Calculate The Ph In 0.110 M Hippuric Acid

Use this premium weak-acid calculator to determine the pH of a 0.110 M hippuric acid solution using the accepted acid dissociation relationship. The calculator supports exact quadratic solving, approximation checks, percent ionization, and a visual concentration chart.

Hippuric Acid pH Calculator

Quick Chemistry Snapshot

Acid type

Weak monoprotic acid

Formula focus

HA ⇌ H⁺ + A^–

Recommended method

Quadratic equation

Default answer

pH ≈ 2.299

How to calculate the pH in 0.110 M hippuric acid

Calculating the pH of a weak acid solution is a classic equilibrium problem in general chemistry. In this case, the solute is hippuric acid and the concentration is 0.110 M. Because hippuric acid is a weak acid, it does not fully dissociate in water. That means you cannot treat the hydrogen ion concentration as equal to the formal acid concentration, which would only be valid for a strong acid like hydrochloric acid. Instead, you need the acid dissociation constant, or Ka, and then solve the equilibrium expression.

For hippuric acid, a commonly used pKa value near room temperature is approximately 3.62. Since pKa and Ka are related by the expression Ka = 10^-pKa, this gives Ka about 2.40 × 10^-4. Once Ka is known, the equilibrium can be written and solved to determine the actual hydrogen ion concentration in a 0.110 M solution. From there, pH follows directly from pH = -log₁₀[H⁺].

The exact answer for 0.110 M hippuric acid with pKa = 3.62 is a pH of about 2.299. That result reflects partial ionization, not complete dissociation. This is why weak-acid problems are equilibrium problems rather than straightforward stoichiometry problems.

Step 1: Write the dissociation equation

Hippuric acid behaves as a weak monoprotic acid, so the equilibrium in water is:

HA ⇌ H⁺ + A^–

Here, HA stands for hippuric acid and A^– is the conjugate base. If the initial concentration is 0.110 M and x dissociates, then the equilibrium concentrations are:

• [HA] = 0.110 – x
• [H⁺] = x
• [A^–] = x

Step 2: Set up the Ka expression

The acid dissociation constant expression is:

Ka = [H⁺][A^–] / [HA]

Substituting the equilibrium values gives:

2.40 × 10^-4 = x² / (0.110 – x)

This equation can be solved either by approximation or exactly with the quadratic formula. Since the concentration and Ka are such that dissociation is not negligibly tiny, the exact approach is the best practice if you want a reliable answer with minimal rounding error.

Step 3: Solve exactly with the quadratic equation

Rearrange:

x² = Ka(0.110 – x)

x² + Kax – Ka(0.110) = 0

Using Ka = 2.40 × 10^-4:

x = [-Ka + √(Ka² + 4KaC)] / 2

where C = 0.110 M. Evaluating the expression gives:

• x = [H⁺] ≈ 0.00502 M
• pH = -log(0.00502) ≈ 2.299

Therefore, the pH in 0.110 M hippuric acid is about 2.30 when rounded appropriately.

Why the weak-acid approximation is close but not always preferred

In many textbook problems, students first try the common approximation that 0.110 – x ≈ 0.110. That simplifies the Ka expression to:

x ≈ √(KaC)

Plugging in the values:

x ≈ √[(2.40 × 10^-4)(0.110)] ≈ 0.00514 M

That gives a pH around 2.289. This is close to the exact answer, but not identical. The reason is that x is not infinitesimally small compared with the initial concentration. Since the percent ionization is around 4.56%, the 5% rule is barely satisfied. In a formal chemistry setting, the exact quadratic solution is the safest and most defensible method.

Method	[H^+] in 0.110 M hippuric acid	Calculated pH	Comment
Exact quadratic	0.00502 M	2.299	Best method for reporting final answer
Weak-acid approximation	0.00514 M	2.289	Close estimate, slightly lower pH
Incorrect strong-acid assumption	0.110 M	0.959	Not valid because hippuric acid is weak

Understanding the chemistry behind the number

A pH near 2.30 tells you the solution is clearly acidic, but it is much less acidic than a 0.110 M strong acid would be. If this were 0.110 M hydrochloric acid, the pH would be about 0.96 because nearly every acid molecule would donate a proton. Hippuric acid behaves differently because equilibrium limits dissociation. Only a modest fraction of molecules ionize in water.

