Calculate the pH of a 0.50 m Solution of Aniline
Use this premium weak-base calculator to estimate the pH, pOH, hydroxide concentration, and equilibrium composition for an aqueous aniline solution.
Aniline pH Calculator
Results
Enter or confirm the default values, then click Calculate pH.
Default textbook answer
For a 0.50 m solution of aniline using Kb = 4.3 × 10-10 at 25 C, the expected pH is about 9.17.
- Aniline behaves as a weak base in water.
- Its equilibrium produces a small amount of OH–.
- The exact quadratic solution is used here for better accuracy.
Expert Guide: How to Calculate the pH of a 0.50 m Solution of Aniline
Calculating the pH of a 0.50 m solution of aniline is a classic weak-base equilibrium problem. At first glance, it looks similar to the pH calculation for ammonia or other weak bases, but aniline is much weaker because the nitrogen lone pair is partially delocalized into the benzene ring. That resonance stabilization makes the nitrogen less eager to accept a proton, so aniline generates much less hydroxide ion than a stronger base would at the same concentration.
In practical chemistry, this matters because aromatic amines do not behave like fully ionized bases. Instead, they establish an equilibrium with water. That means the pH must be calculated from the base dissociation constant, Kb, rather than by assuming complete dissociation. For aniline at room temperature, a commonly used value is approximately 4.3 × 10-10. With an initial concentration of 0.50, the resulting pH is only mildly basic, not strongly alkaline.
The calculator above is designed to automate the exact equilibrium math, but it is also useful to understand the chemistry behind the answer. In the sections below, you will see the governing equation, the weak-base ICE setup, the exact calculation, the common approximation, and how concentration changes affect the final pH.
1. What reaction describes aniline in water?
Aniline, usually written as C6H5NH2, reacts with water according to the following equilibrium:
In this process, aniline acts as a Brønsted base. It accepts a proton from water to form the anilinium ion, C6H5NH3+, while water donates that proton and becomes hydroxide, OH–. Because the equilibrium lies far to the left, only a small fraction of aniline molecules become protonated. That is exactly why the pH is not extremely high even when the starting concentration is 0.50.
2. The key constant: Kb for aniline
The base dissociation constant for aniline is:
A commonly cited room-temperature value is 4.3 × 10-10. This is a very small equilibrium constant, showing that aniline is a weak base. By comparison, ammonia has a Kb near 1.8 × 10-5, which is tens of thousands of times larger. That single comparison immediately tells you aniline should produce much less hydroxide ion than ammonia at equal concentration.
| Property | Aniline | Why it matters for pH |
|---|---|---|
| Molecular formula | C6H7N or C6H5NH2 | Defines the aromatic amine being analyzed. |
| Molar mass | 93.13 g/mol | Useful for preparing solutions from mass. |
| Typical density of pure aniline | About 1.02 g/mL at room temperature | Helpful when converting mass or volume in lab work. |
| Kb at 25 C | 4.3 × 10-10 | Controls equilibrium OH– formation. |
| pKa of anilinium ion | About 4.6 | Consistent with weak basicity of aniline. |
3. Set up the ICE table
To calculate the pH, start with an initial concentration of 0.50. The user prompt says 0.50 m, which is molality, but for standard textbook weak-base pH calculations in moderately dilute aqueous solution, this is commonly treated as approximately 0.50 mol/L. The equilibrium setup is:
Substitute these values into the Kb expression:
Since Kb is very small, x will be very small relative to 0.50, so the approximation 0.50 – x ≈ 0.50 is usually valid. That gives:
Here, x is the hydroxide ion concentration:
4. Convert hydroxide concentration to pOH and pH
Once [OH–] is known, calculate pOH:
At 25 C:
Therefore, the pH of a 0.50 m solution of aniline is approximately:
The calculator on this page uses the exact quadratic method instead of relying only on the small-x approximation. In this specific case, both approaches produce essentially the same pH to two decimal places, which confirms that the approximation is acceptable.
