19 calculate the pH of a 0.36 M CH3COONa solution
Use this premium calculator to determine the pH of a sodium acetate solution from concentration, Ka of acetic acid, and Kw. The tool supports both the common approximation and the exact quadratic method, then visualizes how pH changes with concentration.
For sodium acetate, the acetate ion acts as a weak base in water: CH3COO- + H2O ⇌ CH3COOH + OH-. Because sodium acetate is a salt of a weak acid and a strong base, its aqueous solution is basic.
Calculated Result
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How to calculate the pH of a 0.36 M CH3COONa solution
If you need to solve the problem “19 calculate the pH of a 0.36 M CH3COONa solution,” the key idea is that sodium acetate, CH3COONa, is the salt of a weak acid, acetic acid, and a strong base, sodium hydroxide. In water, the sodium ion is essentially neutral, while the acetate ion undergoes hydrolysis and produces hydroxide ions. That means the solution becomes basic, so the final pH must be greater than 7 at 25 C.
The chemistry begins with complete dissociation of sodium acetate: sodium acetate dissolves to give Na+ and CH3COO-. The sodium ion does not significantly react with water, but the acetate ion does: CH3COO- + H2O ⇌ CH3COOH + OH-. This equilibrium is described by the base dissociation constant Kb of acetate. Since most data tables provide Ka for acetic acid, we compute Kb from the relation Kb = Kw / Ka. At 25 C, Kw is 1.0 × 10^-14, and for acetic acid a common textbook value is Ka = 1.8 × 10^-5. Therefore, Kb for acetate is about 5.56 × 10^-10.
Step by step setup
- Write the hydrolysis equilibrium for acetate: CH3COO- + H2O ⇌ CH3COOH + OH-.
- Compute Kb using Kb = Kw / Ka.
- Use the initial acetate concentration, 0.36 M, as the starting concentration of the weak base.
- Solve for [OH-] either with the approximation or the exact quadratic equation.
- Calculate pOH = -log[OH-].
- Convert to pH using pH = 14 – pOH at 25 C.
Approximation method
In many classroom problems, the hydrolysis of a weak base is treated with the small x approximation. Let x = [OH-] formed. Then:
Kb = x^2 / (0.36 – x)
Because Kb is very small compared with the concentration, x is much smaller than 0.36, so we approximate 0.36 – x ≈ 0.36. That gives:
x = √(Kb × C) = √((5.56 × 10^-10)(0.36)) = √(2.00 × 10^-10) ≈ 1.41 × 10^-5 M
Then:
- pOH = -log(1.41 × 10^-5) ≈ 4.85
- pH = 14.00 – 4.85 ≈ 9.15
So the pH of a 0.36 M CH3COONa solution is approximately 9.15 at 25 C when Ka = 1.8 × 10^-5.
Exact quadratic method
The exact treatment avoids the approximation. Starting from:
Kb = x^2 / (C – x)
Rearranging:
x^2 + Kb x – Kb C = 0
Solve with the quadratic formula:
x = [-Kb + √(Kb^2 + 4KbC)] / 2
With C = 0.36 M and Kb = 5.56 × 10^-10, the exact x value is virtually identical to the approximation for this concentration. That is why the approximation is accepted in most general chemistry and analytical chemistry contexts. The calculator above can show both methods so you can compare them instantly.
Why CH3COONa gives a basic solution
Students often ask why a salt can change pH even though salts seem neutral at first glance. The answer depends on the acid and base from which the salt is formed. Sodium acetate comes from acetic acid, a weak acid, and sodium hydroxide, a strong base. The conjugate base of a weak acid, here acetate, still has enough basic character to accept protons from water. That process generates OH-, which raises the pH.
By contrast, salts from a strong acid and a strong base, such as NaCl, are essentially neutral because neither ion hydrolyzes appreciably. Salts from a strong acid and weak base, such as NH4Cl, produce acidic solutions. Recognizing this pattern lets you predict pH behavior before calculating anything:
- Strong acid + strong base salt: approximately neutral
- Weak acid + strong base salt: basic
- Strong acid + weak base salt: acidic
- Weak acid + weak base salt: depends on relative Ka and Kb
Comparison table: common salt solution behavior at 25 C
| Salt | Parent Acid | Parent Base | Expected Solution Behavior | Typical pH Trend |
|---|---|---|---|---|
| NaCl | HCl, strong | NaOH, strong | Nearly neutral | About 7 |
| CH3COONa | CH3COOH, weak | NaOH, strong | Basic | Greater than 7 |
| NH4Cl | HCl, strong | NH3, weak | Acidic | Less than 7 |
| Na2CO3 | H2CO3, weak | NaOH, strong | More strongly basic | Clearly above 7 |
Data table: acetate concentration vs approximate pH
One useful way to understand sodium acetate chemistry is to see how pH changes with concentration. The values below use Ka = 1.8 × 10^-5 and Kw = 1.0 × 10^-14 at 25 C. These are approximate textbook values and match the same equilibrium logic used in the calculator.