This distinction is important in biochemistry, analytical chemistry, and pharmaceutical chemistry, where weak acids are common. A compound can be present at a relatively high formal concentration yet still yield a much higher pH than a strong acid at the same concentration. That is exactly what happens here.

Percent ionization for 0.110 M hippuric acid

Percent ionization measures how much of the acid actually dissociates:

Percent ionization = ([H⁺] / initial concentration) × 100

Using the exact value:

Percent ionization = (0.00502 / 0.110) × 100 ≈ 4.56%

This means a little under 1 molecule in 20 is dissociated at equilibrium. That is a useful conceptual check because it confirms that hippuric acid is weak, but not so weak that ionization is negligible.

Comparison with other concentrations

The pH of a weak acid depends on both Ka and initial concentration. If the concentration changes, the pH also changes, but not in a simple one-to-one way. For the same pKa value, more dilute weak-acid solutions usually have a higher pH and a larger percent ionization. The table below shows calculated values for hippuric acid at several representative concentrations, using pKa = 3.62 and the exact quadratic method.

Initial concentration (M)	Ka used	Exact [H^+] (M)	pH	Percent ionization
0.010	2.40 × 10^-4	0.00143	2.846	14.31%
0.050	2.40 × 10^-4	0.00335	2.475	6.70%
0.110	2.40 × 10^-4	0.00502	2.299	4.56%
0.250	2.40 × 10^-4	0.00762	2.118	3.05%

These values highlight a pattern that chemistry students should remember: as concentration decreases, weak acids ionize to a greater fraction of their total concentration. That trend explains why the approximation becomes less reliable in dilute solutions and why exact equilibrium calculations matter.

Common mistakes when solving this problem

1. Using concentration directly as [H⁺]. This is the most common mistake. Weak acids do not fully dissociate, so [H⁺] is much smaller than 0.110 M.
2. Confusing pKa and Ka. If pKa = 3.62, that does not mean Ka = 3.62. You must convert using Ka = 10^-pKa.
3. Dropping x without checking. The approximation x << C is useful only when percent ionization stays comfortably small. Here it is borderline acceptable, but the exact solution is better.
4. Incorrect logarithm handling. pH uses the negative base-10 logarithm: pH = -log₁₀[H⁺].
5. Rounding too early. Keep extra digits during the Ka conversion and quadratic solving steps, then round the final pH at the end.

For graded work, report the final pH with appropriate significant figures and mention the method used. If your instructor expects an exact equilibrium result, use the quadratic formula rather than the square-root approximation.

Practical interpretation of the result

A pH of about 2.30 indicates a strongly acidic laboratory solution, though not as aggressive as a strong acid at the same concentration. In practical terms, this matters for solubility, ionization state, and reactivity. For aromatic carboxylic acids such as hippuric acid, protonation and deprotonation behavior influences extraction procedures, buffer response, and chromatographic mobility. If you are working in an academic chemistry lab, understanding the pH helps you predict whether the compound will remain mostly protonated or shift toward the conjugate-base form under different conditions.

The equilibrium result also shows why weak acids are often used in systems where partial ionization is desirable. They provide acidity without the complete dissociation profile of strong mineral acids. In educational settings, hippuric acid makes a useful example because it combines realistic organic acid behavior with a mathematically manageable Ka.

Authoritative references for acid-base data and pH fundamentals

• Chemistry LibreTexts educational resource
• U.S. Environmental Protection Agency acid-base and water chemistry resources
• NIST Chemistry WebBook for authoritative chemical information

Final answer summary

To calculate the pH in 0.110 M hippuric acid, use pKa ≈ 3.62, convert to Ka ≈ 2.40 × 10^-4, write the weak-acid equilibrium expression, and solve for [H⁺] using the quadratic equation. The exact hydrogen ion concentration is approximately 0.00502 M, which gives:

pH = 2.299, or about 2.30

This is the correct chemistry answer because hippuric acid is a weak acid and only partially ionizes in water.