4. Calculate the pH of 0.15 m acetic acid
Use this premium weak-acid calculator to determine the pH, hydrogen ion concentration, percent dissociation, and equilibrium concentrations for acetic acid. For dilute aqueous solutions, 0.15 m is commonly treated as approximately 0.15 M for introductory acid-base calculations.
Default acid
Acetic acid
Typical Ka
1.8 x 10^-5
Typical pKa
4.76
Temperature
25 C
Calculated Results
Enter values and click Calculate pH to see the equilibrium solution, percent ionization, and a concentration comparison chart.
How to calculate the pH of 0.15 m acetic acid
To calculate the pH of 0.15 m acetic acid, you treat acetic acid as a weak acid that only partially ionizes in water. The key equilibrium is:
Because the problem states 0.15 m, many chemistry courses will still solve it using the familiar weak-acid equilibrium framework as if the solution is approximately 0.15 M, especially when the solution is dilute and no activity corrections are requested. Acetic acid has a typical acid dissociation constant of Ka = 1.8 x 10^-5 at 25 C. Since Ka is relatively small, acetic acid dissociates only slightly, which means the hydronium concentration is much lower than the starting acid concentration.
Step 1: Write the equilibrium expression
For a weak acid HA, the equilibrium expression is:
For acetic acid specifically:
If the initial concentration of acetic acid is 0.15 and the amount that dissociates is x, then at equilibrium:
- [CH3COOH] = 0.15 – x
- [H3O+] = x
- [CH3COO-] = x
Step 2: Substitute into the Ka expression
This equation can be solved either with the weak-acid approximation or with the exact quadratic formula. The approximation assumes that x is small compared with 0.15, so 0.15 – x is approximately 0.15.
Since x represents the hydronium concentration, we have:
Step 3: Convert hydronium concentration to pH
So the pH of 0.15 m acetic acid is approximately 2.79 under standard textbook conditions.
Why acetic acid does not have an extremely low pH
Students sometimes expect any acid with a concentration around 0.15 to have a pH near 1. That would be true for a strong acid such as hydrochloric acid, but not for acetic acid. Acetic acid is a weak acid, which means it does not ionize completely. Most of the dissolved acid remains as undissociated CH3COOH molecules. Only a small fraction produces hydronium ions, so the pH is higher than that of a strong acid at the same formal concentration.
This idea is central to acid-base chemistry. The strength of an acid is not determined by concentration alone. It is determined by how extensively the acid ionizes in water. Acetic acid is common in vinegar and laboratory buffers precisely because its moderate strength makes it chemically useful and predictable.
Weak acid versus strong acid at the same concentration
| Acid | Formal concentration | Ka or dissociation behavior | Approximate [H3O+] | Approximate pH |
|---|---|---|---|---|
| Acetic acid, CH3COOH | 0.15 | Ka = 1.8 x 10^-5, partial dissociation | 1.63 x 10^-3 | 2.79 |
| Hydrochloric acid, HCl | 0.15 | Essentially complete dissociation | 0.15 | 0.82 |
| Formic acid, HCOOH | 0.15 | Ka = 1.77 x 10^-4, stronger weak acid | 5.14 x 10^-3 | 2.29 |
The numbers above show why the pH of acetic acid is not as low as a fully dissociated mineral acid. A 0.15 strong acid yields a hydronium concentration near 0.15, while 0.15 acetic acid produces only about 0.00163 hydronium under the same simplified conditions.
Exact quadratic solution for better accuracy
Although the square-root approximation is standard and works well here, the exact approach is even better. Starting from:
Rearrange into quadratic form:
Then solve:
Using C = 0.15 and Ka = 1.8 x 10^-5 gives x ≈ 1.634 x 10^-3 and therefore pH ≈ 2.79. In this case, the approximation and exact solution are almost identical because the acid ionization is only about 1.09 percent of the starting concentration, well under the usual 5 percent guideline.
5 percent rule check
- Approximate x = 1.64 x 10^-3
- Compare x to initial concentration 0.15
- Percent ionization = (1.64 x 10^-3 / 0.15) x 100 ≈ 1.09 percent
Since 1.09 percent is less than 5 percent, the weak-acid approximation is justified.
