Calculate pH of 0.2500 M Solution of CH3COONa
Use this premium calculator to determine the pH of sodium acetate solution by applying weak base hydrolysis chemistry at 25 degrees Celsius. Instant results, step-by-step values, and a dynamic chart are included.
How to calculate the pH of a 0.2500 M solution of CH3COONa
To calculate the pH of a 0.2500 M solution of CH3COONa, you need to recognize that sodium acetate is a salt formed from a strong base, NaOH, and a weak acid, acetic acid, CH3COOH. Because the sodium ion does not significantly affect pH in water, the chemistry is controlled by the acetate ion, CH3COO–. Acetate is the conjugate base of acetic acid, so it reacts with water to produce hydroxide ions. That hydrolysis reaction makes the solution basic.
The governing equilibrium is:
CH3COO– + H2O ⇌ CH3COOH + OH–
This means the pH is not found by treating sodium acetate as a strong base. Instead, you calculate the base dissociation constant of acetate, solve for hydroxide concentration, obtain pOH, and then convert to pH.
Step 1: Identify the known concentration
The problem states that the solution concentration is 0.2500 M. Since sodium acetate dissociates essentially completely in water, the initial acetate concentration is also about 0.2500 M:
- [CH3COO–]initial = 0.2500 M
- [Na+] is spectator for pH purposes
- The solution is expected to be basic because acetate consumes water and generates OH–
Step 2: Convert Ka of acetic acid into Kb for acetate
Most chemistry references list the acid dissociation constant for acetic acid rather than the base constant for acetate. At 25 degrees Celsius, a common value is:
Ka = 1.8 × 10-5
Use the relationship:
Kb = Kw / Ka
With Kw = 1.0 × 10-14,
Kb = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
Step 3: Set up the equilibrium expression
For the hydrolysis of acetate:
CH3COO– + H2O ⇌ CH3COOH + OH–
If x is the amount of acetate that reacts, the equilibrium concentrations are:
- [CH3COO–] = 0.2500 – x
- [CH3COOH] = x
- [OH–] = x
The equilibrium expression is:
Kb = x2 / (0.2500 – x)
Step 4: Apply the weak base approximation
Because Kb is very small, x will be much smaller than 0.2500. That allows the common approximation:
0.2500 – x ≈ 0.2500
Then:
x2 = Kb × 0.2500
x = √(5.56 × 10-10 × 0.2500)
x = √(1.39 × 10-10) = 1.18 × 10-5 M
So the hydroxide concentration is:
[OH–] = 1.18 × 10-5 M
Step 5: Find pOH and pH
Now compute pOH:
pOH = -log(1.18 × 10-5) = 4.93
Finally:
pH = 14.00 – 4.93 = 9.07
Therefore, the pH of a 0.2500 M sodium acetate solution is approximately 9.07 at 25 degrees Celsius.
Why sodium acetate gives a basic solution
This is one of the most important concepts in acid-base chemistry. Salts do not all produce neutral solutions. The pH depends on the acid and base that formed the salt:
- Strong acid + strong base gives a nearly neutral salt
- Weak acid + strong base gives a basic salt
- Strong acid + weak base gives an acidic salt
- Weak acid + weak base depends on relative strengths
Sodium acetate belongs to the second category. Acetic acid is weak, while sodium hydroxide is strong. The acetate ion survives in water as a measurable base and hydrolyzes enough to push the pH above 7. This is why the pH of sodium acetate is basic even though the solution contains no obvious hydroxide source like NaOH.
Approximate method versus exact method
For many homework and laboratory calculations, the approximation method is ideal because it is fast and very accurate when x is tiny compared with the initial concentration. Still, it is good practice to know when the exact quadratic form matters.
| Method | Equation Used | OH- Result for 0.2500 M CH3COONa | Calculated pH | Use Case |
|---|---|---|---|---|
| Approximation | x ≈ √(KbC) | 1.18 × 10^-5 M | 9.07 | Most general chemistry problems |
| Exact quadratic | x = [-Kb + √(Kb² + 4KbC)] / 2 | 1.18 × 10^-5 M | 9.07 | Higher precision checks |
In this case, the exact and approximate methods agree to the displayed precision because the hydrolysis extent is extremely small relative to 0.2500 M. The percent ionization is only a tiny fraction of the total acetate concentration.
