Calculate The Ph Of 25 Ml Of 0.15M Benzoic Acid.

Use this premium weak-acid calculator to determine the pH, hydrogen ion concentration, dissociation extent, and equilibrium species for benzoic acid solutions. The default settings are preloaded for 25 mL of 0.15 M benzoic acid at 25 degrees Celsius.

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Benzoic acid is a weak monoprotic acid. For a solution of known concentration, its pH is determined from the acid dissociation constant, not simply from full ionization.

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How to calculate the pH of 25 mL of 0.15 M benzoic acid

To calculate the pH of 25 mL of 0.15 M benzoic acid, the most important idea is that benzoic acid is a weak acid, not a strong acid. That means it does not fully ionize in water. Instead, only a small fraction of benzoic acid molecules donate a proton to water, so the pH must be found from an equilibrium calculation using the acid dissociation constant, Ka.

The relevant equilibrium is:

C6H5COOH + H2O ⇌ H3O+ + C6H5COO-

At 25 degrees Celsius, benzoic acid has a pKa of about 4.20, which corresponds to a Ka near 6.31 x 10^-5. Because the solution concentration is 0.15 M, the initial concentration of undissociated acid is fairly high compared with Ka. This is exactly the situation where the weak-acid equilibrium method is used.

Step 1: Identify the given values

• Volume = 25 mL = 0.025 L
• Initial concentration of benzoic acid = 0.15 M
• Ka of benzoic acid = 6.31 x 10^-5 at 25 degrees Celsius

Notice that volume helps you calculate the total number of moles present:

moles = M x V = 0.15 x 0.025 = 0.00375 mol

However, for the pH of a simple acid solution with no dilution or titration step, the pH depends on concentration and Ka. The 25 mL volume does not change the pH by itself. It does matter if you want total moles, if you plan to dilute the sample, or if you are connecting the problem to titration chemistry.

Step 2: Set up the ICE table

For a weak acid HA:

HA ⇌ H+ + A-

Let x be the amount that dissociates.

• Initial: [HA] = 0.15, [H+] = 0, [A-] = 0
• Change: [HA] = -x, [H+] = +x, [A-] = +x
• Equilibrium: [HA] = 0.15 – x, [H+] = x, [A-] = x

Substitute into the Ka expression:

Ka = [H+][A-] / [HA] = x^2 / (0.15 – x)

Now insert the value of Ka:

6.31 x 10^-5 = x^2 / (0.15 – x)

Step 3: Solve for x

You can use either the weak-acid approximation or the full quadratic equation. Since this calculator is designed for accuracy, it uses the quadratic method by default.

Rearranging gives:

x^2 + Ka x – Ka C = 0

For C = 0.15 and Ka = 6.31 x 10^-5:

x = (-Ka + sqrt(Ka^2 + 4KaC)) / 2

Solving gives:

• x = [H+] ≈ 3.04 x 10^-3 M

Then:

pH = -log10([H+]) = -log10(3.04 x 10^-3) ≈ 2.52

Rounded to two decimal places, the pH is 2.52. Depending on the Ka value and rounding used by your textbook or instructor, you may also see 2.51.

Final answer: the pH of 25 mL of 0.15 M benzoic acid is approximately 2.52 at 25 degrees Celsius.

Why the 25 mL volume often confuses students

Many chemistry learners see the volume listed in the problem and assume it must directly affect the pH. In this case, it does not. If the concentration remains 0.15 M, then 25 mL, 100 mL, or 500 mL of that same solution all have the same pH. What changes is the total amount of benzoic acid, not the concentration.

Volume becomes essential in the following situations:

1. When calculating total moles of acid present.
2. When preparing the solution from a solid mass of benzoic acid.
3. When mixing with another solution, such as in a neutralization or titration problem.
4. When discussing dilution, because concentration changes after dilution.

Initial moles in 25 mL

For this sample:

• Initial moles of benzoic acid = 0.00375 mol
• Equilibrium moles of H+ = 3.04 x 10^-3 mol/L x 0.025 L ≈ 7.60 x 10^-5 mol
• Percent dissociation ≈ 2.03%

This low percent dissociation confirms that benzoic acid is weak and mostly remains in the undissociated form at equilibrium.

