Calculate the pH of a 0.800 M Aqueous NaCH3CO2 Solution
Use this premium weak-base hydrolysis calculator to find the pH, pOH, hydroxide concentration, and acetate equilibrium values for an aqueous sodium acetate solution. The default setup matches the classic chemistry problem: a 0.800 M NaCH3CO2 solution at 25 degrees C.
NaCH3CO2 fully dissociates into Na+ and CH3CO2-. The acetate ion acts as a weak base in water, producing a mildly basic solution.
Default expected result for 0.800 M sodium acetate with Ka = 1.8 x 10-5: pH is about 9.32. Minor rounding differences can occur depending on constants and method.
How to calculate the pH of a 0.800 M aqueous NaCH3CO2 solution
To calculate the pH of a 0.800 M aqueous sodium acetate solution, you first need to identify what sodium acetate actually does in water. The formula NaCH3CO2 is another way of writing sodium acetate, often written as NaC2H3O2 or CH3COONa. In water, sodium acetate dissociates essentially completely into sodium ions and acetate ions:
NaCH3CO2(aq) → Na+(aq) + CH3CO2–(aq)
The sodium ion is a spectator ion for acid-base behavior, but the acetate ion matters a lot. Acetate is the conjugate base of acetic acid, a weak acid. Because it is the conjugate base of a weak acid, acetate reacts with water to produce hydroxide:
CH3CO2– + H2O ⇌ CH3CO2H + OH–
That hydrolysis reaction is why the solution is basic. The pH is not found by assuming complete hydroxide production, because acetate is a weak base, not a strong base. Instead, you use the base dissociation constant Kb, which is related to the acid dissociation constant Ka of acetic acid through the relationship:
Kb = Kw / Ka
At 25 degrees C, the accepted classroom values are usually Kw = 1.0 x 10-14 and Ka for acetic acid = 1.8 x 10-5. Substituting gives:
Kb = (1.0 x 10-14) / (1.8 x 10-5) = 5.56 x 10-10
Now set up an ICE table for the hydrolysis of acetate. Let the initial acetate concentration be 0.800 M and let x be the amount that reacts:
- Initial: [CH3CO2–] = 0.800, [CH3CO2H] = 0, [OH–] = 0
- Change: minus x, plus x, plus x
- Equilibrium: [CH3CO2–] = 0.800 – x, [CH3CO2H] = x, [OH–] = x
The equilibrium expression is:
Kb = x2 / (0.800 – x)
Because Kb is very small, most general chemistry courses allow the approximation 0.800 – x ≈ 0.800. Then:
x = √(Kb x C) = √[(5.56 x 10-10)(0.800)] = 2.11 x 10-5 M
This x value is the hydroxide concentration. Therefore:
pOH = -log(2.11 x 10-5) = 4.68
pH = 14.00 – 4.68 = 9.32
Final answer: The pH of a 0.800 M aqueous NaCH3CO2 solution at 25 degrees C is approximately 9.32.
If you solve the equilibrium equation with the exact quadratic instead of the approximation, you get essentially the same answer because x is tiny compared with 0.800 M. That is why sodium acetate solutions are basic but only moderately so. They are nowhere near the pH of a strong base such as sodium hydroxide, even at high concentration.
Why sodium acetate is basic in water
The logic behind this calculation becomes easier once you connect it to conjugate acid-base chemistry. Acetic acid is a weak acid, so its conjugate base, acetate, has measurable basicity. In contrast, chloride is the conjugate base of hydrochloric acid, a strong acid, so chloride is essentially nonbasic in water. This comparison explains why sodium chloride solutions are neutral while sodium acetate solutions are basic.
When sodium acetate dissolves, the acetate ion partially removes protons from water, forming acetic acid and hydroxide. The extent of that proton transfer is controlled by Kb. Since Kb is small, only a small fraction of acetate ions hydrolyze, but because the initial concentration here is as large as 0.800 M, the hydroxide concentration still becomes large enough to push the pH above 9.
Key concepts to remember
- NaCH3CO2 is a salt of a strong base and a weak acid.
- Salts of strong bases and weak acids usually produce basic solutions.
- The relevant equilibrium is the hydrolysis of the conjugate base, not the dissociation of the original weak acid.
- You normally start with Ka for the weak acid and convert it to Kb.
- For acetate at ordinary concentrations, the square root approximation is usually excellent.
This chemistry also explains why sodium acetate appears in buffer systems. A mixture of acetic acid and sodium acetate resists pH change because it contains both a weak acid and its conjugate base. In the problem on this page, however, you have only the salt, so the system behaves as a weak base solution rather than a full buffer.
Step by step expert method
- Write the dissociation of the salt. Sodium acetate separates into Na+ and CH3CO2–.
- Identify the acid-base active species. Na+ does not affect pH, but CH3CO2– does.
- Write the hydrolysis reaction. CH3CO2– + H2O ⇌ CH3CO2H + OH–.
- Convert Ka to Kb. Use Kb = Kw / Ka.
- Set up an ICE table. Begin with 0.800 M acetate.
- Solve for x. Either use the exact quadratic or the approximation x = √(KbC).
- Find pOH. pOH = -log[OH–].
- Convert to pH. At 25 degrees C, pH = 14.00 – pOH.
