Calculate the pH in 0.170 M Hippuric Acid
Use this premium weak acid calculator to determine pH, hydrogen ion concentration, percent ionization, and equilibrium species concentrations for hippuric acid solutions.
Default values calculate the pH of 0.170 M hippuric acid using pKa = 3.62, which corresponds to Ka about 2.40 × 10-4.
Expert Guide: How to Calculate the pH in 0.170 M Hippuric Acid
Calculating the pH in a 0.170 M hippuric acid solution is a classic weak acid equilibrium problem. Even though the setup looks simple, it provides an excellent example of how acid strength, concentration, and equilibrium assumptions work together in aqueous chemistry. Hippuric acid is not a strong acid, so it does not dissociate completely in water. That means you cannot simply treat the molarity as equal to the hydrogen ion concentration. Instead, you must use the acid dissociation constant, usually written as Ka, or its logarithmic form, pKa.
For a typical calculation at 25 degrees C, hippuric acid is often represented with a pKa near 3.62, which corresponds to a Ka of about 2.40 × 10-4. With an initial concentration of 0.170 M, the equilibrium pH comes out near 2.20 when solved exactly. This page helps you calculate that result and understand why it is correct.
Quick answer: Using 0.170 M hippuric acid and pKa = 3.62, the exact weak acid equilibrium solution gives a pH of about 2.20. The hydrogen ion concentration is about 6.27 × 10-3 M.
What Is Hippuric Acid?
Hippuric acid is an organic acid with the molecular formula C9H9NO3. In acid base calculations, it behaves as a monoprotic weak acid, meaning each molecule can donate one acidic proton to water. In solution, the equilibrium can be written as:
HA + H2O ⇌ H3O+ + A–
Here, HA is hippuric acid, H3O+ is hydronium, and A– is the hippurate conjugate base. The equilibrium constant expression is:
Ka = [H3O+][A–] / [HA]
Because hippuric acid is weak, only a small fraction ionizes. That is why equilibrium chemistry, not simple stoichiometry alone, is required.
Known Values for the Problem
- Initial concentration of hippuric acid, C = 0.170 M
- Typical pKa of hippuric acid at 25 degrees C = 3.62
- Therefore Ka = 10-3.62 ≈ 2.40 × 10-4
- Goal: find pH = -log[H+]
Step by Step Weak Acid Setup
The standard way to solve this problem is with an ICE table, which tracks Initial, Change, and Equilibrium concentrations.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| HA | 0.170 | -x | 0.170 – x |
| H+ | 0 | +x | x |
| A– | 0 | +x | x |
Substitute these equilibrium values into the Ka expression:
Ka = x2 / (0.170 – x)
Then insert Ka = 2.40 × 10-4:
2.40 × 10-4 = x2 / (0.170 – x)
Multiply both sides through:
x2 + (2.40 × 10-4)x – (4.08 × 10-5) = 0
Now solve using the quadratic formula:
x = [-b + √(b2 – 4ac)] / 2a
With a = 1, b = 2.40 × 10-4, and c = -4.08 × 10-5, the positive root is approximately:
x ≈ 6.27 × 10-3 M
Since x = [H+], the pH is:
pH = -log(6.27 × 10-3) ≈ 2.20
Why the Square Root Shortcut Also Works
For many weak acids, if x is small relative to the initial concentration, we approximate 0.170 – x as 0.170. That gives:
x ≈ √(Ka × C)
Substitute the values:
x ≈ √((2.40 × 10-4)(0.170)) = √(4.08 × 10-5) ≈ 6.39 × 10-3 M
Then:
pH ≈ -log(6.39 × 10-3) ≈ 2.19
This is very close to the exact value of 2.20. The approximation works because the amount ionized is only a few percent of the initial acid concentration. Still, the exact quadratic solution is preferred when you want the most defensible answer.
Percent Ionization
Percent ionization helps describe how much of the acid actually dissociated:
Percent ionization = (x / 0.170) × 100
Using the exact x value of 6.27 × 10-3 M:
Percent ionization ≈ 3.69%
This confirms that hippuric acid is weak in water. Most molecules remain in the undissociated HA form, and only a small fraction becomes H+ and A–.
