0.23, [CH3COOH] = 0, [OH-] = 0
- Change: [CH3COO-] = -x, [CH3COOH] = +x, [OH-] = +x
- Equilibrium: [CH3COO-] = 0.23 – x, [CH3COOH] = x, [OH-] = x
Substituting into the equilibrium expression gives:
Kb = x2 / (0.23 – x)
Since Kb is very small, x is much smaller than 0.23, so the common approximation is:
Kb ≈ x2 / 0.23
Solving for x:
x = √(Kb × 0.23) = √(5.56 × 10-10 × 0.23) ≈ 1.13 × 10-5 M
This x value is the hydroxide concentration:
[OH-] ≈ 1.13 × 10-5 M
Step 3: Convert hydroxide concentration to pOH and pH
Once [OH-] is known, calculate pOH:
pOH = -log[OH-] = -log(1.13 × 10-5) ≈ 4.95
At 25 degrees Celsius, pH + pOH = 14.00. Therefore:
pH = 14.00 – 4.95 = 9.05
Exact vs approximate answer
If you solve the full quadratic equation instead of using the small x approximation, the answer is almost identical for this concentration. The exact expression is:
x = [-Kb + √(Kb2 + 4KbC)] / 2
Here C = 0.23 M and Kb = 5.56 × 10-10. The exact value of x still comes out very close to 1.13 × 10-5 M, so the resulting pH remains about 9.05. This is why the approximation is considered valid in introductory and intermediate equilibrium problems.
Why sodium acetate is basic
A common point of confusion is why a salt can change pH at all. The rule is tied to the strengths of the parent acid and base:
- A salt from a strong acid and strong base is usually neutral.
- A salt from a strong acid and weak base is acidic.
- A salt from a weak acid and strong base is basic.
Sodium acetate belongs to the third category. Acetic acid is weak, while sodium hydroxide is strong. As a result, acetate has enough basic character to pull a proton from water, making hydroxide ions and raising the pH.
Quick worked example for 0.23 M CH3COONa
- Write the hydrolysis reaction: CH3COO- + H2O ⇌ CH3COOH + OH-
- Find Kb from Kw/Ka.
- Use Ka = 1.8 × 10-5, so Kb = 5.56 × 10-10.
- Set x = [OH-] and use x = √(KbC).
- Substitute C = 0.23 M to get [OH-] ≈ 1.13 × 10-5 M.
- Calculate pOH ≈ 4.95.
- Calculate pH ≈ 9.05.
Comparison table: sodium acetate concentration vs pH at 25 degrees Celsius
The table below uses Ka = 1.8 × 10-5 and the weak base relationship to estimate pH values. These values show how the pH slowly rises as the sodium acetate concentration increases.
| CH3COONa concentration (M) | Kb of acetate | Estimated [OH-] (M) | Estimated pOH | Estimated pH at 25 degrees Celsius |
|---|---|---|---|---|
| 0.010 | 5.56 × 10-10 | 2.36 × 10-6 | 5.63 | 8.37 |
| 0.050 | 5.56 × 10-10 | 5.27 × 10-6 | 5.28 | 8.72 |
| 0.100 | 5.56 × 10-10 | 7.45 × 10-6 | 5.13 | 8.87 |
| 0.230 | 5.56 × 10-10 | 1.13 × 10-5 | 4.95 | 9.05 |
| 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 |
Temperature matters more than many students expect
In many classroom examples, pH + pOH is treated as exactly 14.00. That is valid only at 25 degrees Celsius. As temperature changes, the ionic product of water, Kw, changes too. Since pKw = -log(Kw), the pH corresponding to neutrality also shifts. If your instructor specifies a different temperature, you should use the correct pKw value for that temperature rather than automatically subtracting from 14.00.
| Temperature (degrees Celsius) | Approximate pKw | Neutral pH | Why it matters for CH3COONa calculations |
|---|---|---|---|
| 0 | 14.94 | 7.47 | Higher pKw means pH values shift upward for the same pOH framework. |
| 10 | 14.52 | 7.26 | Still above the 25 degree benchmark, so neutrality is not 7.00. |
| 25 | 14.00 | 7.00 | Standard reference used in most textbook sodium acetate problems. |
| 40 | 13.83 | 6.92 | Lower pKw means pH is not found by assuming a fixed 14.00 sum. |
| 60 | 13.26 | 6.63 | Warm water shifts the neutral point significantly lower. |
Common mistakes when solving this problem
- Treating CH3COONa as neutral. It is not neutral because acetate hydrolyzes in water.
- Using Ka directly in the ICE table. The reacting species is acetate, so you need Kb for the base hydrolysis equilibrium.
- Forgetting to convert from pOH to pH. The equilibrium gives [OH-], not [H3O+].
- Assuming pH + pOH = 14.00 at all temperatures. This is true only at 25 degrees Celsius.
- Using the wrong parent acid. Acetate is the conjugate base of acetic acid only, so use acetic acid data.
When the approximation is safe
The approximation x << C is usually safe if the percent ionization is small. For this problem:
Percent hydrolysis = (1.13 × 10-5 / 0.23) × 100 ≈ 0.0049%
This is far below 5%, so the approximation is excellent. In fact, it is more than adequate for almost every educational calculation involving 0.23 M sodium acetate.
What the result means chemically
A pH of about 9.05 means the solution is mildly basic, not strongly caustic. It has more hydroxide than pure water, but the basicity is limited because acetate is only a weak base. This matters in analytical chemistry and buffer preparation because sodium acetate is often combined with acetic acid to make acetate buffers used in laboratories, biochemistry workflows, and titration systems.
Authority links for deeper study
- USGS: pH and Water
- NIST Chemistry WebBook: Acetic Acid Data
- University of Wisconsin: Weak Acids and Bases Tutorial
Bottom line
If you need to calculate the pH of a 0.23 M CH3COONa solution, the most direct path is to identify acetate as a weak base, convert Ka to Kb using Kw/Ka, solve for hydroxide concentration, and then convert to pOH and pH. Using Ka = 1.8 × 10-5 at 25 degrees Celsius, the answer is approximately pH = 9.05. The calculator above automates both the exact and approximate versions, and the chart helps you visualize how concentration affects the pH of sodium acetate solutions.