Calculate The Ph In 0.120 M Hippuric Acid

Use this premium weak-acid calculator to solve the pH of a 0.120 M hippuric acid solution with either the exact quadratic method or the common square-root approximation. The tool also visualizes the acid dissociation results instantly.

Hippuric Acid pH Calculator

Quick Chemistry Notes

• Species: Hippuric acid is a weak monoprotic acid, so the main equilibrium is HA ⇌ H⁺ + A^–.
• Default values: For this calculator, the standard worked example uses 0.120 M concentration and pKa = 3.62.
• Main equation: Ka = [H⁺][A^–] / [HA]
• Exact method: Solve x² / (C – x) = Ka with the quadratic formula.
• Approximation: If dissociation is small, x ≈ √(KaC).
• Expected result: The pH for 0.120 M hippuric acid is about 2.28 using the exact method with pKa 3.62.

How to calculate the pH in 0.120 M hippuric acid

If you need to calculate the pH in 0.120 M hippuric acid, the key idea is that hippuric acid behaves as a weak acid in water. That means it does not dissociate completely the way a strong acid like hydrochloric acid would. Instead, only a fraction of the dissolved acid molecules donate a proton to water, creating hydronium ions and lowering the pH. Because the dissociation is partial, the pH must be determined from the weak-acid equilibrium expression rather than from concentration alone.

For most general chemistry and analytical chemistry problems, the given concentration is the formal concentration of the acid, and you are either given a Ka value or a pKa value. In many textbook and laboratory references, hippuric acid is listed with a pKa around 3.62 at room temperature. Using that value, you can convert to Ka and solve the equilibrium exactly.

Bottom line: For a 0.120 M hippuric acid solution with pKa = 3.62, the exact equilibrium calculation gives a pH of approximately 2.28.

Step 1: Write the acid dissociation equation

Hippuric acid can be represented as HA. In water, the weak-acid equilibrium is:

HA(aq) ⇌ H+(aq) + A-(aq)

If the initial concentration of hippuric acid is 0.120 M, then the standard ICE setup looks like this:

• Initial: [HA] = 0.120, [H⁺] = 0, [A^–] = 0
• Change: [HA] decreases by x, [H⁺] increases by x, [A^–] increases by x
• Equilibrium: [HA] = 0.120 – x, [H⁺] = x, [A^–] = x

Step 2: Convert pKa to Ka

The relationship between Ka and pKa is:

Ka = 10^(-pKa)

With pKa = 3.62:

Ka = 10^(-3.62) = 2.40 × 10^-4

This value tells you that hippuric acid is clearly weak. Its Ka is much smaller than 1, which means only a modest fraction of the 0.120 M solution ionizes.

Step 3: Apply the equilibrium expression

The weak-acid equilibrium expression is:

Ka = [H+][A-] / [HA]

Substitute the ICE values:

2.40 × 10^-4 = x^2 / (0.120 – x)

Now solve for x. Here, x is the equilibrium hydronium concentration, which is also the concentration of the conjugate base formed.

Step 4: Solve exactly with the quadratic formula

Multiply both sides by (0.120 – x):

2.40 × 10^-4 (0.120 – x) = x^2

Expand and rearrange:

2.88 × 10^-5 – 2.40 × 10^-4 x = x^2

x^2 + 2.40 × 10^-4 x – 2.88 × 10^-5 = 0

Then use the quadratic formula:

x = [-b + √(b^2 – 4ac)] / 2a

Using a = 1, b = 2.40 × 10^-4, and c = -2.88 × 10^-5, the physically meaningful root is:

x ≈ 5.25 × 10^-3 M

Since x = [H⁺], calculate pH:

pH = -log10[H+] = -log10(5.25 × 10^-3) ≈ 2.28

Can you use the weak-acid approximation?

Yes. In many classroom settings, you may be allowed to use the approximation that x is small compared with the initial concentration 0.120 M. If that assumption is valid, then 0.120 – x is treated as approximately 0.120. The equilibrium expression becomes:

Ka ≈ x^2 / C

x ≈ √(KaC)

Substitute the values:

x ≈ √((2.40 × 10^-4)(0.120)) = √(2.88 × 10^-5) ≈ 5.37 × 10^-3 M

Then:

pH ≈ -log10(5.37 × 10^-3) ≈ 2.27

This result is very close to the exact answer, differing by only about 0.01 pH unit. The approximation works well because the percent dissociation is modest and the change in acid concentration remains relatively small compared with the initial concentration.

