Calculate the pH of a 0.055 M Solution of CH3COONa
Use this premium calculator to determine the pH, pOH, hydroxide concentration, and hydrolysis behavior of sodium acetate in water using either a quick approximation or an exact quadratic approach.
Interactive pH Calculator
For a 0.055 M sodium acetate solution at 25°C, the result should be mildly basic because acetate is the conjugate base of the weak acid acetic acid.
How to Calculate the pH of a 0.055 M Solution of CH3COONa
To calculate the pH of a 0.055 M solution of CH3COONa, you need to recognize first that sodium acetate is a salt formed from a strong base, sodium hydroxide, and a weak acid, acetic acid. Because the sodium ion does not hydrolyze in water to any meaningful extent, the chemistry is governed by the acetate ion, CH3COO-. Acetate behaves as a weak base in water and reacts with water molecules to produce a small amount of hydroxide ion. That hydroxide production is what makes the solution basic, pushing the pH above 7.
Many students initially think a salt solution should be neutral because it comes from an acid and a base. In reality, the acid and base strengths matter. A salt from a strong acid and strong base is typically neutral. A salt from a weak acid and strong base, such as sodium acetate, is basic. A salt from a strong acid and weak base is acidic. This is one of the central patterns in acid-base equilibrium, and sodium acetate is a classic example used in general chemistry and analytical chemistry.
Step 1: Identify the Relevant Equilibrium
Sodium acetate dissociates essentially completely in water:
CH3COONa → Na+ + CH3COO-
The sodium ion is a spectator ion, so the important equilibrium is:
CH3COO- + H2O ⇌ CH3COOH + OH-
This tells you that acetate removes a proton from water, generating hydroxide ion. Since hydroxide concentration increases, pOH decreases and pH rises above 7.
Step 2: Convert Ka of Acetic Acid into Kb for Acetate
The usual tabulated acid dissociation constant for acetic acid at 25°C is approximately 1.8 × 10-5. The relationship between the acid dissociation constant and the base dissociation constant of its conjugate base is:
Kb = Kw / Ka
Using Kw = 1.0 × 10-14 and Ka = 1.8 × 10-5:
Kb = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
That is a very small Kb, which means acetate is only a weak base. Even so, with a concentration of 0.055 M, it still produces enough hydroxide to make the solution distinctly basic.
Step 3: Set Up an ICE Table
Let the initial acetate concentration be 0.055 M. Let x be the amount that reacts with water:
- Initial: [CH3COO-] = 0.055, [CH3COOH] = 0, [OH-] = 0
- Change: [CH3COO-] = -x, [CH3COOH] = +x, [OH-] = +x
- Equilibrium: [CH3COO-] = 0.055 – x, [CH3COOH] = x, [OH-] = x
Insert those values into the base equilibrium expression:
Kb = [CH3COOH][OH-] / [CH3COO-] = x2 / (0.055 – x)
Step 4: Use the Weak Base Approximation
Because Kb is small and the concentration is much larger than the amount that reacts, x is expected to be very small compared with 0.055. That allows the standard approximation:
0.055 – x ≈ 0.055
So:
x2 / 0.055 = 5.56 × 10-10
x2 = (5.56 × 10-10)(0.055) = 3.06 × 10-11
x = [OH-] = √(3.06 × 10-11) = 5.53 × 10-6 M
Now calculate pOH:
pOH = -log(5.53 × 10-6) = 5.26
Then:
pH = 14.00 – 5.26 = 8.74
Why the Solution Is Basic
The reason this result is greater than 7 is conceptual, not just numerical. Acetate is the conjugate base of a weak acid. Weak acids do not donate protons very strongly, which means their conjugate bases are relatively more willing to accept protons. When acetate is placed in water, it partially accepts a proton from water:
CH3COO- + H2O ⇌ CH3COOH + OH-
The production of OH- makes the solution basic. In contrast, if you dissolved sodium chloride in water, neither Na+ nor Cl- would hydrolyze significantly, and the pH would remain close to 7. If you dissolved ammonium chloride, the NH4+ ion would behave as a weak acid and the solution would become acidic.