With the default values used in this calculator, the numbers work out to approximately [OH–] = 2.11 x 10-5 M, pOH = 4.68, and pH = 9.32. The equilibrium concentration of acetic acid formed is the same as x, while the remaining acetate concentration is still very close to 0.800 M.
| Quantity | Value used | Meaning in the calculation |
|---|---|---|
| Initial sodium acetate concentration | 0.800 M | Starting acetate concentration after full salt dissociation |
| Ka of acetic acid | 1.8 x 10-5 | Used to derive Kb for acetate |
| Kw at 25 degrees C | 1.0 x 10-14 | Water ion product at standard classroom temperature |
| Kb of acetate | 5.56 x 10-10 | Controls hydrolysis and OH– production |
| [OH–] at equilibrium | 2.11 x 10-5 M | Obtained from the weak base equilibrium |
| Final pH | 9.32 | Mildly basic solution |
Exact solution versus approximation
Chemistry students are often told to use the approximation method if x is small relative to the initial concentration. For this problem, that shortcut is very reliable because the hydroxide concentration is only around 2.11 x 10-5 M while the initial acetate concentration is 0.800 M. The percent ionization is therefore tiny.
Using the exact quadratic formula, the equilibrium hydroxide concentration is:
x = [-Kb + √(Kb2 + 4KbC)] / 2
When you substitute Kb = 5.56 x 10-10 and C = 0.800, the exact x differs from the approximation by an amount far smaller than ordinary lab uncertainty for a simple classroom problem.
| Method | [OH–] (M) | pOH | pH | Practical note |
|---|---|---|---|---|
| Approximation x = √(KbC) | 2.108 x 10-5 | 4.676 | 9.324 | Fastest and acceptable for most coursework |
| Exact quadratic | 2.108 x 10-5 | 4.676 | 9.324 | Best for precision and automated calculators |
For this reason, both methods in the calculator on this page lead to nearly identical pH values. The exact mode is useful when concentrations are lower, when constants are larger, or when a teacher specifically asks for a quadratic solution.
How concentration changes the pH of sodium acetate
The pH of a sodium acetate solution increases as concentration increases, but the relationship is not linear. Because weak base hydrolysis follows a square root dependence in the approximation, increasing concentration by a factor of 100 does not increase hydroxide concentration by a factor of 100. Instead, it increases hydroxide concentration by a factor of 10.
This is a useful check on intuition. A 0.800 M sodium acetate solution is more basic than a 0.0800 M solution, but not dramatically more basic. Weak electrolytes and conjugate bases often show this compressed response because equilibrium limits how far the reaction proceeds.
| NaCH3CO2 concentration | Calculated [OH–] | Approximate pOH | Approximate pH |
|---|---|---|---|
| 0.0100 M | 2.36 x 10-6 M | 5.63 | 8.37 |
| 0.100 M | 7.45 x 10-6 M | 5.13 | 8.87 |
| 0.800 M | 2.11 x 10-5 M | 4.68 | 9.32 |
| 1.00 M | 2.36 x 10-5 M | 4.63 | 9.37 |
Notice that going from 0.100 M to 0.800 M increases the pH by only about 0.45 units. That is perfectly normal for weak base chemistry. The logarithmic pH scale compresses changes, and the equilibrium itself prevents complete hydrolysis.
Common mistakes students make
- Treating sodium acetate as a strong base. It is not equivalent to NaOH. Only partial hydrolysis occurs.
- Using Ka directly instead of Kb. Since acetate is the base in solution, you must convert Ka of acetic acid into Kb of acetate.
- Forgetting that sodium is a spectator ion. Na+ does not control pH here.
- Using 0.800 M as the OH– concentration. That would be correct only for a strong base with complete OH– release.
- Mixing up pH and pOH. Once you find hydroxide concentration, compute pOH first, then pH.
- Ignoring temperature dependence. If a problem gives a different temperature, pKw may not be exactly 14.00.
These mistakes usually produce pH values that are far too high. For example, if someone incorrectly treated 0.800 M sodium acetate like 0.800 M sodium hydroxide, they would estimate a pH near 13.9, which is completely unrealistic for acetate.
Authoritative chemistry references
If you want to verify constants, review acid-base fundamentals, or compare pH conventions, these sources are worth consulting:
- NIST Chemistry WebBook entry for acetic acid
- U.S. EPA overview of pH and what the pH scale means
- Purdue University guide to weak acid and weak base equilibrium calculations
These references help ground the classroom calculation in accepted data and standard acid-base methodology. NIST is especially useful when you want property data, while university chemistry pages can be helpful for worked equilibrium strategies.
Bottom line
To calculate the pH of a 0.800 M aqueous NaCH3CO2 solution, recognize that sodium acetate is the salt of a weak acid and a strong base. The acetate ion is the active base, so you calculate Kb from acetic acid’s Ka, solve the hydrolysis equilibrium, find [OH–], convert to pOH, and then convert to pH. Using standard constants at 25 degrees C gives a final pH of about 9.32.
This result is chemically sensible: the solution is basic, but only moderately basic, because acetate is a weak base. The calculator above automates each step, shows the equilibrium values, and plots a chart so you can interpret the result visually as well as numerically.