Calculate the pH of a 5.0 m Solution of Aniline
Use this premium weak-base equilibrium calculator to estimate the pH of an aniline solution. It supports both the common classroom approximation that treats 5.0 m as about 5.0 M and a more rigorous molality-to-molarity conversion using solution density.
Aniline pH Calculator
Calculated Results
Expert Guide: How to Calculate the pH of a 5.0 m Solution of Aniline
Aniline, with formula C6H5NH2, is a classic example of a weak organic base. If you need to calculate the pH of a 5.0 m solution of aniline, the key idea is that aniline does not fully react with water. Instead, it establishes an equilibrium:
Because aniline is a weak base, only a very small fraction of the dissolved molecules accept a proton from water. That means the hydroxide ion concentration is much smaller than the starting analytical concentration. The result is a basic solution, but not an extremely basic one. In most general chemistry problems, the base dissociation constant for aniline at 25 C is taken as approximately Kb = 4.3 × 10-10, which corresponds to pKb ≈ 9.37.
For many classroom and exam problems, a 5.0 m solution is treated approximately as a 5.0 M solution. Strictly speaking, molality and molarity are not the same. Molality is moles of solute per kilogram of solvent, while molarity is moles of solute per liter of solution. If no density is provided, instructors usually expect the simple equilibrium treatment using C = 5.0. Under that approximation, the pH comes out to about 9.67. If you convert molality to molarity using an assumed density of 1.00 g/mL, the effective molarity falls to about 3.41 M and the pH becomes about 9.58. Both results are useful, but they answer slightly different modeling assumptions.
Step 1: Write the base equilibrium expression
For aniline in water, the equilibrium expression is:
Kb = [C6H5NH3+][OH–] / [C6H5NH2]
If the initial concentration of aniline is C and the amount that reacts is x, then at equilibrium:
- [C6H5NH2] = C – x
- [C6H5NH3+] = x
- [OH–] = x
Substituting these values into the equilibrium expression gives:
Kb = x² / (C – x)
Step 2: Use the 5.0 m classroom approximation
If we treat the 5.0 m solution as approximately 5.0 M, then:
- C = 5.0
- Kb = 4.3 × 10-10
Because Kb is very small, x will be much smaller than 5.0, so a common approximation is C – x ≈ C. Then:
x ≈ √(KbC)
Substitute the values:
x ≈ √[(4.3 × 10-10)(5.0)] = √(2.15 × 10-9) ≈ 4.64 × 10-5
Since x = [OH–], then:
- pOH = -log(4.64 × 10-5) ≈ 4.33
- pH = 14.00 – 4.33 = 9.67
That is the standard answer expected in many introductory chemistry settings: the pH of a 5.0 m solution of aniline is approximately 9.67.
Step 3: Understand why the answer is not extremely high
Students are sometimes surprised that a solution with a formal concentration as high as 5.0 still gives a pH under 10. The reason is that aniline is a much weaker base than ammonia and many aliphatic amines. The lone pair on the nitrogen atom is partially delocalized into the benzene ring, reducing its availability to accept a proton. That resonance stabilization makes aniline significantly less basic than a simple alkyl amine.
As a result, even a concentrated aniline solution ionizes only slightly. The percent ionization in the 5.0 M approximation is:
(4.64 × 10-5 / 5.0) × 100 ≈ 0.00093%
That is an extremely small fraction. So the concentration is large, but the equilibrium constant is so small that the hydroxide concentration remains modest.
Step 4: More rigorous treatment when molality matters
If you want to respect the fact that the problem states 5.0 m rather than 5.0 M, you should convert molality to molarity if the density of the solution is known or can be estimated. For aniline, using molar mass 93.13 g/mol, the conversion is:
M = (1000 × m × d) / (1000 + m × molar mass)
where:
- m is molality in mol/kg
- d is solution density in g/mL
- molar mass is in g/mol
If we assume a density of 1.00 g/mL:
M = (1000 × 5.0 × 1.00) / (1000 + 5.0 × 93.13) = 5000 / 1465.65 ≈ 3.41 M
Now apply the weak-base approximation again:
x ≈ √[(4.3 × 10-10)(3.41)] ≈ 3.83 × 10-5
Then:
- pOH ≈ 4.42
- pH ≈ 9.58
This shows that a more rigorous molality-based approach gives a slightly lower pH than the simple classroom answer. The difference is small enough that many textbook problems still accept 9.67 unless density information is supplied.
