Calculate pH of a 0.1 M CH3COOH Solution
Use this interactive weak acid calculator to determine the pH, hydrogen ion concentration, acetate concentration, and percent ionization of acetic acid. The calculator uses the exact equilibrium solution by default and also shows the common weak acid approximation for comparison.
Acetic Acid pH Calculator
Results will appear here
Enter your values and click Calculate pH to see the equilibrium analysis for CH3COOH.
CH3COOH + H2O ⇌ H3O+ + CH3COO-
Ka = [H3O+][CH3COO-] / [CH3COOH]
How to Calculate the pH of a 0.1 M CH3COOH Solution
Calculating the pH of a 0.1 M CH3COOH solution is a classic weak acid equilibrium problem in general chemistry. CH3COOH is acetic acid, the main acidic component associated with vinegar, and unlike strong acids such as hydrochloric acid, it does not fully ionize in water. That single fact changes the entire calculation. Instead of assuming that the acid concentration equals the hydrogen ion concentration, you must use the acid dissociation constant, commonly written as Ka, to determine the equilibrium concentration of hydronium ions. Once that concentration is known, the pH follows from the familiar equation pH = -log[H3O+].
For acetic acid at about 25 C, a commonly used value is Ka = 1.8 × 10-5. If the initial concentration is 0.1 M, the exact solution gives a pH close to 2.88. Many textbooks also teach an approximation method that gives nearly the same value for this case, because acetic acid is weak and only a small percentage of the acid dissociates. The calculator above performs both methods so you can compare them directly.
Step 1: Write the dissociation reaction
The starting point is the equilibrium reaction for acetic acid in water:
CH3COOH + H2O ⇌ H3O+ + CH3COO-
Acetic acid donates a proton to water, producing hydronium and acetate. Because the acid is weak, the equilibrium lies strongly to the left, meaning most acetic acid molecules remain undissociated.
Step 2: Set up the Ka expression
The equilibrium constant expression for this weak acid is:
Ka = [H3O+][CH3COO-] / [CH3COOH]
At 25 C, acetic acid has a pKa around 4.76, corresponding to a Ka near 1.8 × 10-5. These values are standard reference data used in chemistry courses and lab work.
Step 3: Use an ICE setup
An ICE table helps track concentration changes:
- Initial: [CH3COOH] = 0.1 M, [H3O+] ≈ 0, [CH3COO-] = 0
- Change: CH3COOH decreases by x, H3O+ increases by x, CH3COO- increases by x
- Equilibrium: [CH3COOH] = 0.1 – x, [H3O+] = x, [CH3COO-] = x
Substitute these into the Ka expression:
1.8 × 10-5 = x2 / (0.1 – x)
Step 4: Solve for x
There are two common approaches. The first is the exact quadratic solution. Rearranging gives:
x2 + Kax – KaC = 0
where C is the initial acid concentration. Using Ka = 1.8 × 10-5 and C = 0.1 M:
x = [-Ka + √(Ka2 + 4KaC)] / 2
This yields x ≈ 1.33 × 10-3 M. Since x = [H3O+], the pH is:
pH = -log(1.33 × 10-3) ≈ 2.88
The second approach is the weak acid approximation. Because x is much smaller than 0.1, you can often assume that 0.1 – x ≈ 0.1. Then:
Ka ≈ x2 / 0.1
So:
x ≈ √(Ka × 0.1) = √(1.8 × 10-6) ≈ 1.34 × 10-3 M
That gives pH ≈ 2.87 to 2.88, almost identical to the exact result.
