Calculate The Ph Of A 0.15 M Solution Of Aniline

Use this interactive weak base calculator to determine pH, pOH, hydroxide concentration, and conjugate acid formation for aniline in water. The default values are set for a 0.15 M aniline solution at 25 degrees Celsius.

Interactive Aniline pH Calculator

Enter concentration and either the base dissociation constant Kb or pKb. The default chemistry data for aniline is preloaded.

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

Aniline, with molecular formula C₆H₅NH₂, is a classic example of a weak organic base. If you are asked to calculate the pH of a 0.15 M solution of aniline, the key idea is that aniline does not completely ionize in water. Instead, it establishes an equilibrium with water to produce a small amount of hydroxide ion. Because hydroxide concentration controls pOH, and pOH controls pH, the entire problem becomes an equilibrium calculation.

This is exactly the kind of acid-base problem encountered in general chemistry, analytical chemistry, and introductory physical chemistry. The concentration is known, the base is weak, and the equilibrium constant is small. For aniline at 25 degrees Celsius, a commonly cited value is Kb ≈ 4.3 × 10^-10, which corresponds to pKb ≈ 9.37. Those values tell you aniline is much less basic than ammonia. As a result, even a 0.15 M solution will only be mildly basic, not strongly alkaline.

The Fundamental Chemical Equation

In water, aniline behaves as a Brønsted-Lowry base:

C6H5NH2 + H2O ⇌ C6H5NH3+ + OH-

The base dissociation constant expression is:

Kb = [C6H5NH3+][OH-] / [C6H5NH2]

If the initial concentration of aniline is 0.15 M and the amount that reacts is x, then at equilibrium:

• [C₆H₅NH₂] = 0.15 – x
• [C₆H₅NH₃⁺] = x
• [OH^–] = x

Substituting into the equilibrium expression gives:

Kb = x² / (0.15 – x)

Solving the Equilibrium Exactly

Many textbook problems use the weak base approximation, where x is assumed to be much smaller than 0.15. That is usually acceptable here, but an exact quadratic approach is more rigorous and is what this calculator uses. Rearranging:

x² + Kb x – Kb(0.15) = 0

Using Kb = 4.3 × 10^-10:

x² + (4.3 × 10^-10)x – (6.45 × 10^-11) = 0

The physically meaningful root gives:

x = [OH-] ≈ 8.03 × 10^-6 M

Now calculate pOH:

pOH = -log(8.03 × 10^-6) ≈ 5.10

Then calculate pH:

pH = 14.00 – 5.10 ≈ 8.90

Therefore, the pH of a 0.15 M solution of aniline is about 8.90 at 25 degrees Celsius, assuming Kb = 4.3 × 10^-10.

Final result for the standard problem: pH ≈ 8.90. Because aniline is a weak base, the pH is only modestly above neutral.

Why Aniline Is a Weak Base

Aniline contains an amino group attached directly to a benzene ring. That aromatic ring strongly influences the lone pair on nitrogen. In aliphatic amines, the nitrogen lone pair is more available to accept a proton. In aniline, however, that lone pair can interact with the aromatic pi system through resonance. This delocalization lowers the availability of the lone pair, making aniline less basic than many simple amines.

This matters because basicity is tied to the tendency to accept a proton from water. A weak base accepts only a small fraction of protons from water, so only a small concentration of hydroxide is generated. Even when the starting concentration is 0.15 M, the resulting hydroxide ion concentration is only on the order of 10^-6 M.

Approximation Method vs Exact Method

You can also estimate the answer with the weak base shortcut:

x ≈ √(Kb × C)

For aniline:

x ≈ √[(4.3 × 10^-10)(0.15)] = √(6.45 × 10^-11) ≈ 8.03 × 10^-6 M

This is nearly identical to the exact result because x is indeed much smaller than 0.15. That means the approximation is excellent in this case. In a classroom setting, either method may be acceptable unless your instructor explicitly requests the quadratic solution.