Final Equilibrium Concentrations
- [H+] ≈ 6.27 × 10-3 M
- [A–] ≈ 6.27 × 10-3 M
- [HA] ≈ 0.170 – 0.00627 = 0.1637 M
- pH ≈ 2.20
Comparison Table: Exact vs Approximate Method
| Method | [H+] (M) | pH | Difference from Exact |
|---|---|---|---|
| Exact quadratic solution | 6.27 × 10-3 | 2.203 | Baseline |
| Square root approximation | 6.39 × 10-3 | 2.194 | About 0.009 pH units lower |
How Concentration Changes the pH of Hippuric Acid
One of the best ways to build intuition is to compare several concentrations while keeping the same pKa value. More concentrated weak acid solutions have lower pH because there are more acid molecules available to ionize, even though the fraction ionized often decreases.
| Initial Hippuric Acid Concentration (M) | Exact [H+] (M) | Exact pH | Percent Ionization |
|---|---|---|---|
| 0.010 | 1.43 × 10-3 | 2.84 | 14.3% |
| 0.050 | 3.35 × 10-3 | 2.47 | 6.70% |
| 0.100 | 4.78 × 10-3 | 2.32 | 4.78% |
| 0.170 | 6.27 × 10-3 | 2.20 | 3.69% |
| 0.500 | 1.08 × 10-2 | 1.97 | 2.16% |
Common Mistakes Students Make
- Treating hippuric acid like a strong acid. If you incorrectly set [H+] = 0.170 M, you would get pH 0.77, which is far too low.
- Using pKa directly in the equilibrium equation. You must convert pKa to Ka first unless your calculator does the conversion automatically.
- Ignoring the exact method when needed. Approximations are helpful, but exact calculations are more reliable.
- Forgetting that x affects the denominator. In weak acid problems, [HA] at equilibrium becomes C – x.
- Reporting too many significant figures. For most educational settings, pH 2.20 is the appropriately rounded result.
When Should You Use Ka and When Should You Use pKa?
Ka is the direct equilibrium constant and is required in the equation. pKa is simply a compact way to express acid strength on a logarithmic scale. You can move between them with:
- Ka = 10-pKa
- pKa = -log(Ka)
In textbooks and laboratory references, you may see one or the other. If your source gives pKa = 3.62, your first step is to convert it to Ka ≈ 2.40 × 10-4.
How Reliable Is the Result?
The final pH depends slightly on the exact acid dissociation constant used. Different reference sources may report a small range of values depending on temperature, ionic strength, or methodology. For routine general chemistry work, using pKa around 3.62 is reasonable and gives a pH near 2.20 for a 0.170 M solution.
If you are working in analytical chemistry, biochemistry, or a regulated laboratory environment, you may also need to account for activity effects rather than using concentration alone. At higher ionic strengths, the measured pH can deviate modestly from the idealized classroom calculation. Still, for standard educational problem solving, the exact concentration based equilibrium calculation is the accepted method.
Useful Formula Summary
- Ka = [H+][A–] / [HA]
- pH = -log[H+]
- Ka = 10-pKa
- Approximation for a weak acid: [H+] ≈ √(KaC)
- Percent ionization = ([H+] / C) × 100
Authority References and Further Reading
If you want to verify constants, review acid base principles, or explore the compound in more detail, these authoritative resources are useful:
- PubChem, National Institutes of Health: Hippuric Acid Compound Record
- University of California Davis educational chemistry material on weak acid equilibria
- NIST reference material for pH related measurements and standards
Bottom Line
To calculate the pH in 0.170 M hippuric acid, treat it as a weak monoprotic acid and solve the equilibrium using Ka. With pKa = 3.62, Ka is about 2.40 × 10-4. Solving the equilibrium equation gives [H+] ≈ 6.27 × 10-3 M and a final pH of about 2.20. That answer reflects partial ionization, which is exactly what you expect for a weak acid. Use the calculator above to test other concentrations, compare the exact and approximate methods, and visualize how the species distribute at equilibrium.
Educational note: real laboratory pH measurements can vary slightly due to temperature, ionic strength, and electrode calibration. The calculator above is designed for standard aqueous equilibrium calculations.