Calculate the pH of 0.1 M CH3COOH
Use this premium weak acid calculator to find pH, hydrogen ion concentration, percent ionization, and acetate concentration for acetic acid solutions. Default values are set for a 0.100 M CH3COOH solution at 25 degrees Celsius.
For 0.100 M CH3COOH using Ka = 1.8 × 10^-5, the exact pH is about 2.88 at 25 degrees Celsius.
How to calculate the pH of 0.1 M CH3COOH
To calculate the pH of 0.1 M CH3COOH, you need to remember that acetic acid is a weak acid, not a strong acid. That means it does not dissociate completely in water. Instead, only a small fraction of the acid molecules donate protons to water. The equilibrium is written as CH3COOH + H2O ⇌ H3O+ + CH3COO-. Because the dissociation is partial, you cannot simply assume that the hydrogen ion concentration is equal to the original acid concentration. Instead, you use the acid dissociation constant, Ka, for acetic acid.
At 25 degrees Celsius, a commonly used Ka value for acetic acid is 1.8 × 10-5. Starting with an initial concentration of 0.100 M, the ICE table approach gives initial concentrations of 0.100 M acetic acid, 0 M acetate, and approximately 0 M hydrogen ion coming from the acid itself. Let x be the amount of acetic acid that dissociates. At equilibrium, [H+] = x, [CH3COO–] = x, and [CH3COOH] = 0.100 – x.
1.8 x 10^-5 = x^2 / (0.100 – x)
Because acetic acid is weak and x is small relative to 0.100, many introductory chemistry courses use the approximation 0.100 – x ≈ 0.100. Then the equation becomes x2 = (1.8 × 10-5)(0.100), so x = square root of 1.8 × 10-6, which is about 1.34 × 10-3 M. Since x equals the hydrogen ion concentration, pH = -log[H+] = -log(1.34 × 10-3) ≈ 2.87.
If you use the exact quadratic method, the result is nearly the same. Solving x2 + Ka x – KaC = 0 for x gives x = (-Ka + square root of Ka2 + 4KaC) / 2. Substituting Ka = 1.8 × 10-5 and C = 0.100 gives x ≈ 1.332 × 10-3 M, which leads to a pH of about 2.88. The approximate and exact answers differ only slightly because the percent ionization is low.
Final answer for 0.1 M acetic acid
The pH of 0.1 M CH3COOH is approximately 2.88 when you use Ka = 1.8 × 10-5 at 25 degrees Celsius. In many textbooks, you may also see 2.87 because of different rounding or because Ka is sometimes listed as 1.75 × 10-5 or 1.76 × 10-5. Both values are chemically reasonable if the underlying constant and significant figures are stated clearly.
Quick step-by-step summary
- Write the equilibrium equation: CH3COOH ⇌ H+ + CH3COO–.
- Use Ka = 1.8 × 10-5 for acetic acid at 25 degrees Celsius.
- Let x be the amount that dissociates from 0.100 M initial concentration.
- Set up the Ka expression: 1.8 × 10-5 = x2 / (0.100 – x).
- Solve exactly or use the weak acid approximation.
- Find [H+] ≈ 1.33 × 10-3 M.
- Compute pH = -log[H+] ≈ 2.88.
Why acetic acid does not behave like a strong acid
Students often make the mistake of calculating the pH of 0.1 M acetic acid as if it were a strong monoprotic acid. If that were true, [H+] would be 0.1 M and the pH would be 1.00. That is far too acidic for acetic acid because most CH3COOH molecules remain undissociated in water. The value of Ka tells you how far the equilibrium lies toward products. For acetic acid, Ka is small, so the dissociation is limited.
This difference between strong and weak acids has practical implications in chemistry, food science, environmental science, and biochemistry. Vinegar, for example, contains acetic acid and is acidic enough to taste sour and influence chemical reactions, but it is not nearly as acidic as a strong acid of equal molarity. That is one reason weak acid calculations are a core part of acid-base chemistry.
Common student errors
- Treating acetic acid as fully dissociated and using pH = 1 for a 0.1 M solution.
- Forgetting to use the Ka expression.
- Using pKa directly without converting carefully when concentration is given.
- Applying the approximation without checking whether x is small compared with the initial concentration.
- Rounding too early and getting pH 2.8, 2.9, or 3.0 without justification.
Exact method versus approximation
For weak acids, the approximation method is popular because it saves time. However, the exact quadratic method is more rigorous and always valid if applied correctly. For 0.1 M CH3COOH, the approximation works very well because the degree of ionization is low. The five percent rule says the approximation is acceptable if x is less than five percent of the initial concentration. Here, x is about 1.33 × 10-3 M, which is only about 1.33 percent of 0.100 M, so the approximation is acceptable.
| Method | Expression Used | [H+] for 0.100 M CH3COOH | Calculated pH | Comment |
|---|---|---|---|---|
| Exact quadratic | x = (-Ka + √(Ka² + 4KaC)) / 2 | 1.332 × 10-3 M | 2.875 | Most rigorous classroom answer |
| Approximation | x ≈ √(KaC) | 1.342 × 10-3 M | 2.872 | Very close because ionization is low |
| Incorrect strong acid assumption | [H+] = 0.100 M | 1.000 × 10-1 M | 1.000 | Not valid for acetic acid |
Percent ionization of 0.1 M CH3COOH
Percent ionization tells you what fraction of the original acid molecules actually dissociate. It is calculated as ([H+] / initial concentration) × 100. Using the exact result, percent ionization = (1.332 × 10-3 / 0.100) × 100 ≈ 1.33 percent. That small value explains why the weak acid approximation works well. It also illustrates an important concept: even though the solution is definitely acidic, most acetic acid molecules remain in their protonated form at equilibrium.