Step by Step Procedure You Can Use on Exams

1. Write the base ionization equation for aniline in water.
2. Set up an ICE table with initial, change, and equilibrium concentrations.
3. Write the Kb expression.
4. Substitute equilibrium concentrations into the expression.
5. Solve for x, which equals [OH^–].
6. Calculate pOH using pOH = -log[OH-].
7. Calculate pH using pH = 14 – pOH.

ICE Table for 0.15 M Aniline

Species	Initial (M)	Change (M)	Equilibrium (M)
C6H5NH2	0.15	-x	0.15 – x
C6H5NH3^+	0	+x	x
OH^–	0	+x	x

Comparison Table: Aniline vs Other Weak Bases

To understand whether your answer is reasonable, it helps to compare aniline with other familiar bases. A stronger weak base creates more hydroxide and therefore a higher pH at the same concentration. The table below shows representative 25 degrees Celsius values commonly taught in chemistry courses.

Base	Representative Kb at 25 degrees Celsius	Representative pKb	Relative Basicity
Aniline	4.3 × 10^-10	9.37	Weak
Ammonia	1.8 × 10^-5	4.75	Much stronger than aniline
Pyridine	1.7 × 10^-9	8.77	Slightly stronger than aniline
Methylamine	4.4 × 10^-4	3.36	Far stronger than aniline

Notice the huge difference between aniline and ammonia. The Kb of ammonia is roughly five orders of magnitude larger than that of aniline. That difference is why ammonia solutions are noticeably more basic than comparable aniline solutions.

How pH Changes with Concentration

For weak bases, pH rises as concentration rises, but not in a simple linear way. Because the hydroxide concentration for a weak base often follows an approximate square root dependence, increasing the base concentration by a factor of 100 does not increase hydroxide by a factor of 100. Instead, the increase is more 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
0.150	8.03 × 10^-6	5.10	8.90
0.500	1.47 × 10^-5	4.83	9.17

The 0.15 M entry fits smoothly into the expected trend. That is a good self-check. If your answer were something like pH 11.5, you would know immediately that it is unrealistic for a base as weak as aniline.

Common Mistakes Students Make

• Using Ka instead of Kb. Aniline is acting as a base in water, so use Kb unless the problem is framed using the conjugate acid.
• Forgetting to convert pKb to Kb. If pKb is given, use Kb = 10^-pKb.
• Confusing pOH and pH. The equilibrium gives hydroxide, so pOH comes first, then pH.
• Assuming complete dissociation. Aniline is not a strong base, so [OH^–] is not equal to 0.15 M.
• Dropping the negative sign in the logarithm. Use pOH = -log[OH-], not simply log[OH-].

Interpretation of the Result

A pH of about 8.90 means the solution is basic, but only mildly so. In practical terms, aniline in water is not remotely as alkaline as sodium hydroxide, potassium hydroxide, or even concentrated ammonia. This result reflects the relatively low proton-accepting ability of aniline.

Because aniline is an aromatic amine, its behavior is chemically important in dye chemistry, pharmaceutical synthesis, and industrial organic chemistry. However, from an acid-base perspective, its equilibrium in water is straightforward: low Kb, small hydroxide concentration, pOH just above 5, and pH just below 9.

Useful Reference Sources

If you want to verify acid-base concepts, equilibrium constants, or solution chemistry methods, these authoritative educational and government resources are excellent:

• Chemistry LibreTexts educational resource
• NIST Chemistry WebBook
• NIH PubChem entry for aniline
• University of Wisconsin chemistry resources

Final Answer Summary

To calculate the pH of a 0.15 M solution of aniline, write the weak base equilibrium, use the base dissociation constant for aniline, solve for hydroxide concentration, and convert that value into pOH and then pH. With Kb = 4.3 × 10^-10, the equilibrium hydroxide concentration is about 8.03 × 10^-6 M, the pOH is about 5.10, and the pH is about 8.90.

This is the expected result for a weak aromatic amine. If you are solving homework, studying for an exam, or checking a lab pre-calculation, that is the number you should expect under standard 25 degrees Celsius conditions.