Calculate The Ph Of A 0.15 M Solution Of Aniline

Use this premium weak-base calculator to determine the pH, pOH, hydroxide concentration, and percent ionization for an aqueous aniline solution. The default values are set for 0.15 M aniline at 25 C using a typical literature Kb of 4.3 × 10^-10.

Aniline pH Calculator

Results

Quick Notes

• Aniline is a weak base because the nitrogen lone pair is partly delocalized into the aromatic ring.
• A weak base does not fully dissociate, so the pH is only mildly basic.
• At 0.15 M, aniline gives a pH close to 8.9 at 25 C using Kb = 4.3 × 10^-10.

Expert Guide: How to Calculate the pH of a 0.15 M Solution of Aniline

To calculate the pH of a 0.15 M solution of aniline, you treat aniline as a weak Brønsted base in water. The key idea is that aniline does not react completely with water. Instead, only a very small fraction of the dissolved base accepts a proton, generating hydroxide ions. Because pH depends on hydrogen ion concentration and weak bases are easier to analyze through hydroxide production first, the standard route is to calculate pOH and then convert it to pH.

Aniline, formula C₆H₅NH₂, is an aromatic amine. Compared with aliphatic amines such as methylamine, it is much less basic. The reason is structural: the lone pair on nitrogen can interact with the benzene ring through resonance. That delocalization lowers the tendency of nitrogen to accept a proton. In practical terms, this means the base dissociation constant, Kb, for aniline is small, often cited around 4.3 × 10^-10 at 25 C. That small value is the central reason the pH of a 0.15 M solution is only modestly above 7.

Step 1: Write the equilibrium equation

For a weak base B in water, the generic equilibrium is:

B + H₂O ⇌ BH⁺ + OH^–

For aniline, the equation becomes:

C₆H₅NH₂ + H₂O ⇌ C₆H₅NH₃⁺ + OH^–

The equilibrium expression is:

Kb = [C₆H₅NH₃⁺][OH^–] / [C₆H₅NH₂]

Step 2: Set up the ICE table

Suppose the initial concentration of aniline is 0.15 M and initially there is no appreciable anilinium ion or hydroxide from the base itself. Let x be the amount of aniline that reacts.

• Initial: [C₆H₅NH₂] = 0.15, [C₆H₅NH₃⁺] = 0, [OH^–] = 0
• Change: -x, +x, +x
• Equilibrium: 0.15 – x, x, x

Substitute into the Kb expression:

4.3 × 10^-10 = x² / (0.15 – x)

Step 3: Solve for x, the hydroxide concentration

Because Kb is very small, x will be tiny compared with 0.15. You can often use the approximation:

0.15 – x ≈ 0.15

This gives:

4.3 × 10^-10 = x² / 0.15

x² = 6.45 × 10^-11

x = 8.03 × 10^-6 M

That means:

[OH^–] = 8.03 × 10^-6 M

If you solve the quadratic equation exactly, you get practically the same answer for this case, which confirms that the approximation is valid.

Step 4: Convert hydroxide concentration to pOH

Now use the pOH definition:

pOH = -log[OH^–]

pOH = -log(8.03 × 10^-6) = 5.10 approximately

Step 5: Convert pOH to pH

At 25 C, use:

pH + pOH = 14.00

pH = 14.00 – 5.10 = 8.90 approximately

Final answer: The pH of a 0.15 M solution of aniline is approximately 8.91 at 25 C when Kb = 4.3 × 10^-10.

Why the pH is not very high

Students often expect any amine solution to be strongly basic, but aniline is a classic exception. Aromatic amines are less basic than many simple alkyl amines because resonance stabilizes the neutral base. The nitrogen lone pair is not as available for bonding to a proton. As a result:

• Only a small fraction of aniline molecules form anilinium ions.
• The hydroxide concentration remains low.
• The final pH is basic, but only mildly basic.

