Calculate the pH in 0.150 M Hippuric Acid
Use this premium weak-acid calculator to solve the pH of a 0.150 M hippuric acid solution using either the exact quadratic method or the common weak-acid approximation. The default dissociation constant is set for hippuric acid, but you can adjust it if your textbook or lab manual uses a different value.
Chart shows the equilibrium breakdown of total acid into undissociated HA and dissociated species H+ and A–.
How to calculate the pH in 0.150 M hippuric acid
To calculate the pH in 0.150 M hippuric acid, you treat hippuric acid as a weak monoprotic acid that only partially dissociates in water. That means you cannot assume the hydrogen ion concentration is equal to the initial acid concentration the way you would for a strong acid such as hydrochloric acid. Instead, you use the acid dissociation equilibrium:
Ka = [H+][A-] / [HA]
In this expression, HA represents hippuric acid, H+ is the hydrogen ion concentration generated by dissociation, and A– is the hippurate conjugate base. For many classroom and analytical chemistry problems, hippuric acid is assigned a pKa near 3.62 at room temperature, which corresponds to a Ka of about 2.40 × 10-4. Once Ka is known, the pH can be found from the equilibrium concentration of H+.
Step-by-step setup for 0.150 M hippuric acid
Start with an ICE table, which tracks Initial, Change, and Equilibrium concentrations. Let the initial concentration of hippuric acid be 0.150 M, and let x be the amount that dissociates.
Change: [HA] = -x, [H+] = +x, [A-] = +x
Equilibrium: [HA] = 0.150 – x, [H+] = x, [A-] = x
Substitute those equilibrium expressions into the Ka expression:
If you use Ka = 2.40 × 10-4, the exact equation becomes:
Rearranging gives a quadratic:
Solving the quadratic yields x ≈ 0.00588 M. Since x equals [H+], the pH is:
So, the pH of a 0.150 M hippuric acid solution is approximately 2.23 when pKa = 3.62 is used. Depending on the source for Ka or pKa, your answer may vary slightly, usually by a few hundredths of a pH unit.
Why hippuric acid is treated as a weak acid
Hippuric acid is an organic acid, structurally related to aromatic carboxylic acid chemistry. It contains a carboxylic acid functional group, and like many carboxylic acids, it ionizes only partially in water. This partial ionization is exactly why weak-acid equilibrium methods are required. If a student incorrectly assumes full dissociation, they would estimate the pH as:
That is far too low for a weak acid of this strength. The correct pH is much higher because only a small fraction of the 0.150 M acid actually ionizes. In the equilibrium solution above, the percent dissociation is:
This result is chemically reasonable. A few percent dissociation is typical for a weak acid at moderate concentration and with a pKa in the mid-3 range. That also tells you something important about the approximation method: because dissociation is under 5%, the simplified weak-acid square-root approach is acceptable here.
Exact method versus approximation method
There are two standard ways to solve weak-acid pH problems. The exact method uses the quadratic equation. The approximation method assumes x is small compared with the starting concentration, allowing you to replace 0.150 – x with 0.150. For hippuric acid at 0.150 M, both methods give very similar answers.
x = √(KaC)
Using Ka = 2.40 × 10-4 and C = 0.150:
Then:
The exact result was about 2.23, while the approximation gives about 2.22. The difference is negligible for many homework, quiz, and introductory laboratory purposes. However, if your instructor requests rigorous equilibrium work, the quadratic method is the safer choice.
| Method | Ka used | [H+] (M) | Calculated pH | Percent dissociation |
|---|---|---|---|---|
| Exact quadratic | 2.40 × 10-4 | 0.00588 | 2.230 | 3.92% |
| Weak-acid approximation | 2.40 × 10-4 | 0.00600 | 2.222 | 4.00% |
| Incorrect strong-acid assumption | Not applicable | 0.150 | 0.824 | 100% |
How concentration changes the pH of hippuric acid
One of the most useful insights in weak-acid chemistry is that the pH of a solution depends not just on acid identity, but also on concentration. Because weak acids only partially dissociate, a more dilute solution generally shows a greater fraction dissociated, although its total hydrogen ion concentration can still be lower. This is why a 0.150 M solution of hippuric acid has a lower pH than a 0.010 M solution, even though the dilute sample dissociates to a larger percentage.