Comparison of exact and approximate results

The following table shows why both methods are often discussed together in chemistry courses. The exact method is always more rigorous, but the approximation is often fast and reasonably accurate for a weak acid at moderate concentration.

Method	Ka used	[H^+] produced	Calculated pH	Comment
Exact quadratic	2.40 × 10^-4	5.25 × 10^-3 M	2.28	Best choice when full precision is expected.
Square-root approximation	2.40 × 10^-4	5.37 × 10^-3 M	2.27	Fast estimate; excellent agreement in this case.

Percent dissociation in 0.120 M hippuric acid

Percent dissociation tells you what fraction of the initial acid molecules actually ionized. It is calculated as:

% dissociation = ([H+] / initial concentration) × 100

Using the exact result:

% dissociation = (0.00525 / 0.120) × 100 ≈ 4.37%

That means more than 95% of the original hippuric acid remains in its protonated HA form at equilibrium. This is exactly the sort of behavior expected for a weak acid.

Equilibrium quantity	Value for 0.120 M hippuric acid	Interpretation
Initial [HA]	0.120 M	Formal concentration before dissociation.
Equilibrium [H^+]	5.25 × 10^-3 M	Determines pH directly.
Equilibrium [A^–]	5.25 × 10^-3 M	Equal to [H^+] for a monoprotic acid in this setup.
Equilibrium [HA]	0.11475 M	Most acid remains undissociated.
Percent dissociation	4.37%	Confirms weak-acid behavior.

Why the pH is not simply 0.92

Students sometimes mistakenly treat every acid like a strong acid and assume [H⁺] = 0.120 M. If that were true, the pH would be:

pH = -log10(0.120) ≈ 0.92

But that result would apply only if the acid dissociated essentially completely. Hippuric acid does not do that. Its finite Ka means the equilibrium lies far to the left compared with a strong acid. As a result, the real hydronium concentration is only around 0.00525 M, not 0.120 M, and the actual pH is much higher at about 2.28.

When to choose the exact calculation over the approximation

In practice, the exact calculation is preferred when:

• Your instructor explicitly asks for an ICE-table solution with no approximation.
• You need maximum accuracy for a lab report or graded problem set.
• The acid is not especially weak or the concentration is very low, making x not negligible.
• You are using software or a calculator anyway, so there is no reason to avoid the quadratic formula.

The approximation is often acceptable when:

• You need a quick estimate.
• The acid is weak and concentrated enough that x is much smaller than the initial concentration.
• You are checking your work by hand before using a more exact method.

Common mistakes when solving hippuric acid pH problems

1. Using pKa directly in the Ka equation. You must convert pKa to Ka first.
2. Treating the acid as strong. Weak acids require equilibrium treatment.
3. Choosing the negative quadratic root. Concentrations cannot be negative.
4. Forgetting that x equals [H⁺]. Once solved, x is the hydronium concentration in this simple weak-acid setup.
5. Mixing up M and mM. A concentration entered as 120 mM must be converted to 0.120 M.

Practical interpretation of a pH near 2.28

A pH of about 2.28 indicates a distinctly acidic solution. It is far more acidic than neutral water at pH 7, but not as acidic as a strong acid solution of the same formal concentration. This difference matters in analytical chemistry, buffer calculations, and any biological context where weak organic acids are discussed. Organic acids like hippuric acid often appear in discussions of metabolism, urine chemistry, and acid-base equilibrium, so being able to calculate their pH accurately is a valuable skill.

Recommended authoritative references

For additional chemistry background and reliable academic support, see these resources:

• LibreTexts Chemistry for broad acid-base and equilibrium explanations.
• U.S. Environmental Protection Agency for pH fundamentals and water chemistry context.
• University of California, Berkeley Chemistry for academic chemistry learning resources.

Final answer for calculate the pH in 0.120 M hippuric acid

Using a typical room-temperature value of pKa = 3.62 for hippuric acid, the acid dissociation constant is Ka = 2.40 × 10^-4. Solving the weak-acid equilibrium exactly for a 0.120 M solution gives [H⁺] ≈ 5.25 × 10^-3 M, so the pH is approximately 2.28. The quick approximation method gives about 2.27, which is extremely close, but the rigorous reported answer is pH = 2.28.