Calculate The Ph Of 0.10 M Nac2H3O2

Use this premium calculator to find the pH of sodium acetate solutions by converting the acid dissociation constant of acetic acid into the base dissociation constant of acetate, then solving the equilibrium exactly.

• Handles Ka or pKa input modes
• Uses the exact quadratic equilibrium solution
• Displays pH, pOH, Kb, hydroxide concentration, and percent hydrolysis
• Generates a concentration chart with Chart.js

How to Calculate the pH of 0.10 M NaC2H3O2

Sodium acetate, written as NaC₂H₃O₂, is the sodium salt of acetic acid. In water, it dissociates essentially completely into Na⁺ and acetate ions, C₂H₃O₂^–. The sodium ion is a spectator ion for this calculation, but the acetate ion matters a great deal because it is the conjugate base of a weak acid. That means the solution is not neutral. Instead, acetate reacts with water and produces a small amount of hydroxide, making the solution basic.

If your chemistry problem asks you to calculate the pH of 0.10 M NaC₂H₃O₂, the key idea is that you are solving a weak base equilibrium. Many students initially think a salt should always be neutral, but salts formed from a strong base and a weak acid usually produce basic solutions. Sodium acetate is one of the classic examples.

Step 1: Identify the Acid-Base Chemistry

First, write the dissociation of sodium acetate and then the hydrolysis of acetate:

NaC2H3O2 → Na+ + C2H3O2- C2H3O2- + H2O ⇌ HC2H3O2 + OH-

The second equation is the important one for pH. Acetate acts as a weak base, accepting a proton from water. Because hydroxide is formed, the pH rises above 7.00.

Step 2: Convert Ka to Kb

Most textbooks provide the acid dissociation constant for acetic acid, not the base dissociation constant for acetate. So you use the conjugate relationship:

Kb = Kw / Ka

At 25 C, a standard value is:

• K_w = 1.0 × 10^-14
• K_a for acetic acid ≈ 1.8 × 10^-5

Substituting gives:

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

This value is small, which tells you that acetate is a weak base. Even so, in a 0.10 M solution, it still generates enough OH^– to produce a clearly basic pH.

Step 3: Set Up an ICE Table

Now use an equilibrium setup for the hydrolysis reaction:

C2H3O2- + H2O ⇌ HC2H3O2 + OH-

Initial concentrations for a 0.10 M sodium acetate solution are approximately:

• [C₂H₃O₂^–] = 0.10 M
• [HC₂H₃O₂] = 0
• [OH^–] = 0

Let x be the amount of acetate that reacts:

• Change: acetate = -x, acetic acid = +x, hydroxide = +x
• Equilibrium: [C₂H₃O₂^–] = 0.10 – x, [HC₂H₃O₂] = x, [OH^–] = x

Then plug into the K_b expression:

Kb = [HC2H3O2][OH-] / [C2H3O2-] = x^2 / (0.10 – x)

Step 4: Solve for x

Since K_b is very small, many instructors allow the approximation 0.10 – x ≈ 0.10. That gives:

x^2 / 0.10 = 5.56 × 10^-10 x^2 = 5.56 × 10^-11 x = 7.45 × 10^-6 M

Because x equals [OH^–] at equilibrium, you can now calculate pOH:

pOH = -log(7.45 × 10^-6) = 5.13 pH = 14.00 – 5.13 = 8.87

So the pH of 0.10 M sodium acetate is approximately 8.87 at 25 C when using K_a = 1.8 × 10^-5.

Final answer: a 0.10 M NaC2H3O2 solution has a pH of about 8.87 at 25 C, assuming standard values for acetic acid and water.

Why the Solution Is Basic

This is an important conceptual point. Sodium acetate comes from sodium hydroxide, which is a strong base, and acetic acid, which is a weak acid. The sodium ion does not affect pH very much, but the acetate ion does. Because acetate is the conjugate base of a weak acid, it has measurable basicity in water.

In practical terms, that means the acetate ion pulls protons from water molecules just enough to create a nontrivial hydroxide concentration. The effect is modest, not extreme, because acetate is still a weak base. That is why the pH lands in the upper eights instead of near 12 or 13.

