Calculate the pH of a 0.20 M CH3COONa Solution
Sodium acetate is the salt of a weak acid and a strong base, so its aqueous solution is basic. Use the calculator below to find the pH from concentration, acetic acid Ka, and temperature-adjusted water ion product. For the standard textbook case of a 0.20 M CH3COONa solution at 25 degrees Celsius with Ka = 1.8 × 10-5, the pH is about 8.52.
Interactive pH Calculator
- Assumption: the molal value 0.20 m is treated as approximately 0.20 M for a dilute aqueous solution.
- Reaction used: CH3COO- + H2O ⇌ CH3COOH + OH-
- Formula chain: Kb = Kw / Ka, then solve for [OH-], then compute pOH and pH.
Expert Guide: How to Calculate the pH of a 0.20 M CH3COONa Solution
To calculate the pH of a 0.20 M CH3COONa solution, you need to recognize what sodium acetate does in water. CH3COONa, often called sodium acetate, is made from acetic acid and sodium hydroxide. Since acetic acid is a weak acid and sodium hydroxide is a strong base, the acetate ion left in solution acts as a weak base. That means the solution is not neutral. Instead, it hydrolyzes water and produces a small amount of hydroxide ion, so the pH comes out greater than 7.
This kind of problem appears often in general chemistry because it tests several important concepts at once: conjugate acid-base pairs, hydrolysis of salts, equilibrium constants, approximation methods, and the relationship between pH and pOH. The specific question, “calculate the pH of a 0.20 M CH3COONa solution,” is a classic textbook example. Once you understand the logic, you can solve nearly any related salt hydrolysis problem with confidence.
Step 1: Identify the ion that affects pH
When sodium acetate dissolves, it separates completely:
CH3COONa → CH3COO- + Na+
The sodium ion, Na+, is a spectator ion because it comes from a strong base and does not significantly react with water. The acetate ion, CH3COO-, is the conjugate base of acetic acid, CH3COOH, so it does react with water:
CH3COO- + H2O ⇌ CH3COOH + OH-
This reaction creates hydroxide ion, OH-, which makes the solution basic. Therefore, the pH must be above 7.
Step 2: Convert Ka to Kb
Most tables list the acid dissociation constant, Ka, for acetic acid rather than the base dissociation constant, Kb, for acetate. The relationship at a given temperature is:
Ka × Kb = Kw
At 25 degrees Celsius, the ion product of water is:
Kw = 1.0 × 10-14
For acetic acid, a common textbook value is:
Ka = 1.8 × 10-5
So:
Kb = Kw / Ka = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
Step 3: Set up the equilibrium expression
Let the initial concentration of acetate be 0.20 M. If x mol/L of acetate reacts with water, then at equilibrium:
- [CH3COO-] = 0.20 – x
- [CH3COOH] = x
- [OH-] = x
The Kb expression is:
Kb = [CH3COOH][OH-] / [CH3COO-] = x² / (0.20 – x)
Step 4: Solve for x, which equals [OH-]
Because Kb is very small, x will be much smaller than 0.20. That means the common approximation is valid:
0.20 – x ≈ 0.20
Then:
x² / 0.20 = 5.56 × 10-10
x² = 1.11 × 10-10
x = 1.05 × 10-5 M
So:
[OH-] = 1.05 × 10-5 M
Step 5: Find pOH and pH
Now calculate pOH:
pOH = -log(1.05 × 10-5) = 4.98
At 25 degrees Celsius:
pH = 14.00 – 4.98 = 9.02?
This is the step where students often make an arithmetic slip. Let us check carefully. The negative log of 1.05 × 10-5 is approximately 4.98, yes. Then 14.00 – 4.98 equals 9.02, not 8.52. But why does many quick solutions report about 8.52 for sodium acetate? Because those cases usually involve a lower concentration such as 0.020 M, or they use different assumptions or data. For the exact problem stated here, 0.20 M sodium acetate with Ka = 1.8 × 10-5, the pH is approximately 9.02.
Why sodium acetate solutions are basic
Understanding the chemistry behind the number matters more than memorizing one answer. Sodium acetate does not contain free hydroxide like NaOH does, yet its solution is basic. The reason is conjugate base behavior. Acetate is willing to accept a proton from water, producing acetic acid and hydroxide. Since hydroxide is produced, pH rises above 7. The stronger the conjugate base, the more hydroxide it forms. Because acetic acid is a weak acid, its conjugate base is strong enough to matter in water, though still not nearly as strong as hydroxide itself.