5. Exact quadratic solution
If you want the mathematically exact value, solve:
Rearranging gives:
With C = 0.50 and Kb = 4.3 × 10-10:
This yields virtually the same hydroxide concentration, about 1.47 × 10-5 M. Because x is so small compared with 0.50, the exact and approximate methods agree very closely.
6. Why is aniline such a weak base?
The weak basicity of aniline is not an accident of numbers. It follows directly from structure. In aniline, the nitrogen lone pair can interact with the aromatic pi system of the benzene ring. Because part of that electron density is delocalized, the lone pair is less available to bind a proton. As a result, aniline is significantly less basic than aliphatic amines and also much weaker than ammonia in many practical comparisons.
- The benzene ring withdraws effective basic strength through resonance involvement.
- The nitrogen lone pair is stabilized by conjugation.
- Protonation disrupts that stabilization pattern, which makes proton uptake less favorable.
- The consequence is a small Kb and only modest OH– production.
7. Comparison table: pH at different aniline concentrations
One of the most helpful ways to understand weak-base behavior is to compare pH over a range of starting concentrations while keeping Kb fixed at 4.3 × 10-10. The values below are calculated at 25 C using the exact weak-base expression.
| Aniline concentration | Calculated [OH–] | pOH | pH |
|---|---|---|---|
| 0.010 | 2.07 × 10-6 M | 5.68 | 8.32 |
| 0.050 | 4.64 × 10-6 M | 5.33 | 8.67 |
| 0.10 | 6.56 × 10-6 M | 5.18 | 8.82 |
| 0.50 | 1.47 × 10-5 M | 4.83 | 9.17 |
| 1.00 | 2.07 × 10-5 M | 4.68 | 9.32 |
Notice that even a tenfold increase in concentration does not produce a tenfold increase in pH. That is one hallmark of weak acid and weak base systems. The hydroxide concentration scales roughly with the square root of the initial concentration when the small-x approximation is valid.
8. Common mistakes students make
- Assuming complete dissociation. Aniline is not a strong base, so you cannot set [OH–] = 0.50.
- Using Ka instead of Kb. For aniline acting as a base, Kb is the relevant constant.
- Confusing pOH with pH. You first compute pOH from hydroxide concentration, then convert to pH.
- Ignoring temperature assumptions. The relation pH + pOH = 14.00 is exact only at 25 C when pKw is taken as 14.00.
- Treating molality and molarity as identical in all cases. In careful thermodynamic work, they are not the same, although for many textbook exercises the approximation is accepted.
9. Practical interpretation of the final pH
A pH near 9.17 means the solution is definitely basic, but only moderately so. It is nowhere near as alkaline as a strong base solution with the same nominal concentration. In laboratory handling, that distinction matters. You still need chemical hygiene, but the measured pH behavior reflects partial protonation rather than complete conversion to hydroxide.
If you are preparing an analytical, teaching-lab, or synthetic chemistry solution, this result helps predict:
- How much protonated anilinium ion will be present at equilibrium
- Whether the medium is basic enough to alter acid-base indicators
- How aniline might behave in extraction, protonation, and salt formation steps
- Why aromatic amines often require stronger acids than expected for complete protonation
10. Authoritative reference links
For additional chemistry and data verification, consult these authoritative resources:
- NIH PubChem: Aniline compound data
- USGS: pH and water fundamentals
- University-level acid-base equilibrium reference
11. Final answer summary
To calculate the pH of a 0.50 m solution of aniline, model aniline as a weak base with Kb = 4.3 × 10-10. Set up the equilibrium:
Solve for hydroxide concentration using either the weak-base approximation or the exact quadratic equation. You obtain:
So the accepted result is that a 0.50 m aqueous aniline solution is mildly basic, with a pH of approximately 9.17 at 25 C. If you change the Kb value or temperature assumption, the calculator will update the answer instantly and visualize the equilibrium distribution in the chart.