Calculate the pH of a 2.50×10-3 M Aniline Solution
Use this interactive weak-base calculator to find pH, pOH, hydroxide concentration, conjugate acid concentration, and percent ionization for aniline in water using an exact equilibrium approach.
Aniline pH Calculator
How to calculate the pH of a 2.50×10-3 M aniline solution
To calculate the pH of a 2.50×10-3 M aniline solution, you treat aniline, C6H5NH2, as a weak base in water. Unlike a strong base such as sodium hydroxide, aniline does not ionize completely. Instead, it reacts only slightly with water to form the anilinium ion and hydroxide ion:
Because pH depends on the hydroxide ion concentration generated by this equilibrium, the key constant is the base dissociation constant, Kb. For aniline at 25 degrees C, a typical textbook value is approximately 4.3×10-10. This is a very small Kb, which tells you that aniline is a weak base and that only a tiny fraction of the original molecules accept a proton from water.
Step 1: Identify the known values
- Initial aniline concentration, C = 2.50×10-3 M
- Kb for aniline = 4.3×10-10
- Temperature assumption = 25 degrees C, so pH + pOH = 14.00
Step 2: Set up the equilibrium expression
If x is the amount of hydroxide formed at equilibrium, then:
Plugging in the Kb value gives:
Step 3: Solve for x
Since aniline is weak and Kb is very small, the change x is much smaller than the starting concentration. That means the common approximation 2.50×10-3 – x ≈ 2.50×10-3 works very well. Then:
So the hydroxide concentration is approximately:
Step 4: Convert hydroxide concentration to pOH
Step 5: Convert pOH to pH
Therefore, the pH of a 2.50×10-3 M aniline solution is about 8.02 at 25 degrees C when using Kb = 4.3×10-10.
Why aniline is only weakly basic
Students often expect any amine to produce a strongly basic solution, but aniline is much less basic than many aliphatic amines. The reason is resonance. The nitrogen lone pair in aniline can interact with the aromatic benzene ring, which delocalizes some of the electron density that would otherwise be available to accept a proton. In practical terms, that lowers the tendency of aniline to react with water and generate hydroxide ions.
Compare this with ammonia or methylamine, where the lone pair is more available. Those bases generally have larger Kb values and therefore produce more OH– at the same starting concentration. This is why the pH of aniline solutions tends to be modestly above 7 rather than dramatically high.
Exact solution versus approximation
In weak acid and weak base chemistry, the approximation method is frequently used because it is fast and usually accurate when the dissociation is very small relative to the initial concentration. For aniline in this problem, the approximation is excellent. Still, it is useful to understand the exact quadratic method.
Using C = 2.50×10-3 and Kb = 4.3×10-10, the exact result is essentially the same as the approximation: x is still very close to 1.04×10-6 M. The percent ionization is tiny:
Since only about 0.04% of the dissolved aniline is protonated, the approximation is justified. A common rule is that if the percent ionization is under 5%, the simplification is generally valid.
Common mistakes when solving this problem
- Using Ka instead of Kb: Aniline is acting as a base here, so the correct equilibrium constant is Kb.
- Forgetting to calculate pOH first: Since the equilibrium gives OH–, the direct logarithm gives pOH, not pH.
- Treating aniline like a strong base: You cannot assume [OH–] = 2.50×10-3 M.
- Mishandling scientific notation: 2.50×10-3 M equals 0.00250 M, not 0.0250 M.
- Ignoring temperature assumptions: The relation pH + pOH = 14.00 is standard at 25 degrees C.
Comparison table: aniline versus other common weak bases
The table below gives typical room-temperature Kb values and shows why aniline behaves as a much weaker base than several common nitrogen bases. These values can vary slightly by source and temperature, but they illustrate the general trend well.
| Base | Typical Kb | Relative basic strength | Chemical note |
|---|---|---|---|
| Aniline | 4.3×10-10 | Very weak | Lone pair delocalized into aromatic ring |
| Ammonia | 1.8×10-5 | Much stronger than aniline | No aromatic resonance withdrawal |
| Methylamine | 4.4×10-4 | Far stronger than aniline | Alkyl group donates electron density |
| Pyridine | 1.7×10-9 | Slightly stronger than aniline | Nitrogen lone pair outside aromatic sextet |
How concentration affects the pH of aniline solutions
Because aniline is a weak base, the pH rises gradually as concentration increases. The increase is not linear. Doubling the concentration does not double the pH change. Instead, the hydroxide concentration depends on the square root of KbC when the weak-base approximation holds. That means pH changes more slowly than concentration.
| Initial aniline concentration (M) | Approximate [OH–] (M) | Approximate pOH | Approximate pH |
|---|---|---|---|
| 1.00×10-4 | 2.07×10-7 | 6.684 | 7.316 |
| 1.00×10-3 | 6.56×10-7 | 6.183 | 7.817 |
| 2.50×10-3 | 1.04×10-6 | 5.984 | 8.016 |
| 1.00×10-2 | 2.07×10-6 | 5.684 | 8.316 |
Worked logic in plain language
If you want the shortest conceptual path to the answer, here it is. Start by recognizing that aniline is a weak base, so it produces a small amount of OH– in water. Use Kb and the starting concentration in an ICE setup. Solve for x, which equals the equilibrium OH– concentration. Take the negative log to get pOH. Then subtract from 14 to find pH. For this exact problem, that process gives a pH just over 8.
That final value also passes the chemical reasonableness test. If the pH were 11 or 12, it would imply a much stronger base than aniline really is. If it were exactly 7, that would imply essentially no basic effect at all. A pH around 8 is chemically sensible for a dilute solution of a weak aromatic amine.
When to use Ka instead
Sometimes you are given the anilinium ion, C6H5NH3+, instead of aniline. In that case, the species in water behaves as a weak acid, and you would use Ka rather than Kb. The two constants are related by:
For aniline with Kb = 4.3×10-10, the corresponding Ka for the anilinium ion is:
This is useful when solving buffer or salt-hydrolysis problems involving aniline and its conjugate acid.
Authoritative references for equilibrium chemistry and pH
For deeper study, consult authoritative educational and government resources:
- LibreTexts Chemistry for weak acid and weak base equilibrium explanations.
- U.S. Environmental Protection Agency for reliable background on pH, water chemistry, and acid-base concepts.
- MIT Chemistry for university-level chemistry learning resources.
Bottom line
To calculate the pH of a 2.50×10-3 M aniline solution, use the weak-base equilibrium expression with the Kb of aniline. Solving the equilibrium gives an OH– concentration of about 1.04×10-6 M, a pOH of about 5.98, and a final pH of about 8.02. The result is only slightly basic because aniline is a weak base whose nitrogen lone pair is partially delocalized into the benzene ring.