Calculate the pH of a 0.23 M CH3COONa Solution
Use this premium sodium acetate pH calculator to determine the alkalinity of a 0.23 M CH3COONa solution at 25 degrees Celsius. The tool applies weak base hydrolysis, converts Ka to Kb, and calculates pOH and pH with either an exact quadratic solution or the common square root approximation.
Sodium Acetate pH Calculator
Enter the molar concentration of sodium acetate.
Default value corresponds to acetic acid near 25 degrees Celsius.
Use 1.0 × 10-14 at 25 degrees Celsius.
Exact is more rigorous. Approximation is usually excellent for weak hydrolysis.
The calculator uses the constants you enter. This selector is informational and helps document your assumptions.
Results
Click Calculate pH to compute the pH of the sodium acetate solution and generate a comparison chart.
How to calculate the pH of a 0.23 M CH3COONa solution
Sodium acetate, written as CH3COONa, is the sodium salt of acetic acid. When it dissolves in water, it dissociates almost completely into sodium ions and acetate ions. The sodium ion is a spectator ion under these conditions, but the acetate ion acts as the conjugate base of a weak acid. That means acetate reacts with water to produce a small amount of hydroxide ion, making the solution basic. If you want to calculate the pH of a 0.23 M CH3COONa solution, the chemistry is a classic weak base hydrolysis problem.
The key idea is that acetate comes from acetic acid, which is a weak acid. Because acetic acid does not ionize fully, its conjugate base has measurable basicity. The hydrolysis reaction is:
To calculate pH, we convert the acid dissociation constant of acetic acid, Ka, into the base dissociation constant of acetate, Kb. At 25 degrees Celsius, a commonly used value for acetic acid is Ka = 1.8 × 10-5, and the ion product of water is Kw = 1.0 × 10-14. The relationship is:
Substituting the standard values:
Now set up the hydrolysis equilibrium. If the initial concentration of acetate is 0.23 M and x is the amount that reacts with water, then at equilibrium:
- [CH3COO–] = 0.23 – x
- [CH3COOH] = x
- [OH–] = x
The equilibrium expression becomes:
Because Kb is very small, x is tiny compared with 0.23, so many textbooks use the approximation 0.23 – x ≈ 0.23. That gives:
Since x equals the hydroxide concentration, the pOH is:
Then:
So the pH of a 0.23 M CH3COONa solution is approximately 9.05 at 25 degrees Celsius using the standard constant for acetic acid. If you solve the quadratic equation exactly instead of using the approximation, the answer changes only in the fourth or fifth decimal place, which confirms that the approximation is reliable here.
Why sodium acetate solutions are basic
Students often ask why a salt can make water acidic, neutral, or basic. The answer depends on the strengths of the parent acid and base that formed the salt. Sodium acetate is formed from sodium hydroxide, a strong base, and acetic acid, a weak acid. The sodium ion does not affect pH much, but the acetate ion does. Because acetate is the conjugate base of a weak acid, it accepts protons from water to create hydroxide ions. This raises the pH above 7.
By contrast, a salt such as sodium chloride comes from a strong acid and a strong base, so neither ion hydrolyzes significantly and the solution stays near neutral. A salt such as ammonium chloride comes from a weak base and a strong acid, so it produces an acidic solution. Understanding this pattern lets you predict pH behavior even before doing the math.
Quick rule for salts in water
- Strong acid + strong base salt: usually neutral
- Weak acid + strong base salt: usually basic
- Strong acid + weak base salt: usually acidic
Step by step method for this exact problem
- Identify CH3COONa as sodium acetate, a salt that produces acetate ions in water.
- Recognize acetate as the conjugate base of acetic acid.
- Use the relation Kb = Kw / Ka.
- Set up the ICE table for acetate hydrolysis.
- Solve for [OH–] using either the quadratic equation or the square root approximation.
- Calculate pOH from hydroxide concentration.
- Convert pOH to pH using pH + pOH = 14 at 25 degrees Celsius.
Exact calculation versus approximation
For weak acids and weak bases, the approximation method is popular because it is fast and usually very accurate. In this case, the exact equation is:
Rearranging gives:
Solving this quadratic gives x ≈ 1.13037 × 10-5 M. The approximation gave x ≈ 1.13039 × 10-5 M. The difference is practically negligible for most classroom and laboratory applications. That is why chemistry instructors often accept the approximation as long as the assumption is justified.
| Method | Hydroxide concentration, [OH–] | pOH | pH | Comment |
|---|---|---|---|---|
| Exact quadratic | 1.13037 × 10-5 M | 4.9469 | 9.0531 | Most rigorous calculation for the stated constants |
| Square root approximation | 1.13039 × 10-5 M | 4.9469 | 9.0531 | Difference is far below typical reporting precision |
Comparison data: how concentration affects sodium acetate pH
One important concept is that the pH of a sodium acetate solution depends on concentration. More concentrated solutions of the same salt generally produce a slightly higher pH because more acetate is available to hydrolyze and generate hydroxide. However, the change is not linear, because pH is a logarithmic scale and hydrolysis is governed by an equilibrium constant.