As the solution becomes more dilute, percent ionization rises. That does not necessarily mean the solution becomes more acidic in terms of pH. In fact, the pH rises as concentration drops, but the fraction of molecules that dissociate can increase. This is a subtle point in equilibrium chemistry and often appears on exams.
| Initial CH3COOH Concentration | Approximate [H+], using Ka = 1.8 × 10-5 | Approximate pH | Approximate Percent Ionization |
|---|---|---|---|
| 1.0 M | 4.24 × 10-3 M | 2.37 | 0.42% |
| 0.100 M | 1.34 × 10-3 M | 2.87 | 1.34% |
| 0.0100 M | 4.24 × 10-4 M | 3.37 | 4.24% |
| 0.00100 M | 1.34 × 10-4 M | 3.87 | 13.4% |
Using pKa to estimate pH
You may also see acetic acid described by its pKa instead of Ka. Since pKa = -log Ka, the pKa of acetic acid at 25 degrees Celsius is about 4.74 to 4.76 depending on the source. The pKa is especially useful when discussing buffers and the Henderson-Hasselbalch equation. However, for a simple solution containing only acetic acid in water, you still need to connect pKa back to the actual equilibrium concentration of hydrogen ions. The pKa value confirms that acetic acid is weak, but by itself it does not immediately give the pH unless you set up the equilibrium relationship correctly.
Real-world context for acetic acid solutions
Acetic acid is one of the most familiar weak acids. It is the principal acidic component of vinegar, though household vinegar is usually reported in percent acidity by mass or by volume rather than simple molarity. Laboratory acetic acid solutions are used in titrations, buffer preparation, and demonstrations of weak acid behavior. In biological and industrial settings, acetate chemistry also matters because acetic acid and acetate participate in buffering and reaction pathways.
When someone asks for the pH of 0.1 M CH3COOH, they are usually working a standard equilibrium problem in general chemistry or analytical chemistry. The expected skill is not merely producing a number but also understanding why the weak acid model applies, why the solution is not fully dissociated, and how to justify the approximation or exact method. For that reason, it is a very common exam question.
Authoritative chemistry references
For readers who want to verify weak acid data and acid-base methods, these educational and government resources are excellent references:
- LibreTexts Chemistry educational resource
- NIST Chemistry WebBook
- U.S. Environmental Protection Agency
Detailed worked example for students
Suppose your instructor gives you the problem exactly as follows: calculate the pH of 0.1 M CH3COOH. You begin by identifying the acid as acetic acid, a weak monoprotic acid. Next, you write the dissociation equation. Then you look up or are given Ka = 1.8 × 10-5. Construct the ICE table: initial concentrations are 0.100, 0, and 0. The change is -x, +x, and +x. At equilibrium, the concentrations are 0.100 – x, x, and x. Substitute into the Ka expression:
If you want the exact answer, rearrange:
Using the quadratic formula gives x ≈ 0.001332 M. Then:
Rounded appropriately, the pH is 2.88. If your class allows the approximation, then x = √(1.8 × 10-6) ≈ 0.001342 M, giving pH ≈ 2.87. Either route shows that the solution is acidic but much less acidic than a fully dissociated 0.1 M strong acid solution.
Key takeaways
- The pH of 0.1 M CH3COOH is about 2.88 at 25 degrees Celsius when Ka = 1.8 × 10-5.
- Acetic acid is weak, so only a small fraction dissociates.
- The exact quadratic and approximate square-root methods give nearly the same answer here.
- Percent ionization is about 1.33 percent for this solution.
- If your result is near pH 1, you accidentally treated acetic acid as a strong acid.
Frequently asked questions
Is the answer 2.87 or 2.88?
Both are acceptable depending on the Ka value used and the number of significant figures retained. If Ka = 1.8 × 10-5 and the exact quadratic method is used, the result is about 2.875, usually rounded to 2.88.
Why is the pH not 1 for a 0.1 M acid solution?
That rule applies only to a strong monoprotic acid that dissociates completely. Acetic acid is weak and dissociates only partially.
Do I need the quadratic formula every time?
Not always. For 0.1 M acetic acid, the weak acid approximation is valid because the dissociated amount is small compared with the initial concentration. Still, the quadratic method is safest and most accurate.
What happens if the solution is more dilute?
The pH increases because the hydrogen ion concentration falls, but the percent ionization generally increases because dilution favors dissociation of a weak acid.