Why the approximation works here
The approximation works because acetic acid is weak and the amount that ionizes is small compared with the starting concentration. For a 0.1 M solution, percent ionization is only about 1.3 percent. Since this is well below 5 percent, the simplification is generally acceptable for introductory chemistry calculations. However, the exact quadratic is always more rigorous and is especially useful at lower concentrations, with larger Ka values, or whenever you need high precision.
| Parameter | Typical value for acetic acid at 25 C | Why it matters |
|---|---|---|
| Ka | 1.8 × 10-5 | Controls the extent of dissociation in water. |
| pKa | 4.76 | Useful for buffer calculations and acid strength comparisons. |
| 0.1 M CH3COOH pH | About 2.88 | Shows moderate acidity, but far weaker than a strong acid of the same concentration. |
| Percent ionization at 0.1 M | About 1.3% | Explains why the approximation method is valid. |
Comparing acetic acid to a strong acid at the same concentration
A very common point of confusion is assuming that equal molarity means equal pH. That is false when comparing weak and strong acids. A 0.1 M strong monoprotic acid like HCl essentially fully dissociates, so [H3O+] ≈ 0.1 M and pH ≈ 1.00. Acetic acid, by contrast, gives [H3O+] only around 1.33 × 10-3 M, making the solution nearly 76 times lower in hydronium concentration than 0.1 M HCl. This is why pH changes so dramatically even when the formal molarity is the same.
| Solution | Formal concentration | Approximate [H3O+] | Approximate pH |
|---|---|---|---|
| HCl | 0.1 M | 0.1 M | 1.00 |
| CH3COOH | 0.1 M | 1.33 × 10-3 M | 2.88 |
| Difference | Same formal molarity | Strong acid is about 75 to 76 times higher | About 1.88 pH units lower for HCl |
How percent ionization is calculated
Once x is known, percent ionization follows from:
Percent ionization = (x / initial concentration) × 100
For 0.1 M acetic acid:
Percent ionization = (1.33 × 10-3 / 0.1) × 100 ≈ 1.33%
This value has practical importance because it describes how weak the acid actually is in water. A low percent ionization means most molecules remain as CH3COOH rather than dissociating into ions.
Common mistakes students make
- Treating acetic acid like a strong acid. This leads to an incorrect pH of 1 instead of about 2.88.
- Forgetting the equilibrium denominator. Ka always depends on both products and remaining undissociated acid.
- Using pKa incorrectly. pKa is not the pH. It is a constant that describes acid strength.
- Ignoring the 5 percent rule. The approximation should be checked if precision matters.
- Mixing up concentration and amount. Molarity is moles per liter, not simply moles.
When the pH of acetic acid changes
The pH of an acetic acid solution depends mainly on concentration and Ka. If concentration decreases, the weak acid generally ionizes to a greater percentage, though total [H3O+] may still decrease. If temperature changes, Ka can shift, which changes the final pH. That is why most reference values are quoted at a standard temperature, usually 25 C. In laboratory settings, this distinction matters when comparing measured pH values with textbook predictions.
Practical interpretation of a pH near 2.88
A pH of about 2.88 means the solution is definitely acidic, but not nearly as aggressive as a strong acid with the same nominal molarity. In acid base chemistry, acetic acid is widely used because it is strong enough to provide meaningful proton donation while still retaining substantial undissociated acid, which makes it useful in buffer systems when combined with acetate salts. This is one reason the acetic acid and acetate pair appears so often in chemistry and biochemistry labs.
Authoritative references for acetic acid equilibrium data
If you want to verify constants or review foundational acid base chemistry, these sources are highly reliable:
- LibreTexts Chemistry for weak acid equilibrium explanations and worked examples.
- U.S. Environmental Protection Agency for water chemistry and pH background resources.
- NIST Chemistry WebBook for authoritative chemical data and reference values.
Final answer for a 0.1 M CH3COOH solution
Using Ka = 1.8 × 10-5 at 25 C, the exact equilibrium calculation gives:
- [H3O+] ≈ 1.33 × 10-3 M
- pH ≈ 2.88
- [CH3COO-] ≈ 1.33 × 10-3 M
- [CH3COOH] remaining ≈ 0.0987 M
- Percent ionization ≈ 1.33%
This result is a great example of how equilibrium controls the chemistry of weak acids. Even though the starting concentration is fairly high, acetic acid ionizes only slightly, so the pH remains much higher than the pH of a strong acid at the same molarity. If you want to explore how pH changes for other acetic acid concentrations, simply enter a new value in the calculator above and compare the exact and approximate methods.