Exact Method vs Approximation

The approximation method is usually acceptable because x is much smaller than 0.10 M. In fact, the percent hydrolysis is tiny. However, an exact solution can also be found by solving the quadratic form of the equilibrium equation:

x^2 + Kb x – KbC = 0

For this problem, the exact and approximate values are nearly identical. That is why chemistry students are often taught the square-root shortcut for weak acid and weak base calculations when the equilibrium shift is very small relative to the starting concentration.

Quantity	Typical Value at 25 C	Meaning for This Problem
Ka of acetic acid	1.8 × 10^-5	Used to determine the basicity of acetate
pKa of acetic acid	4.76	Alternative way to represent acid strength
Kb of acetate	5.56 × 10^-10	Controls OH^– formation in solution
[OH^–] in 0.10 M NaC2H3O2	7.45 × 10^-6 M	Equilibrium hydroxide concentration
pOH	5.13	Negative log of hydroxide concentration
pH	8.87	Final basic solution pH

How Concentration Changes the pH

The pH of sodium acetate depends on concentration. More concentrated sodium acetate solutions produce more hydroxide and therefore higher pH values. The relationship is not linear, because pH is logarithmic. Below is a comparison table using the same K_a value of acetic acid and standard 25 C conditions.

NaC2H3O2 Concentration	Calculated [OH^–]	Calculated pOH	Calculated pH
0.001 M	7.45 × 10^-7 M	6.13	7.87
0.010 M	2.36 × 10^-6 M	5.63	8.37
0.10 M	7.45 × 10^-6 M	5.13	8.87
0.50 M	1.67 × 10^-5 M	4.78	9.22
1.00 M	2.36 × 10^-5 M	4.63	9.37

Common Mistakes Students Make

1. Treating sodium acetate as neutral. It is not neutral because acetate is a weak base.
2. Using K_a directly in the ICE table. The reacting species is acetate, so you need K_b.
3. Forgetting the conjugate relationship. Always use K_aK_b = K_w.
4. Confusing pH and pOH. You calculate [OH^–] first, then pOH, then pH.
5. Ignoring temperature assumptions. If K_w differs from 1.0 × 10^-14, your pH can shift slightly.

Shortcut Formula You Can Use

For weak bases where the approximation is valid, there is a useful compact route:

[OH-] ≈ √(Kb × C)

Here, C is the initial acetate concentration. For 0.10 M sodium acetate:

[OH-] ≈ √((5.56 × 10^-10)(0.10)) = 7.45 × 10^-6 M

This shortcut is mathematically identical to the ICE table approximation and is very popular in exam settings. Still, it is only safe when the ionization is small compared with the starting concentration. The calculator above uses the exact quadratic result, so it remains reliable across a wider range of values.

Real-World Relevance of Sodium Acetate pH

Sodium acetate appears in buffer preparation, analytical chemistry, biochemistry labs, and industrial formulations. A sodium acetate solution by itself is mildly basic, but when mixed with acetic acid it becomes part of the well-known acetate buffer system. That buffer is useful because it resists pH changes near the pK_a of acetic acid, which is about 4.76.

Although a pure sodium acetate solution has a pH near 8.87 at 0.10 M, buffer mixtures containing both acetate and acetic acid can be formulated over a range of acidic pH values using the Henderson-Hasselbalch equation. That distinction is important: a salt solution and a buffer solution are not the same thing. Many homework errors happen because students mix those ideas together.

Quick Summary Procedure

• Recognize NaC₂H₃O₂ as a salt of a weak acid and strong base.
• Write the acetate hydrolysis reaction with water.
• Convert K_a of acetic acid to K_b of acetate using K_w/K_a.
• Set up the equilibrium expression K_b = x² / (C – x).
• Solve for x = [OH^–].
• Calculate pOH, then convert to pH.

Authoritative Sources for Further Study

NIST Chemistry WebBook (.gov)
U.S. Environmental Protection Agency pH Reference (.gov)
University of Wisconsin Acid-Base Equilibria Resource (.edu)

Bottom Line

To calculate the pH of 0.10 M NaC₂H₃O₂, treat acetate as a weak base, convert the known K_a of acetic acid into K_b, solve for the hydroxide concentration, and then convert to pH. Using K_a = 1.8 × 10^-5 and K_w = 1.0 × 10^-14, you obtain a hydroxide concentration of about 7.45 × 10^-6 M, a pOH of 5.13, and a final pH of 8.87. That result makes chemical sense because sodium acetate is the salt of a weak acid and therefore produces a mildly basic solution in water.