What does the symbol m mean here?
In chemistry, lowercase m usually means molality, while uppercase M means molarity. Molality is moles of solute per kilogram of solvent, and molarity is moles of solute per liter of solution. For many basic classroom acid-base questions involving relatively dilute aqueous solutions, instructors may expect you to proceed as if the numerical value can be used in the Ka setup directly. That is what has been done here.
In advanced physical chemistry, there is a difference between concentration-based equilibrium expressions and activity-based treatments. At higher ionic strengths or when precision matters, activities and activity coefficients become important. However, for a standard general chemistry problem asking for the pH of 0.15 m acetic acid, the accepted result is usually around 2.79.
Common mistakes when solving this problem
- Assuming complete dissociation. Acetic acid is weak, so you cannot set [H3O+] equal to 0.15.
- Using pKa incorrectly. pKa is related to Ka through pKa = -log10(Ka), but you still need equilibrium logic to find pH from a starting acid concentration.
- Forgetting the ICE setup. Initial, change, equilibrium tables help organize weak-acid calculations and reduce algebra mistakes.
- Dropping units carelessly. Whether the problem states M or m, you should note the assumption you are making if you use concentration directly.
- Rounding too early. Keep a few guard digits through the calculation so the final pH is reported accurately.
Acetic acid data table at 25 C
| Initial concentration | Calculated [H3O+] | Calculated pH | Percent ionization |
|---|---|---|---|
| 0.010 | 4.15 x 10^-4 | 3.38 | 4.15% |
| 0.050 | 9.40 x 10^-4 | 3.03 | 1.88% |
| 0.100 | 1.33 x 10^-3 | 2.88 | 1.33% |
| 0.150 | 1.63 x 10^-3 | 2.79 | 1.09% |
| 0.500 | 2.99 x 10^-3 | 2.52 | 0.60% |
This comparison highlights an important trend: as the formal concentration increases, the pH drops, but the fraction of molecules that ionize becomes smaller. That is characteristic of weak acids. The equilibrium shifts in a way that limits full dissociation.
Shortcut methods and when to use them
Square-root shortcut
For a weak acid with low dissociation, use:
For this problem:
Quadratic method
Use the exact expression when:
- The acid is not very weak
- The concentration is low enough that ionization is not negligible
- Your instructor specifically asks for the exact result
- You want to verify the approximation
Interpreting the result chemically
A pH of 2.79 means the solution is clearly acidic, but still far less acidic than a strong acid of the same concentration. This is consistent with the chemistry of acetic acid in vinegar-like systems. Pure household vinegar typically contains much more acetic acid than 0.15, often around 5 percent by mass, but the same weak-acid reasoning still applies. The equilibrium concentration of hydronium is controlled by Ka, not just by the amount of acid added.
In practical terms, this pH explains why acetic acid can sting and preserve food while still being much less corrosive than strong mineral acids at comparable analytical concentration. It is also why acetic acid and acetate are useful in buffer systems. Once acetate is present, the pH becomes even more governed by the Henderson-Hasselbalch relationship rather than by the pure weak-acid equation alone.
Authority sources for acid-base constants and solution chemistry
- NIST Chemistry WebBook for reference thermochemical and chemical data from a U.S. government source.
- University of California educational chemistry material on Ka and acid dissociation.
- U.S. Environmental Protection Agency for broader background on pH and aqueous chemistry in environmental systems.
Worked summary
- Start with acetic acid dissociation: CH3COOH ⇌ H3O+ + CH3COO-
- Use Ka = 1.8 x 10^-5 at 25 C
- Let x = [H3O+] formed
- Set up: 1.8 x 10^-5 = x^2 / (0.15 – x)
- Approximate or solve exactly to get x ≈ 1.63 x 10^-3
- Calculate pH = -log10(1.63 x 10^-3) ≈ 2.79
If you are answering a homework or exam prompt that simply asks, calculate the pH of 0.15 m acetic acid, the concise final answer is:
This calculator above can also help you test related values, compare weak acids with different Ka values, and visualize how much acetic acid remains undissociated at equilibrium.