Exact versus approximate solution

For weak bases, the square root approximation is commonly taught because it is fast and usually accurate when the percent ionization is small. Here the percent ionization is:

(8.03 × 10^-6 / 0.15) × 100 = 0.0054%

That is far below 5%, so the approximation is excellent. Still, a high-quality calculator should be able to solve the quadratic exactly, which is what the tool above does when you select the exact method.

Comparison table: aniline versus other weak bases

The table below helps place aniline in context. These values are typical 25 C data used for general chemistry comparisons, and the pH values shown are approximate calculated values for 0.15 M solutions.

Base	Typical Kb	pKb	Approximate pH at 0.15 M	Interpretation
Aniline	4.3 × 10^-10	9.37	8.91	Very weak aromatic amine base
Pyridine	1.7 × 10^-9	8.77	9.20	Weak base, but stronger than aniline
Ammonia	1.8 × 10^-5	4.74	11.22	Much stronger weak base
Methylamine	4.4 × 10^-4	3.36	11.90	Strongly basic relative to aniline

This table shows how unusually weak aniline is compared with common introductory chemistry examples. Even pyridine, which is also aromatic, is noticeably more basic than aniline. The resonance explanation is a major concept tested in acid-base chemistry and organic chemistry courses.

Concentration effects for aniline

Increasing concentration raises pH, but not dramatically for such a weak base. Because hydroxide concentration scales approximately with the square root of concentration for a weak base, doubling the concentration does not double the pH rise. The relationship is real but moderate.

Aniline Concentration (M)	Approximate [OH-] (M)	Approximate pOH	Approximate pH
0.010	2.07 × 10^-6	5.68	8.32
0.050	4.64 × 10^-6	5.33	8.67
0.100	6.56 × 10^-6	5.18	8.82
0.150	8.03 × 10^-6	5.10	8.91
0.300	1.14 × 10^-5	4.94	9.06

Common mistakes to avoid

1. Using Ka instead of Kb. Aniline is usually treated as a weak base in water, so Kb is the most direct constant.
2. Forgetting to calculate pOH first. Weak bases produce OH-, so pOH comes before pH.
3. Assuming full dissociation. Aniline is far too weak for that assumption.
4. Ignoring temperature context. The relation pH + pOH = 14.00 is strictly tied to 25 C.
5. Rounding too early. Keep several significant digits until the final step.

How this problem appears in exams

In general chemistry, this question usually tests equilibrium setup, Kb manipulation, ICE tables, and pH conversion. In organic chemistry, the same problem may also test conceptual basicity trends. Professors often pair aniline with ammonia, cyclohexylamine, pyridine, or p-nitroaniline to see whether students understand how resonance and substituent effects change base strength.

Why aniline is weaker than aliphatic amines

Alkyl groups tend to donate electron density toward nitrogen, making the lone pair more available and increasing basicity. In aniline, the aromatic ring allows delocalization of the lone pair. That lowers proton affinity. This structural effect is strong enough that a 0.15 M solution of methylamine is much more basic than a 0.15 M solution of aniline, as the comparison table clearly shows.

Best formula summary

• Kb = x² / (C – x)
• If x is small: x ≈ √(KbC)
• pOH = -log[OH-]
• pH = 14.00 – pOH at 25 C

Authoritative references for deeper study

NIH PubChem: Aniline
NIST Chemistry WebBook: Aniline data
University of Wisconsin acid-base tutorial

Bottom line

If you need to calculate the pH of a 0.15 M solution of aniline, the process is straightforward once you remember that aniline is a weak base. Start with the weak-base equilibrium, use Kb to find hydroxide concentration, convert to pOH, and then convert to pH. With a typical Kb of 4.3 × 10^-10 at 25 C, the answer is about pH 8.91. That number is entirely consistent with aniline being only slightly basic in water.

Educational note: equilibrium constants can vary slightly by source, ionic strength, and temperature. For classroom problems, always use the Kb value provided by your instructor or textbook when available.