Exact quadratic solution versus approximation
For weak bases, you can either use the square root shortcut or solve the full quadratic equation. The exact relationship is:
x² + Kb x – KbC = 0
The physically meaningful root is:
x = [-Kb + √(Kb² + 4KbC)] / 2
For aniline at these concentrations, the exact solution and the shortcut are practically identical because x is tiny relative to C. That means the usual approximation is excellent here.
Comparison table: basicity of aniline versus other weak bases
The following comparison helps explain why aniline gives a fairly modest pH even at high concentration.
| Base | Approximate Kb at 25 C | Approximate pKb | Relative basicity comment |
|---|---|---|---|
| Aniline | 4.3 × 10-10 | 9.37 | Weak aromatic amine; lone pair is less available due to resonance |
| Pyridine | 1.7 × 10-9 | 8.77 | Weak base, but somewhat stronger than aniline |
| Ammonia | 1.8 × 10-5 | 4.75 | Much stronger base than aniline |
| Methylamine | 4.4 × 10-4 | 3.36 | Strongly basic compared with aromatic amines |
This table makes the chemistry clear: aniline is orders of magnitude less basic than ammonia or methylamine. So even concentrated aniline solutions do not produce extremely high hydroxide levels.
Concentration dependence for aniline
Another useful way to understand this system is to compare pH across several concentrations, assuming 25 C and the same Kb value.
| Assumed concentration of aniline (M) | Calculated [OH-] (M) | Approximate pOH | Approximate pH |
|---|---|---|---|
| 0.10 | 6.56 × 10-6 | 5.18 | 8.82 |
| 0.50 | 1.47 × 10-5 | 4.83 | 9.17 |
| 1.00 | 2.07 × 10-5 | 4.68 | 9.32 |
| 3.41 | 3.83 × 10-5 | 4.42 | 9.58 |
| 5.00 | 4.64 × 10-5 | 4.33 | 9.67 |
Notice the pattern: increasing concentration does raise pH, but slowly. For weak bases, pH often changes with the logarithm of the square root of concentration, not in a one-to-one linear way. That is why a large jump in concentration produces a more modest change in pH than many students initially expect.
Common mistakes when solving aniline pH problems
- Using Ka instead of Kb. Aniline is a base, so the equilibrium with water produces OH–, not H+.
- Forgetting the final pH step. After calculating [OH–], you first find pOH, then convert to pH.
- Treating aniline as a strong base. It is weak, so you must use an equilibrium calculation.
- Confusing molality with molarity. If density is absent, the approximation is usually acceptable, but rigorous work should distinguish them.
- Ignoring resonance effects. The aromatic ring reduces aniline basicity, which is why Kb is so small.
When should you use the exact answer 9.67?
You should use pH ≈ 9.67 when the problem is from a standard general chemistry context and gives only 5.0 m aniline with no density or activity data. In that setting, most instructors want the idealized weak-base treatment using the formal concentration as if it were molarity. If the problem specifically emphasizes solution nonideality, density, or activity coefficients, then you should switch to a more advanced treatment.
Authority sources for chemistry data and acid-base background
If you want to verify physical properties and review acid-base concepts from reliable institutions, these are strong references:
- NIH PubChem: Aniline
- NIST Chemistry WebBook: Aniline
- MIT OpenCourseWare: acid-base equilibrium resources
Final answer and interpretation
For the standard textbook approximation, the pH of a 5.0 m solution of aniline is about 9.67. This result comes from treating aniline as a weak base with Kb = 4.3 × 10-10 and solving for the small equilibrium concentration of hydroxide ions. If you instead convert 5.0 m to molarity using an assumed density of 1.00 g/mL, the pH becomes about 9.58. Both values show the same chemical truth: aniline is basic, but only weakly so.