| CH3COONa concentration (M) | Calculated [OH–] (M) | Calculated pOH | Calculated pH at 25 degrees Celsius |
|---|---|---|---|
| 0.010 | 2.36 × 10-6 | 5.63 | 8.37 |
| 0.050 | 5.27 × 10-6 | 5.28 | 8.72 |
| 0.100 | 7.45 × 10-6 | 5.13 | 8.87 |
| 0.230 | 1.13 × 10-5 | 4.95 | 9.05 |
| 0.500 | 1.67 × 10-5 | 4.78 | 9.22 |
| 1.000 | 2.36 × 10-5 | 4.63 | 9.37 |
The table shows a clear pattern. Doubling concentration does not double pH because pH is tied to the logarithm of hydroxide concentration. Instead, the pH rises gradually. This is an important insight for students solving weak base hydrolysis problems and for anyone using sodium acetate in buffers, laboratory preparations, or process chemistry.
Common mistakes when solving this problem
1. Treating sodium acetate as a strong base
CH3COONa is not sodium hydroxide. It is a salt that produces a weak basic effect through acetate hydrolysis. If you assume complete hydroxide release, your pH will be far too high.
2. Using Ka directly instead of converting to Kb
The species in solution that controls pH is acetate, the conjugate base. That means you need Kb, not Ka. Always convert with Kb = Kw / Ka.
3. Forgetting to calculate pOH first
Because hydrolysis generates OH–, you first compute pOH and only then convert to pH. This is one of the most common sources of sign and logic errors.
4. Ignoring temperature assumptions
The values of Kw and even reported Ka can vary slightly with temperature and source. Most textbook answers assume 25 degrees Celsius. If your instructor or laboratory gives a different temperature, use the constants supplied for that condition.
When to use the Henderson-Hasselbalch equation instead
If your problem contains both acetic acid and sodium acetate together, you are dealing with a buffer rather than a pure salt solution. In that case, the Henderson-Hasselbalch equation is often the preferred method:
For a pure 0.23 M sodium acetate solution without added acetic acid, hydrolysis is the right framework. Some learners confuse these two scenarios because both involve acetate, but they are not solved the same way. Buffers depend on the ratio of acid to conjugate base. Pure salt hydrolysis depends on Kb and the initial salt concentration.
Real world uses of sodium acetate and why pH matters
Sodium acetate appears in analytical chemistry, biochemistry, food processing, and industrial applications. In laboratories, it is frequently used in buffer preparation and precipitation protocols. In biochemical workflows, acetate-based buffers can help maintain a desired pH range. In manufacturing and food systems, sodium acetate may serve flavor, preservation, or process control roles. In each case, understanding pH matters because acidity affects solubility, reaction rates, enzyme behavior, and product stability.
A 0.23 M sodium acetate solution is moderately concentrated, so its pH near 9.05 tells you the solution is clearly basic but not strongly caustic. That distinction matters. A pH around 9 is enough to influence many acid-sensitive systems, but it behaves very differently from a strong base such as 0.23 M NaOH, which would have a much higher pH.
Authoritative references for acid-base constants and water chemistry
For students, teachers, and professionals who want trusted reference material, the following sources are useful starting points:
- National Institute of Standards and Technology (NIST) for scientific reference data and chemical measurement standards.
- University of Wisconsin Chemistry acid-base equilibrium resources for educational treatment of weak acid and weak base equilibria.
- University of California, Berkeley Chemistry for advanced chemistry instruction and supporting educational material.
FAQ: calculate the pH of a 0.23 M CH3COONa solution
Is the solution acidic, neutral, or basic?
It is basic. The acetate ion hydrolyzes in water and produces hydroxide ion, raising the pH above 7.
What is the final pH?
Using Ka = 1.8 × 10-5 and Kw = 1.0 × 10-14 at 25 degrees Celsius, the pH is approximately 9.05.
Why do we use Kb if the given constant is Ka?
Because the species controlling the pH is acetate, which is a base. Acetate is the conjugate base of acetic acid, so you convert Ka to Kb first.
Can I use the square root shortcut safely here?
Yes. The hydrolysis is weak enough that the approximation is excellent. The exact and approximate results are essentially identical at standard reporting precision.
Would the answer change at another temperature?
Yes. Kw changes with temperature, and reported Ka values can vary slightly as well. If your course or process uses a different temperature, use the corresponding constants.
Bottom line
To calculate the pH of a 0.23 M CH3COONa solution, recognize sodium acetate as the salt of a weak acid and strong base. Convert the acid dissociation constant of acetic acid to the base dissociation constant of acetate, write the hydrolysis equilibrium, solve for hydroxide concentration, determine pOH, and finally calculate pH. With standard constants at 25 degrees Celsius, the result is pH ≈ 9.05. That makes the solution mildly basic, exactly as expected for a sodium acetate solution in water.