Approximation Versus Exact Calculation
For weak acids and weak bases, chemistry courses often use the square-root approximation because it is fast and usually very accurate. However, the exact method is also easy to handle with a calculator or software tool. Starting from:
Kb = x2 / (0.055 – x)
Rearranging gives:
x2 + Kb x – Kb(0.055) = 0
Solving that quadratic gives virtually the same answer as the approximation because the degree of hydrolysis is tiny. The percentage ionization is:
(5.53 × 10-6 / 0.055) × 100 ≈ 0.010%
Since this is far below 5%, the approximation is fully justified.
| Method | Equation Used | [OH-] for 0.055 M CH3COONa | pH | Practical Use |
|---|---|---|---|---|
| Approximation | x = √(KbC) | 5.53 × 10-6 M | 8.74 | Best for quick homework and exam work |
| Exact quadratic | x = [-Kb + √(Kb2 + 4KbC)] / 2 | 5.53 × 10-6 M | 8.74 | Best when checking approximation validity |
Important Constants and Their Typical Values
Results can shift slightly depending on the Ka value you use, because textbooks and databases may report acetic acid dissociation constants with small differences due to temperature, ionic strength, and rounding. Most general chemistry problems assume 25°C and use values near 1.8 × 10-5. Since pH is logarithmic, small changes in Ka create only small changes in the final pH.
| Quantity | Typical Value at 25°C | Role in This Calculation | Effect if Value Increases |
|---|---|---|---|
| Ka for CH3COOH | 1.8 × 10-5 | Used to find Kb of acetate | Larger Ka gives smaller Kb and slightly lower pH |
| Kw for water | 1.0 × 10-14 | Relates Ka and Kb | Larger Kw gives larger Kb and slightly higher pH |
| Initial CH3COONa concentration | 0.055 M | Starting base concentration | Higher concentration gives more OH- and higher pH |
| Computed Kb for CH3COO- | 5.56 × 10-10 | Determines hydrolysis extent | Higher Kb gives more hydrolysis and higher pH |
Common Mistakes Students Make
- Treating sodium acetate as a strong base. It is not. Sodium acetate contains acetate, which is only a weak base.
- Using Ka directly to find pH. You must first convert Ka to Kb because the reacting species in solution is acetate, not acetic acid.
- Forgetting to calculate pOH first. Since the base reaction generates OH-, you find pOH before converting to pH.
- Using 0.055 as x. The hydrolysis amount is tiny relative to the initial concentration.
- Ignoring temperature assumptions. If the problem does not state otherwise, 25°C and Kw = 1.0 × 10-14 are generally assumed.
Short Comparison With Similar Salts
Understanding sodium acetate becomes easier if you compare it to other salts. Sodium chloride gives a nearly neutral solution because both ions come from strong parent species. Ammonium chloride gives an acidic solution because NH4+ is the conjugate acid of a weak base. Sodium acetate gives a basic solution because acetate is the conjugate base of a weak acid. This classification framework is very useful in problem solving.
- NaCl: neutral, pH about 7
- NH4Cl: acidic, pH less than 7
- CH3COONa: basic, pH greater than 7
Real-World Relevance of Sodium Acetate pH
Sodium acetate is not just a classroom chemical. It is used in buffer preparation, food processing, textile applications, and laboratory chemistry. When sodium acetate is combined with acetic acid, it creates the well-known acetate buffer system. The pH behavior of the acetate ion is central in biochemistry, pharmaceutical formulations, and analytical procedures where moderate pH control is needed. A 0.055 M sodium acetate solution by itself is only mildly basic, but in combination with acetic acid it becomes part of a powerful and widely used buffer pair.
In analytical chemistry, understanding the hydrolysis of salts helps predict whether a sample matrix will shift pH during dilution or dissolution. In educational laboratories, sodium acetate is often chosen precisely because it illustrates weak-base hydrolysis without requiring dangerous strong bases. In industrial contexts, the acetate system can influence corrosion behavior, microbial stability, and compatibility with formulations.
Quick Worked Summary
- Write the hydrolysis reaction: CH3COO- + H2O ⇌ CH3COOH + OH-
- Use Ka for acetic acid: 1.8 × 10-5
- Convert to Kb: Kb = 1.0 × 10-14 / 1.8 × 10-5 = 5.56 × 10-10
- Use x = √(KbC) = √[(5.56 × 10-10)(0.055)] = 5.53 × 10-6
- Compute pOH = 5.26
- Compute pH = 14.00 – 5.26 = 8.74
Authoritative References for Acid-Base Data
Final Conclusion
The pH of a 0.055 M solution of CH3COONa is approximately 8.74 under standard 25°C conditions. The key idea is that sodium acetate dissociates to form acetate, and acetate is the conjugate base of the weak acid acetic acid. It hydrolyzes in water to produce OH-, making the solution basic. Once you know how to convert Ka to Kb and apply either the square-root approximation or the quadratic formula, this type of problem becomes straightforward and highly predictable.
If you use the calculator above, you can also explore how the result changes if the concentration, Ka, or Kw are varied. That makes it useful not just for obtaining the answer, but for understanding the full equilibrium behavior of weak-base salt solutions.