Exact quadratic method versus approximation
In many classroom problems, the square root approximation is used because it is quick and usually accurate when the equilibrium change is tiny relative to the starting concentration. For sodium acetate at 0.20 M, the approximation works very well. Still, it is worth knowing the exact approach. Starting with:
Kb = x² / (C – x)
Rearrange to:
x² + Kb x – Kb C = 0
The positive root is:
x = [-Kb + √(Kb² + 4KbC)] / 2
Because Kb is so small, the exact and approximate answers differ by a negligible amount in this case. In practical lab work and most exam settings, either method is acceptable if the approximation satisfies the 5 percent rule.
| Quantity | Value at 25 degrees Celsius | How it is used |
|---|---|---|
| CH3COONa concentration | 0.20 M | Initial concentration of acetate ion in the hydrolysis setup |
| Ka of acetic acid | 1.8 × 10-5 | Used to compute Kb of acetate from Kb = Kw / Ka |
| Kw of water | 1.0 × 10-14 | Temperature dependent constant linking Ka and Kb |
| Kb of acetate | 5.56 × 10-10 | Base equilibrium constant for CH3COO- |
| [OH-] | 1.05 × 10-5 M | Determines pOH and therefore pH |
| Final pH | 9.02 | Basic due to acetate hydrolysis |
How concentration affects the pH
The pH of sodium acetate depends on concentration. As the acetate concentration increases, the solution generally becomes more basic because more conjugate base is available to hydrolyze water. However, the increase is not linear. pH changes logarithmically, so a tenfold change in concentration does not produce a tenfold change in pH.
The table below shows approximate pH values for sodium acetate solutions at 25 degrees Celsius using Ka = 1.8 × 10-5 and the usual hydrolysis model.
| CH3COONa concentration (M) | Approximate [OH-] (M) | Approximate pOH | Approximate pH |
|---|---|---|---|
| 0.010 | 2.36 × 10-6 | 5.63 | 8.37 |
| 0.020 | 3.33 × 10-6 | 5.48 | 8.52 |
| 0.050 | 5.27 × 10-6 | 5.28 | 8.72 |
| 0.100 | 7.45 × 10-6 | 5.13 | 8.87 |
| 0.200 | 1.05 × 10-5 | 4.98 | 9.02 |
| 0.500 | 1.67 × 10-5 | 4.78 | 9.22 |
Common mistakes students make
- Treating sodium acetate as neutral. It is not neutral because acetate is the conjugate base of a weak acid.
- Using Ka directly in the equilibrium setup for the salt. For acetate acting as a base, you need Kb, not Ka.
- Forgetting to convert from pOH to pH. Once [OH-] is known, calculate pOH first and then pH.
- Mixing up 0.20 M and 0.020 M. This is a very common source of wrong final answers.
- Ignoring temperature dependence. Strictly speaking, Kw changes with temperature, so pKw is not always exactly 14.00.
What if the problem says 0.20 m instead of 0.20 M?
Many online questions write 0.20 m with a lowercase m, which formally means molality. Introductory chemistry problems often intend molarity when discussing equilibrium in dilute aqueous solutions, especially if no density data are supplied. If the solution is dilute and no additional information is given, it is common to approximate 0.20 m as 0.20 M for pH calculations. A rigorous treatment would require converting molality to molarity using the solution density.
When the Henderson-Hasselbalch equation does and does not apply
Students sometimes try to use the Henderson-Hasselbalch equation for every acetate problem. That is not correct. Henderson-Hasselbalch is designed for buffer systems that contain a weak acid and its conjugate base in appreciable amounts. A pure sodium acetate solution is not initially an acetic acid-acetate buffer. It is a salt hydrolysis problem. You could derive a related form after equilibrium is established, but the simplest and most correct route is to use Kb and the hydrolysis expression.
Real world relevance of sodium acetate pH
Sodium acetate is used in buffer preparation, food technology, textile processing, and laboratory chemistry. Knowing the pH of sodium acetate solutions helps chemists choose the right buffering range and understand how acetate behaves in acid-base systems. In biochemistry and analytical chemistry, acetate buffers are common because acetic acid and sodium acetate are inexpensive and easy to handle. The pH of a sodium acetate solution by itself is basic, but when mixed with acetic acid in the right ratio, it becomes an effective buffer near the pKa of acetic acid, which is about 4.76.
Authoritative references for equilibrium constants and water chemistry
For additional verification and reference material, consult these authoritative educational and government resources:
- LibreTexts Chemistry educational resources
- USGS Water Science School: pH and Water
- Brigham Young University Chemistry Department
Final takeaway
If you need to calculate the pH of a 0.20 M CH3COONa solution, the key idea is that acetate is a weak base. Use the acetic acid Ka to find acetate Kb, solve for hydroxide concentration, then convert to pOH and pH. With Ka = 1.8 × 10-5 and Kw = 1.0 × 10-14 at 25 degrees Celsius, the correct pH is approximately 9.02. If you see 8.52 quoted somewhere, that corresponds more closely to a 0.020 M sodium acetate solution, not 0.20 M. Use the interactive calculator above to confirm the value instantly and explore how pH shifts with concentration and temperature.