Calculate the pH in 0.160 M Hippuric Acid
Use this premium weak-acid calculator to find the hydrogen ion concentration, pH, percent ionization, and equilibrium concentrations for a 0.160 M hippuric acid solution using the acid dissociation constant and the full quadratic solution.
Chart compares the starting acid concentration with the equilibrium concentrations of HA, H+, and A–.
How to calculate the pH in 0.160 M hippuric acid
To calculate the pH in 0.160 M hippuric acid, you treat hippuric acid as a weak monoprotic acid that partially dissociates in water. The governing equilibrium is:
HA ⇌ H+ + A–
Here, HA is hippuric acid, H+ is the hydrogen ion produced by dissociation, and A– is the conjugate base of hippuric acid. Because hippuric acid is not a strong acid, it does not dissociate completely. That means the pH cannot be found by simply taking the negative logarithm of the initial concentration. Instead, you must use the acid dissociation constant, Ka, or its logarithmic form, pKa.
Short answer: using a typical Ka = 2.4 × 10-4 for hippuric acid at about 25 C, the pH of a 0.160 M solution is approximately 2.22 when solved with the quadratic equation.
Step 1: Write the Ka expression
For a weak acid HA placed in water at an initial concentration of 0.160 M, the equilibrium expression is:
Ka = [H+][A–] / [HA]
If x represents the amount of acid that dissociates, then at equilibrium:
- [H+] = x
- [A–] = x
- [HA] = 0.160 – x
Substitute these values into the expression:
Ka = x2 / (0.160 – x)
Step 2: Insert the Ka of hippuric acid
A commonly used pKa value for hippuric acid is about 3.62, which corresponds to a Ka near 2.4 × 10-4. Substituting that value gives:
2.4 × 10-4 = x2 / (0.160 – x)
Now rearrange into quadratic form:
x2 + (2.4 × 10-4)x – (3.84 × 10-5) = 0
Solving for the positive root gives:
x ≈ 0.00608 M
That is the equilibrium hydrogen ion concentration.
Step 3: Convert hydrogen ion concentration into pH
Once [H+] is known, pH follows from the usual formula:
pH = -log[H+]
So:
pH = -log(0.00608) ≈ 2.22
This is the pH of a 0.160 M hippuric acid solution under typical room-temperature conditions using the full equilibrium treatment.
Why the quadratic method is preferred here
Students are often taught the weak-acid approximation:
x ≈ √(Ka × C)
Using that shortcut with Ka = 2.4 × 10-4 and C = 0.160 M gives:
x ≈ √(3.84 × 10-5) ≈ 0.00620 M
This leads to a pH of about 2.21, which is very close to the more accurate quadratic result. The approximation works reasonably well because the amount dissociated is only a small fraction of the initial acid concentration. However, for a polished chemistry calculation, especially one intended for publication, tutoring, or exam review, the quadratic method is better because it avoids assuming x is negligible.
| Method | Equation used | [H+] result | Calculated pH | Difference from quadratic |
|---|---|---|---|---|
| Quadratic solution | x = (-Ka + √(Ka² + 4KaC)) / 2 | 0.00608 M | 2.22 | Baseline |
| Weak-acid approximation | x ≈ √(KaC) | 0.00620 M | 2.21 | About 0.01 pH unit |
Understanding what the number means chemically
A pH of about 2.22 means the solution is distinctly acidic, but not because every hippuric acid molecule ionizes. In fact, most of the acid remains in the protonated HA form. Only a small percentage dissociates. This is the defining behavior of a weak acid.
The percent ionization can be found from:
% ionization = (x / C) × 100
Substituting the values:
(0.00608 / 0.160) × 100 ≈ 3.80%
So even though the pH is low, only about 3.8% of the hippuric acid molecules donate a proton at equilibrium in this concentration range. That is a useful reminder that pH alone does not tell you whether an acid is strong or weak. Strong and weak refer to degree of ionization, while pH reflects the resulting hydrogen ion concentration.