ICE table approach for CH3COONa
Students often find the ICE method helpful because it organizes initial, change, and equilibrium values clearly.
- Initial: [CH3COO–] = 0.2500, [CH3COOH] = 0, [OH–] = 0
- Change: -x, +x, +x
- Equilibrium: 0.2500 – x, x, x
- Substitute into Kb: Kb = x² / (0.2500 – x)
- Solve for x
Once x is found, that value is the hydroxide ion concentration generated by acetate hydrolysis. The rest of the solution follows from the standard logarithmic formulas for pOH and pH.
Comparison table: pH of sodium acetate at different concentrations
One useful way to build intuition is to compare pH values across a range of sodium acetate concentrations while keeping Ka = 1.8 × 10^-5 and temperature = 25 degrees Celsius. Higher concentration leads to a somewhat higher pH, but the rise is not dramatic because the hydroxide concentration depends on the square root of concentration.
| CH3COONa Concentration (M) | Kb of Acetate | Approx. [OH-] (M) | pOH | pH |
|---|---|---|---|---|
| 0.0100 | 5.56 × 10^-10 | 2.36 × 10^-6 | 5.63 | 8.37 |
| 0.0500 | 5.56 × 10^-10 | 5.27 × 10^-6 | 5.28 | 8.72 |
| 0.1000 | 5.56 × 10^-10 | 7.45 × 10^-6 | 5.13 | 8.87 |
| 0.2500 | 5.56 × 10^-10 | 1.18 × 10^-5 | 4.93 | 9.07 |
| 0.5000 | 5.56 × 10^-10 | 1.67 × 10^-5 | 4.78 | 9.22 |
Common mistakes when solving this problem
- Using Ka directly to find pH. You need Kb for acetate, not Ka for acetic acid by itself.
- Treating sodium acetate as neutral. The acetate ion hydrolyzes and raises pH above 7.
- Assuming CH3COONa is a strong base. It is not. The base behavior is weak and equilibrium based.
- Forgetting the pOH to pH conversion. The hydrolysis calculation gives OH–, so pOH comes first.
- Ignoring temperature. The values of Kw and Ka can shift slightly with temperature.
Real chemistry context and why this matters
Sodium acetate appears in introductory chemistry because it teaches a powerful general principle: conjugate bases of weak acids create basic solutions. That same concept appears in buffer design, analytical chemistry, biochemistry, environmental chemistry, and industrial formulation.
For example, acetate chemistry is used in acetate buffers, common biochemical media, and practical pH control systems. Although a pure sodium acetate solution is basic, a mixture of sodium acetate and acetic acid forms a classic buffer. In that case, the Henderson-Hasselbalch equation becomes useful. But for the problem on this page, where the solute is sodium acetate alone, the proper route is hydrolysis equilibrium.
Authoritative references for acid-base constants and water chemistry
If you want to verify values or read more deeply, these sources are excellent starting points:
Among these, NIST and EPA are authoritative scientific resources, and Chemistry LibreTexts is hosted by educational institutions and widely used in chemistry instruction. If you are studying equilibrium constants, pH scales, or aqueous systems, these are dependable references to consult.
Quick summary for exam use
- Recognize CH3COONa as a salt of a weak acid and strong base.
- Write hydrolysis: CH3COO– + H2O ⇌ CH3COOH + OH–.
- Compute Kb = Kw / Ka = 1.0 × 10^-14 / 1.8 × 10^-5 = 5.56 × 10^-10.
- Use x ≈ √(KbC) = √[(5.56 × 10^-10)(0.2500)] = 1.18 × 10^-5 M.
- pOH = 4.93.
- pH = 14.00 – 4.93 = 9.07.
If you remember that sodium acetate is basic because acetate is the conjugate base of a weak acid, this problem becomes much easier. The final pH is not extreme, but it is definitely greater than 7. That is the expected chemical behavior for a moderately concentrated sodium acetate solution in water.