Approximation method versus exact quadratic method

Students are often taught the weak-acid shortcut:

[H+] ≈ sqrt(Ka x C)

Using Ka = 6.31 x 10^-5 and C = 0.15:

[H+] ≈ sqrt(6.31 x 10^-5 x 0.15) ≈ 3.08 x 10^-3 M

That gives a pH of about 2.51, which is very close to the exact answer because dissociation is only around 2%. Since the 5% rule is satisfied, the approximation is acceptable. Still, the exact method is better for a calculator because it remains accurate across a wider range of weak-acid concentrations and Ka values.

Method	[H+] (M)	Calculated pH	Comment
Quadratic equation	3.04 x 10^-3	2.52	Most accurate for this problem
Weak-acid approximation	3.08 x 10^-3	2.51	Acceptable because dissociation is about 2%
Incorrect strong-acid assumption	0.15	0.82	Far too acidic because benzoic acid does not fully ionize

Benzoic acid compared with other common weak acids

Benzoic acid is stronger than acetic acid but weaker than some mineral acids by a huge margin. Comparing pKa values helps show where it sits chemically. Lower pKa means a stronger acid.

Acid	Approximate pKa at 25 degrees Celsius	Approximate Ka	Relative strength note
Formic acid	3.75	1.78 x 10^-4	Stronger than benzoic acid
Benzoic acid	4.20	6.31 x 10^-5	Moderately weak organic acid
Acetic acid	4.76	1.74 x 10^-5	Weaker than benzoic acid
Phenol	9.95	1.12 x 10^-10	Much weaker acid

These comparison values are useful because they show why benzoic acid produces a pH lower than acetic acid at the same concentration. Its larger Ka means more hydrogen ions are released into the solution.

Common mistakes when solving this problem

• Treating benzoic acid as a strong acid. This gives a wildly incorrect pH.
• Forgetting to convert mL to L when calculating moles from molarity.
• Using pKa directly as Ka. If pKa is given, convert using Ka = 10^-pKa.
• Ignoring equilibrium. Weak acids require an equilibrium expression.
• Assuming volume changes pH at fixed concentration. It does not, unless dilution or mixing occurs.

When the Henderson-Hasselbalch equation does and does not apply

The Henderson-Hasselbalch equation is excellent for buffers containing both benzoic acid and benzoate:

pH = pKa + log10([A-] / [HA])

But for a pure benzoic acid solution with no added benzoate, Henderson-Hasselbalch is not the right starting point. You first need the weak-acid equilibrium calculation. If benzoic acid were mixed with sodium benzoate, then buffer chemistry would apply.

Practical context for benzoic acid chemistry

Benzoic acid and benzoate systems matter in food preservation, pharmaceutical chemistry, and laboratory acid-base studies. The benzoic acid and benzoate pair is often used to teach weak-acid equilibria because its pKa is well characterized and its chemistry is straightforward. In practical formulations, pH strongly influences the ratio of benzoic acid to benzoate, which in turn affects solubility, preservative action, and partitioning behavior.

If you want to explore reliable property data and chemistry references, these sources are useful:

• NIH PubChem benzoic acid record
• NIST Chemistry WebBook entry for benzoic acid
• LibreTexts Chemistry educational resource

Short worked example summary

1. Write the weak-acid dissociation reaction for benzoic acid.
2. Use an ICE table with initial concentration 0.15 M.
3. Set Ka = x^2 / (0.15 – x).
4. Solve for x using Ka = 6.31 x 10^-5.
5. Find [H+] ≈ 3.04 x 10^-3 M.
6. Calculate pH = -log10(3.04 x 10^-3) ≈ 2.52.

Final answer

The pH of 25 mL of 0.15 M benzoic acid is approximately 2.52 at 25 degrees Celsius, assuming Ka = 6.31 x 10^-5. The 25 mL volume corresponds to 0.00375 moles of benzoic acid, but the pH itself is determined by the acid concentration and equilibrium constant rather than the total sample size.

Data values shown above are standard approximate reference values commonly used in general chemistry at 25 degrees Celsius. Small differences may occur among sources because of rounding and temperature dependence.