Worked example with the full ICE approach
- Start with the reaction: HA ⇌ H+ + A–
- Initial concentrations: [HA] = 0.160, [H+] = 0, [A–] = 0
- Change: -x, +x, +x
- Equilibrium: [HA] = 0.160 – x, [H+] = x, [A–] = x
- Substitute into Ka: 2.4 × 10-4 = x2 / (0.160 – x)
- Solve the quadratic: x ≈ 0.00608
- Compute pH: -log(0.00608) ≈ 2.22
This stepwise structure is the cleanest method for homework, exam preparation, and laboratory report writing because it shows the physical meaning behind each term.
Comparison with other weak acids
Hippuric acid is a moderately weak organic acid. It is stronger than acetic acid, but weaker than some aromatic carboxylic acids that have more electron-withdrawing substituents. Comparing pKa values can help put the result in context. Lower pKa means stronger acid and generally lower pH at the same concentration.
| Acid | Typical pKa at 25 C | Approximate Ka | Predicted acidity trend at equal concentration |
|---|---|---|---|
| Benzoic acid | 4.20 | 6.3 × 10-5 | Weaker than hippuric acid |
| Acetic acid | 4.76 | 1.7 × 10-5 | Clearly weaker than hippuric acid |
| Hippuric acid | 3.62 | 2.4 × 10-4 | Intermediate organic weak acid |
| Formic acid | 3.75 | 1.8 × 10-4 | Slightly weaker than hippuric acid |
What can change the exact pH value?
While 2.22 is a strong working answer for most educational contexts, the exact pH of a real solution can shift slightly depending on several factors:
- Temperature: Ka values depend on temperature, so pH is not perfectly fixed across all conditions.
- Ionic strength: At higher solute concentrations, activities differ from concentrations. Advanced calculations may use activity coefficients instead of raw molarity.
- Reference source: Published pKa values for organic acids can differ slightly depending on solvent conditions and measurement methods.
- Concentration basis: Some sources use molarity, while others may mention molality or buffered systems. For dilute aqueous chemistry, molarity is typically assumed.
Because your question specifically asks for the pH in 0.160 M hippuric acid, the standard classroom interpretation is an aqueous solution at about 25 C using accepted literature Ka or pKa data.
Common mistakes to avoid
- Treating hippuric acid as a strong acid. If you assume complete dissociation, you would calculate pH as -log(0.160) = 0.80, which is far too low.
- Using pKa directly as pH. pKa is a property of the acid, not the solution pH.
- Forgetting the quadratic option. The shortcut is often acceptable, but the full equation is more rigorous.
- Ignoring units. Ka expressions depend on concentration terms, so consistency matters.
- Using the wrong logarithm base. pH uses base-10 logarithms, not natural logarithms.
Quick interpretation of the equilibrium concentrations
At equilibrium for 0.160 M hippuric acid with Ka = 2.4 × 10-4:
- [H+] ≈ 0.00608 M
- [A–] ≈ 0.00608 M
- [HA] ≈ 0.15392 M
This tells you that the protonated acid remains the dominant species in solution. The conjugate base is present, but at a much lower concentration. That is exactly what we expect from a weak acid with a modest Ka value.
Practical uses of this calculation
Knowing how to calculate the pH of hippuric acid matters in analytical chemistry, biochemistry, pharmaceutical chemistry, and acid-base instructional settings. Hippuric acid is relevant in metabolic and urinary chemistry, and pH calculations help chemists understand:
- solution speciation
- buffering potential when combined with its conjugate base
- reaction conditions for extraction or chromatography
- ionization-dependent solubility and separation behavior
Even when a problem appears straightforward, solving it correctly reinforces the larger concept that weak-acid systems are governed by equilibrium, not complete dissociation assumptions.
Authoritative references for acid-base data and chemistry fundamentals
- NIH PubChem: Hippuric Acid
- LibreTexts Chemistry hosted by educational institutions
- National Institute of Standards and Technology
Final answer
If you are asked to calculate the pH in 0.160 M hippuric acid, the standard answer is:
pH ≈ 2.22
This result assumes an aqueous solution near 25 C and a hippuric acid Ka of approximately 2.4 × 10-4.