Calculate the pH in 8.70 × 10-2 M CH3CO2H
Use this premium weak-acid calculator to find the hydrogen ion concentration, pH, percent dissociation, and equilibrium composition for acetic acid. The default setup matches the problem statement: 8.70 × 10-2 M CH3CO2H.
Chart compares the initial acid concentration with equilibrium amounts of H+, CH3CO2–, and undissociated CH3CO2H.
How to calculate the pH in 8.70 × 10-2 M CH3CO2H
To calculate the pH in 8.70 × 10-2 M CH3CO2H, you treat acetic acid as a weak acid, not as a strong acid that dissociates completely. That is the key idea. Acetic acid, commonly written as CH3CO2H or HC2H3O2, only partially ionizes in water:
Because the ionization is incomplete, the pH is not equal to the negative log of the initial acid concentration. Instead, you must use the acid dissociation constant, Ka. For acetic acid at room temperature, a common textbook value is approximately 1.8 × 10-5. Once you know the concentration and the Ka value, you can solve for the hydrogen ion concentration and then calculate pH.
Final answer for the default problem
For a solution that is 8.70 × 10-2 M acetic acid, using Ka = 1.8 × 10-5, the equilibrium hydrogen ion concentration is about 1.24 × 10-3 M. Therefore:
This value makes chemical sense. The pH is acidic, but not nearly as low as a strong acid of the same concentration would be. A strong monoprotic acid at 0.0870 M would have pH close to 1.06, while acetic acid has a pH near 2.91 because it dissociates only slightly.
Step by step solution
The safest way to solve this problem is with an ICE table and the Ka expression. Let the initial concentration of acetic acid be C = 0.0870 M, and let x represent the amount that dissociates.
Change: [CH3CO2H] = -x, [H3O+] = +x, [CH3CO2-] = +x
Equilibrium: [CH3CO2H] = 0.0870 – x, [H3O+] = x, [CH3CO2-] = x
Now substitute into the acid dissociation expression:
Since Ka for acetic acid is 1.8 × 10-5, the equation becomes:
You can solve this either by the weak-acid approximation or by the quadratic formula. The approximation assumes x is much smaller than 0.0870, so:
Then calculate pH:
If you use the exact quadratic solution, you get essentially the same answer but with slightly higher accuracy:
Substituting the values gives x ≈ 1.242 × 10-3 M, so pH ≈ 2.906. Rounded properly, the pH is 2.91.
Why you should not treat acetic acid like a strong acid
One of the most common mistakes in introductory chemistry is to see a concentration like 8.70 × 10-2 M and immediately compute pH = -log(0.0870). That would give a pH around 1.06, which is far too acidic for acetic acid. The reason is simple: acetic acid does not fully ionize. Its Ka value is small, meaning the equilibrium strongly favors the undissociated acid.
- Strong acid assumption: complete ionization, [H+] = initial concentration
- Weak acid reality: partial ionization, [H+] must be solved from Ka
- For CH3CO2H: only a small fraction dissociates
In this problem, the percent dissociation is only about 1.43%. That means more than 98% of the acetic acid remains in molecular form at equilibrium.
Comparison table: weak acid result versus strong acid assumption
| Scenario | Initial acid concentration | Estimated [H+] | Calculated pH | Interpretation |
|---|---|---|---|---|
| Acetic acid solved correctly as a weak acid | 0.0870 M | 1.24 × 10-3 M | 2.91 | Uses Ka and equilibrium |
| Incorrect strong acid assumption | 0.0870 M | 8.70 × 10-2 M | 1.06 | Too acidic because it assumes 100% ionization |
| Difference | Same starting concentration | About 70 times more H+ predicted by strong-acid error | About 1.85 pH units lower | Shows why weak-acid treatment matters |
What the chemistry means in practical terms
A pH near 2.91 indicates a distinctly acidic solution, but it is still substantially less acidic than a strong acid at the same molarity. This is why Ka matters so much in acid-base chemistry. Acetic acid is the main acidic component of vinegar, although household vinegar is typically around 5% acetic acid by mass and not exactly the same concentration as the textbook problem here. Real solution pH can also shift somewhat with temperature, ionic strength, and the exact Ka value selected from a reference table.
In general chemistry classes, acetic acid is often used to teach:
- weak acid equilibrium setup
- ICE table organization
- Ka and pKa interpretation
- approximation versus exact quadratic methods
- buffer chemistry involving acetate and acetic acid
Data table: how concentration changes the pH of acetic acid
The following table uses the same Ka value, 1.8 × 10-5, and shows representative pH values for acetic acid solutions of different concentrations. These values are based on the weak-acid equilibrium relationship and illustrate a real and useful trend: as concentration decreases, pH rises and percent dissociation increases.
| Acetic acid concentration (M) | Approximate [H+] (M) | Approximate pH | Percent dissociation |
|---|---|---|---|
| 1.00 | 4.23 × 10-3 | 2.37 | 0.42% |
| 0.100 | 1.33 × 10-3 | 2.88 | 1.33% |
| 0.0870 | 1.24 × 10-3 | 2.91 | 1.43% |
| 0.0100 | 4.15 × 10-4 | 3.38 | 4.15% |
| 0.00100 | 1.25 × 10-4 | 3.90 | 12.5% |
When the approximation works and when it does not
Many students are taught to use the shortcut x ≈ √(KaC). This is extremely useful, but it depends on x being small relative to the initial concentration. A common classroom rule is that the approximation is acceptable when the percent dissociation is below about 5%. For the 0.0870 M acetic acid problem, x is only around 1.43% of the initial concentration, so the approximation is very good.
- If percent dissociation is under 5%, the approximation is generally acceptable.
- If the acid is very dilute, percent dissociation can rise and the quadratic method becomes safer.
- The calculator above lets you compare both methods instantly.
Common mistakes when solving this exact pH problem
- Using pH = -log(0.0870) as if the acid were strong.
- Using the wrong Ka for acetic acid. Many textbooks use 1.8 × 10-5, though some references give slightly different values.
- Forgetting that x appears in the denominator when setting up the equilibrium expression.
- Rounding too early, which can slightly distort the final pH.
- Ignoring units. Ka is unitless in many classroom contexts, but concentration values must still be in molarity.
How this connects to pKa and buffers
Acetic acid has a pKa near 4.74 to 4.76 at room temperature, depending on the reference source and conditions. Since the calculated pH of the solution here is around 2.91, the solution is well below the pKa, meaning the protonated acid form strongly dominates over acetate. That fits what we found from the equilibrium calculation: most of the acetic acid remains undissociated.
If sodium acetate were added, the solution would become a buffer and the Henderson-Hasselbalch equation would become relevant. But for a solution containing only acetic acid in water, the Ka equilibrium method is the correct starting point.
Reliable chemistry references
If you want to review pH, weak acid behavior, and water chemistry from authoritative educational and government sources, these references are useful:
Bottom line
To calculate the pH in 8.70 × 10-2 M CH3CO2H, you must treat acetic acid as a weak acid and use its dissociation constant. Set up the equilibrium, solve for the hydrogen ion concentration, and then take the negative log. With Ka = 1.8 × 10-5, the exact pH is about 2.906, which rounds to 2.91.
Use the calculator on this page if you want to test alternative Ka values, compare approximation versus quadratic methods, or visualize how much acetic acid remains undissociated at equilibrium. For this specific question, however, the headline result is straightforward: