Calculate The Ph Of A 0.32 M Ch3Coona Solution

This interactive calculator determines the pH of sodium acetate solutions by using acetate hydrolysis, the acetic acid dissociation constant, and either the exact quadratic method or the standard weak-base approximation.

Calculator

Core chemistry

CH₃COONa dissociates completely into Na⁺ and CH₃COO^–. The acetate ion hydrolyzes in water:

CH₃COO^– + H₂O ⇌ CH₃COOH + OH^–

Then Kb = Kw / Ka.

How to calculate the pH of a 0.32 M CH3COONa solution

To calculate the pH of a 0.32 M CH₃COONa solution, you need to recognize that sodium acetate is a salt made from a strong base, NaOH, and a weak acid, CH₃COOH. That immediately tells you the solution will be basic, not neutral. Sodium ions do not significantly affect pH, but acetate ions do. The acetate ion acts as a weak base in water and generates hydroxide ions, which raises the pH above 7.

This is one of the most common equilibrium problems in general chemistry because it tests several core ideas at once: salt hydrolysis, conjugate acid-base pairs, the relationship between Ka and Kb, and the link between pOH and pH. If you can solve sodium acetate correctly, you can also solve many similar problems involving salts of weak acids and strong bases.

Step 1: Identify the species that controls pH

When CH₃COONa dissolves in water, it dissociates essentially completely:

CH₃COONa → CH₃COO^– + Na⁺

The sodium ion is a spectator ion for acid-base purposes. The acetate ion, however, can react with water:

CH₃COO^– + H₂O ⇌ CH₃COOH + OH^–

That means the solution is basic because hydroxide is produced.

Step 2: Convert Ka of acetic acid into Kb of acetate

Most textbooks list the acid dissociation constant of acetic acid rather than the base dissociation constant of acetate. At 25°C, a commonly used value is:

• Ka for acetic acid = 1.8 × 10^-5
• Kw for water = 1.0 × 10^-14

Because acetate is the conjugate base of acetic acid, use the relationship:

Kb = Kw / Ka

Substitute the values:

Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10

So acetate is a weak base with a small Kb, but because the sodium acetate concentration is moderately high at 0.32 M, enough hydroxide still forms to produce a clearly basic pH.

Step 3: Set up the equilibrium expression

Let the initial acetate concentration be 0.32 M. Before hydrolysis begins:

• [CH₃COO^–] = 0.32 M
• [CH₃COOH] = 0
• [OH^–] ≈ 0 for setup purposes

If x is the amount of acetate that reacts, then at equilibrium:

• [CH₃COO^–] = 0.32 – x
• [CH₃COOH] = x
• [OH^–] = x

The equilibrium expression becomes:

Kb = x² / (0.32 – x)

Step 4: Use the weak-base approximation

Because Kb is small, x is usually much smaller than the initial concentration. That lets you simplify 0.32 – x to approximately 0.32:

5.56 × 10^-10 = x² / 0.32

So:

x² = (5.56 × 10^-10)(0.32) = 1.78 × 10^-10

x = 1.33 × 10^-5 M

Since x = [OH^–], the hydroxide concentration is 1.33 × 10^-5 M.

Step 5: Convert [OH-] into pOH and pH

Now calculate pOH:

pOH = -log(1.33 × 10^-5) = 4.88

Then use:

pH = 14.00 – 4.88 = 9.12

That is the standard answer for the pH of a 0.32 M CH₃COONa solution at 25°C when Ka = 1.8 × 10^-5.

Step 6: Check whether the approximation is valid

A good chemistry habit is to verify that x is indeed small compared with the original concentration:

(1.33 × 10^-5 / 0.32) × 100 = 0.0042%

That is far below the typical 5% cutoff, so the approximation is excellent. If you solve the full quadratic equation, you get essentially the same pH to standard reporting precision.

Why the solution is basic instead of neutral

Students often wonder why a salt solution can have a pH different from 7 if salts are formed from acids and bases. The answer depends on the strength of the parent acid and base. Sodium acetate comes from NaOH, a strong base, and acetic acid, a weak acid. The strong base contributes a neutral spectator ion, Na⁺, but the weak acid leaves behind a conjugate base strong enough to react with water. This is why sodium acetate solutions are basic.