| [CH3COONa] (M) | Kb of acetate | Approx [OH-] (M) | Approx pOH | Approx pH |
|---|---|---|---|---|
| 0.010 | 5.56 × 10^-10 | 2.36 × 10^-6 | 5.63 | 8.37 |
| 0.050 | 5.56 × 10^-10 | 5.27 × 10^-6 | 5.28 | 8.72 |
| 0.100 | 5.56 × 10^-10 | 7.45 × 10^-6 | 5.13 | 8.87 |
| 0.360 | 5.56 × 10^-10 | 1.41 × 10^-5 | 4.85 | 9.15 |
| 1.000 | 5.56 × 10^-10 | 2.36 × 10^-5 | 4.63 | 9.37 |
Common mistakes when solving this problem
1. Treating sodium acetate as a strong base
CH3COONa itself is not a strong base like NaOH. Only the acetate ion acts as a weak base through hydrolysis. So you cannot simply set [OH-] equal to 0.36 M. Doing that would produce a wildly incorrect pH.
2. Using Ka directly instead of converting to Kb
The reacting species in water is acetate, which is a base. Therefore you need Kb for acetate, not Ka for acetic acid. The correct relation is Kb = Kw / Ka. This one conversion is often the whole difference between a right and wrong answer.
3. Forgetting to calculate pOH first
Since the hydrolysis reaction produces OH-, your equilibrium calculation gives [OH-]. You then calculate pOH from that concentration and only after that convert to pH. Some students accidentally compute pH directly from [OH-], which mixes up the acid and base definitions.
4. Ignoring temperature assumptions
The equation pH + pOH = 14.00 is exact only at 25 C when Kw = 1.0 × 10^-14. If the temperature changes, Kw changes too. That is why this calculator allows a Kw input, making it useful for more advanced equilibrium work.
Detailed conceptual explanation for students and exam preparation
A problem such as “19 calculate the pH of a 0.36 M CH3COONa solution” is a classic acid base equilibrium question because it tests several connected skills at once. First, you must classify the salt. Second, you must identify which ion hydrolyzes. Third, you must connect conjugate acid base constants using Kw. Finally, you must translate equilibrium concentrations into pOH and pH.
On a test, the reasoning should move quickly. The presence of Na+ tells you nothing important about pH in this context because sodium is the cation of a strong base. The acetate ion, however, is the conjugate base of acetic acid. Conjugate bases of weak acids are basic, so the solution must have pH above 7. Next, because Ka for acetic acid is known, you compute Kb. Since Kb is small and the formal concentration is fairly large, the square root approximation is valid and gives a fast answer. You should still mentally check whether x is less than 5 percent of the starting concentration. Here x is about 1.41 × 10^-5 M and the initial concentration is 0.36 M, so the approximation is excellent.
Another useful perspective is to compare sodium acetate with an acetic acid solution of the same formal concentration. A 0.36 M acetic acid solution would be acidic, while 0.36 M sodium acetate is basic. This reversal happens because one species donates protons and the other, its conjugate base, accepts them. In buffer problems, these two species often appear together, but in this standalone salt solution only the acetate hydrolysis controls the pH.
When to use the exact method instead of the approximation
The exact method becomes more important when the concentration is very low, when Kb is not extremely small relative to concentration, or when high precision is needed. In many introductory chemistry settings, the approximation is acceptable if x is less than 5 percent of the initial concentration. At 0.36 M sodium acetate, that condition is clearly satisfied. Even so, modern calculators and software can solve the exact quadratic instantly, so there is no downside to checking the exact result.
That is why the calculator above offers both methods. If you are studying for school, use the approximation first to practice hand solving. Then compare it with the exact result to confirm your understanding.
Authoritative references for acid base and equilibrium constants
If you need only the short answer, the accepted textbook result for this problem is: pH ≈ 9.15.