| Initial hippuric acid concentration (M) | Approximate [H+] using exact method (M) | Approximate pH | Approximate percent dissociation |
|---|---|---|---|
| 0.500 | 0.01084 | 1.965 | 2.17% |
| 0.150 | 0.00588 | 2.230 | 3.92% |
| 0.050 | 0.00335 | 2.475 | 6.70% |
| 0.010 | 0.00144 | 2.842 | 14.44% |
The numbers in the table show a classic weak-acid pattern. As the initial concentration drops, pH increases, but the percent dissociation grows. Students often confuse these ideas, so it helps to keep them separate: lower concentration means less total hydrogen ion concentration, which raises pH, but it can still mean a larger fraction of molecules dissociate.
Common mistakes when solving the pH of 0.150 M hippuric acid
- Using the strong-acid formula and assuming complete dissociation.
- Forgetting to convert pKa to Ka before using an ICE table.
- Using pH = -log(Ka), which is not valid for this type of problem.
- Ignoring the subtraction term in the denominator when the percent dissociation is too large for approximation.
- Rounding [H+] too early, which can shift the final pH noticeably.
- Mixing up molarity with molality if a problem statement uses a different concentration unit.
How to decide whether the approximation is valid
The 5% rule is commonly used. If x divided by the initial concentration is below 5%, the approximation is generally acceptable. For 0.150 M hippuric acid, x is about 0.00588 M, so:
Because 3.92% is below 5%, the approximation passes the test. That is why both methods agree closely in this case. At lower concentrations, however, the approximation becomes less reliable because dissociation becomes a larger fraction of the total acid concentration.
Interpreting the chemistry behind the answer
A pH of about 2.23 means the solution is distinctly acidic, but not nearly as acidic as a strong acid at the same formal concentration. It also means the equilibrium strongly favors the undissociated acid relative to the conjugate base. If the equilibrium concentrations are approximately [HA] = 0.144 M and [A–] = 0.00588 M, then most dissolved hippuric acid remains protonated. That is consistent with the pKa value, which indicates a moderate weak acid.
In practical analytical contexts, hippuric acid may appear in discussions of biological metabolites, aromatic carboxylic acids, and urinary chemistry. However, in a standard general chemistry or analytical chemistry problem, the objective is usually narrower: apply equilibrium principles correctly, calculate hydrogen ion concentration, and report pH with proper significant figures.
Worked summary in ordered steps
- Write the weak-acid dissociation equation: HA ⇌ H+ + A–.
- Use the given concentration, 0.150 M, as the initial [HA].
- Convert pKa to Ka if needed: Ka = 10-pKa.
- Set up the ICE table with x as the amount dissociated.
- Substitute into Ka = x² / (0.150 – x).
- Solve exactly with the quadratic or approximately using x = √(KaC).
- Calculate pH from pH = -log[H+].
- Optionally verify the 5% rule to justify the approximation.
Authoritative chemistry references
If you want to verify acid-base definitions, weak-acid methods, or compound identity information, these authoritative resources are useful:
- NIH PubChem: Hippuric Acid
- LibreTexts Chemistry is popular, but for a strict .gov or .edu reference set use the two links below as well.
- NIST Chemistry WebBook
- Purdue University General Chemistry Acid-Base Review
Bottom line
To calculate the pH in 0.150 M hippuric acid, model the compound as a weak monoprotic acid and solve its dissociation equilibrium. With a typical pKa of 3.62, the exact pH comes out to about 2.23. The weak-acid approximation also works well here and gives essentially the same value. If your assignment provides a slightly different pKa or Ka, your final answer may shift modestly, but the method stays exactly the same.