By contrast, a salt such as NaCl comes from a strong acid and strong base. Neither ion hydrolyzes significantly, so the solution remains nearly neutral. The parent acid-base strengths matter more than the fact that the compound is called a salt.

Exact vs approximate calculation methods

For this problem, the approximation is more than good enough, but professional work often compares both methods. The exact approach solves the quadratic equation generated by:

Kb = x² / (C – x)

Rearranging gives:

x² + Kb x – Kb C = 0

Using the positive root:

x = [-Kb + √(Kb² + 4KbC)] / 2

For C = 0.32 M and Kb = 5.56 × 10^-10, the exact value of x is practically identical to the approximation. The difference appears only in later decimal places, so the pH still rounds to about 9.12.

Parameter	Value used	Chemical meaning
Sodium acetate concentration	0.32 M	Initial acetate concentration after complete salt dissociation
Ka of acetic acid	1.8 × 10^-5	Acid strength of CH3COOH at 25°C
Kw of water	1.0 × 10^-14	Ion-product constant of water at 25°C
Kb of acetate	5.56 × 10^-10	Base strength of CH3COO^–
[OH^–]	1.33 × 10^-5 M	Hydroxide formed by hydrolysis
pOH	4.88	Negative log of hydroxide concentration
pH	9.12	Final basicity of the solution

Comparison with other common acetate solution strengths

The pH of sodium acetate depends on concentration because the hydroxide concentration from a weak base scales roughly with the square root of concentration. That means pH rises as concentration increases, but not linearly. Doubling the concentration does not double the pH shift. This subtle point matters in analytical chemistry, buffer prep, and lab calculations.

CH3COONa concentration	Approximate [OH^–]	Approximate pOH	Approximate pH at 25°C
0.010 M	2.36 × 10^-6 M	5.63	8.37
0.050 M	5.27 × 10^-6 M	5.28	8.72
0.100 M	7.45 × 10^-6 M	5.13	8.87
0.320 M	1.33 × 10^-5 M	4.88	9.12
0.500 M	1.67 × 10^-5 M	4.78	9.22
1.000 M	2.36 × 10^-5 M	4.63	9.37

Common mistakes when solving this problem

1. Treating sodium acetate as a strong base. It is not. The basicity comes from acetate hydrolysis, not from complete OH^– release.
2. Using Ka directly in the ICE table. Since acetate acts as a base, use Kb. If only Ka is given, convert it with Kb = Kw / Ka.
3. Forgetting sodium ion is a spectator. Na⁺ does not change pH in ordinary aqueous solutions.
4. Confusing molarity with moles. The problem gives concentration already, so use 0.32 M directly in the equilibrium setup.
5. Stopping at pOH. Many students calculate [OH^–] correctly, find pOH, and forget to convert to pH.

When temperature and data source matter

The final numerical pH depends slightly on the chosen constants. Different textbooks may list Ka for acetic acid as 1.75 × 10^-5, 1.8 × 10^-5, or 1.74 × 10^-5. Also, Kw changes with temperature. For classroom work at 25°C, using Ka = 1.8 × 10^-5 and Kw = 1.0 × 10^-14 is usually expected. In advanced analytical or environmental chemistry, it is better to use the exact temperature-dependent values from a reliable source.

Practical interpretation of the answer

A pH of about 9.12 means the solution is mildly basic. It is nowhere near as basic as a strong base solution of the same concentration, but it is definitely above neutral. This behavior explains why sodium acetate is frequently used in buffer systems with acetic acid. When both are present together, the solution can resist pH changes effectively around the pKa of acetic acid, roughly 4.76. However, sodium acetate alone is not a buffer by itself in the strict practical sense unless paired with an appreciable amount of acetic acid.

Authoritative references for further study

If you want validated background data for acid-base constants, water chemistry, and equilibrium methods, these sources are useful:

• NIST Chemistry WebBook for trusted chemical property data.
• LibreTexts Chemistry for equilibrium derivations and worked acid-base examples.
• U.S. Environmental Protection Agency water quality resources for pH and aqueous chemistry context.

Final answer

Using standard 25°C constants, the pH of a 0.32 M CH₃COONa solution is approximately 9.12. The key steps are to recognize acetate as a weak base, compute Kb = Kw / Ka, solve for the hydroxide concentration, and then convert from pOH to pH.

Note: Numerical answers may differ slightly by a few hundredths depending on the exact Ka and Kw values specified by your instructor or textbook.