Calculate the pH of a 0.10 M CH3CO2H Solution with Ka = 1.8 × 10-5
Use this premium weak acid calculator to solve acetic acid equilibrium, estimate hydrogen ion concentration, and visualize how dissociation compares with the initial acid concentration.
Weak Acid pH Calculator
How to calculate the pH of a 0.10 M CH3CO2H solution when Ka = 1.8 × 10-5
If you are trying to calculate the pH of a 0.10 M CH3CO2H solution with Ka = 1.8 × 10-5, you are solving a classic weak acid equilibrium problem. CH3CO2H is acetic acid, one of the most commonly discussed weak acids in general chemistry. Unlike strong acids, acetic acid does not ionize completely in water. That means you cannot simply say the hydrogen ion concentration equals the starting molarity. Instead, you must use the acid dissociation constant, or Ka, to determine how much acid dissociates at equilibrium.
The reaction is:
CH3CO2H + H2O ⇌ H3O+ + CH3CO2-
Because acetic acid is weak, only a small fraction of the 0.10 M initial concentration becomes hydronium ions. The pH is therefore higher than 1.00, which would be the pH of a fully dissociated 0.10 M strong monoprotic acid. This distinction between strong and weak acids is one of the core ideas in acid-base chemistry, and understanding it helps with buffer calculations, titration curves, laboratory work, and exam problems.
Step-by-step setup using an ICE table
The cleanest way to begin is with an ICE table, which tracks the Initial, Change, and Equilibrium concentrations.
1. Write the equilibrium expression
For acetic acid dissociation:
Ka = [H3O+][CH3CO2-] / [CH3CO2H]
Given:
- Initial acetic acid concentration = 0.10 M
- Ka = 1.8 × 10-5
- Initial [H3O+] and [CH3CO2-] from the acid are taken as approximately 0
2. Build the ICE table
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| CH3CO2H | 0.10 | -x | 0.10 – x |
| H3O+ | 0 | +x | x |
| CH3CO2- | 0 | +x | x |
3. Substitute into Ka
Plug the equilibrium concentrations into the equilibrium expression:
1.8 × 10-5 = x2 / (0.10 – x)
Since Ka is small, many textbooks first try the weak acid approximation, assuming x is small compared with 0.10. That gives:
1.8 × 10-5 ≈ x2 / 0.10
So:
x2 = 1.8 × 10-6
x = 1.34 × 10-3 M
Because x = [H3O+], the pH is:
pH = -log(1.34 × 10-3) ≈ 2.87
Exact quadratic solution
For a more rigorous answer, solve the full equation without approximation:
1.8 × 10-5 = x2 / (0.10 – x)
Rearranging:
x2 + (1.8 × 10-5)x – 1.8 × 10-6 = 0
Apply the quadratic formula:
x = [-Ka + √(Ka2 + 4KaC)] / 2
where C = 0.10.
Using the given values:
- Ka = 1.8 × 10-5
- C = 0.10
- x ≈ 1.332 × 10-3 M
- pH = -log(1.332 × 10-3) ≈ 2.88
The exact and approximate methods agree very closely. That is why the approximation is usually considered acceptable for acetic acid at this concentration.
Why the weak acid approximation works here
A useful chemistry check is the 5% rule. If the amount ionized, x, is less than 5% of the initial concentration, then replacing 0.10 – x with 0.10 is generally acceptable.
Here:
percent ionization = (1.332 × 10-3 / 0.10) × 100 ≈ 1.33%
Since 1.33% is well below 5%, the approximation is justified.
| Quantity | Approximation Method | Exact Quadratic Method | Difference |
|---|---|---|---|
| [H3O+] | 1.342 × 10^-3 M | 1.332 × 10^-3 M | 0.010 × 10^-3 M |
| pH | 2.87 | 2.88 | About 0.01 pH unit |
| Percent ionization | 1.34% | 1.33% | Negligible for most coursework |
How this compares with strong acids and more dilute acetic acid
Students often confuse concentration with strength. A 0.10 M acid is not automatically a “very acidic” solution in the same way for every compound. Strength depends on how completely it ionizes. Acetic acid is weak, so a 0.10 M solution gives a pH near 2.88, not 1.00.
| Solution | Typical Acid Type | [H3O+] Estimate | Approximate pH |
|---|---|---|---|
| 0.10 M HCl | Strong acid | 0.10 M | 1.00 |
| 0.10 M CH3CO2H | Weak acid | 1.33 × 10^-3 M | 2.88 |
| 0.010 M CH3CO2H | Weak acid | 4.15 × 10^-4 M | 3.38 |
| Pure water at 25 °C | Neutral reference | 1.0 × 10^-7 M | 7.00 |
Common mistakes when solving this problem
- Using Ka = 1.8 instead of 1.8 × 10^-5: this is the biggest error. The exponent matters.
- Treating acetic acid like a strong acid: if you set [H3O+] = 0.10 M, you will incorrectly get pH = 1.
- Forgetting the logarithm sign: pH is -log[H3O+], not just log[H3O+].
- Dropping x without checking: the approximation should be justified by the 5% rule.
- Confusing Ka with pKa: acetic acid has pKa around 4.74 to 4.76 at room temperature, not pH 4.76 in this problem.
What the answer means chemically
A pH near 2.88 tells you the solution is acidic, but not nearly as acidic as an equal concentration of a strong acid. Most of the acetic acid remains in the molecular form CH3CO2H, while a small portion forms H3O+ and CH3CO2-. This is typical of weak acid equilibria: the equilibrium lies mostly to the left, and the Ka value quantifies that tendency.
In practical terms, this behavior is important in vinegar chemistry, biochemical buffering, environmental chemistry, and analytical chemistry. Acetic acid and acetate together form one of the most familiar buffer systems in the introductory lab. Once you know how to solve the simple acid-only case, you are well prepared to handle Henderson-Hasselbalch buffer questions later.
Useful formulas to remember
- Ka = [H3O+][A-] / [HA]
- For a weak acid only: x ≈ √(KaC)
- pH = -log[H3O+]
- Percent ionization = ([H3O+] / initial acid concentration) × 100
- pKa = -log(Ka)
Authority sources for acid-base chemistry
For reliable chemistry reference material, review these authoritative educational resources:
- LibreTexts Chemistry for equilibrium and weak acid problem-solving tutorials.
- National Institute of Standards and Technology (NIST) for trusted scientific standards and chemistry-related reference information.
- University of California, Berkeley Chemistry for university-level chemistry teaching resources and concepts.
Final answer
To calculate the pH of a 0.10 M CH3CO2H solution with Ka = 1.8 × 10-5, set up the weak acid equilibrium, solve for [H3O+], and then take the negative logarithm. The exact hydrogen ion concentration is approximately 1.332 × 10-3 M, which gives:
pH ≈ 2.88
This page calculator automates the process, checks the approximation, and visualizes the equilibrium concentrations so you can move from memorizing formulas to actually understanding what the chemistry means.