Calculate the pH of 0.150 M Aniline
This premium calculator solves the pH of an aqueous aniline solution using weak-base equilibrium chemistry. Enter the concentration, adjust the base dissociation constant if needed, and compare the approximation method with the exact quadratic solution.
Aniline pH Calculator
Use the default values to solve the pH of 0.150 M aniline instantly.
How to Calculate the pH of 0.150 M Aniline
Aniline, with the formula C₆H₅NH₂, is a weak base. That one fact controls the entire calculation. Unlike a strong base such as sodium hydroxide, aniline does not fully react with water. Instead, it establishes an equilibrium:
C₆H₅NH₂ + H₂O ⇌ C₆H₅NH₃⁺ + OH⁻
Because aniline is only weakly basic, the hydroxide concentration produced is much smaller than the starting concentration of aniline. To calculate the pH of a 0.150 M solution correctly, you use the base dissociation constant, Kb. A common value for aniline at room temperature is about 4.3 × 10-10. That makes it far weaker than ammonia, which is why the pH rises only modestly above neutral.
Quick Answer for 0.150 M Aniline
Using Kb = 4.3 × 10-10 and a starting concentration of 0.150 M, the hydroxide concentration is approximately 8.03 × 10-6 M. That gives:
- pOH ≈ 5.095
- pH ≈ 8.905
So, the pH of 0.150 M aniline is about 8.91 at 25 degrees Celsius.
Step-by-Step Equilibrium Setup
To solve this in a formal chemistry problem, start with an ICE table. Let the amount of OH⁻ formed be x.
- Initial: [C₆H₅NH₂] = 0.150, [C₆H₅NH₃⁺] = 0, [OH⁻] = 0
- Change: -x, +x, +x
- Equilibrium: 0.150 – x, x, x
The equilibrium expression is:
Kb = ([C₆H₅NH₃⁺][OH⁻]) / ([C₆H₅NH₂]) = x² / (0.150 – x)
Substitute the Kb value:
4.3 × 10-10 = x² / (0.150 – x)
Since the base is weak, x is tiny compared with 0.150, so many courses allow the approximation 0.150 – x ≈ 0.150. Then:
x² = (4.3 × 10-10)(0.150)
x² = 6.45 × 10-11
x = 8.03 × 10-6 M
Since x = [OH⁻], calculate pOH:
pOH = -log(8.03 × 10-6) ≈ 5.095
Finally:
pH = 14.000 – 5.095 = 8.905
Why Aniline Is a Weaker Base Than Ammonia
A lot of acid-base intuition comes from comparing compounds. Ammonia and aniline both contain nitrogen with a lone pair, yet ammonia is much more basic. The reason is resonance. In aniline, the lone pair on nitrogen can interact with the aromatic ring, reducing its availability to accept a proton. In ammonia, the lone pair is more localized and therefore more available for bonding to H⁺.
This structural effect is why aromatic amines often behave differently from simple aliphatic amines. If you are comparing weak bases in an exam or lab context, always think about whether electron density on nitrogen is stabilized, withdrawn, or delocalized.
| Base | Typical Kb at 25 degrees Celsius | pKb | Basicity Trend |
|---|---|---|---|
| Aniline | 4.3 × 10-10 | 9.37 | Weak due to resonance delocalization of the lone pair |
| Ammonia | 1.8 × 10-5 | 4.74 | Much stronger weak base than aniline |
| Methylamine | 4.4 × 10-4 | 3.36 | Stronger because alkyl substitution donates electron density |
The table shows that aniline is roughly 42,000 times weaker than ammonia in terms of Kb ratio, and over 1,000,000 times weaker than methylamine. That numerical comparison helps explain why a 0.150 M aniline solution still only reaches a pH near 8.9 instead of 11 or 12.
Approximation Versus Exact Quadratic Solution
For most textbook problems, the approximation is more than acceptable. But a premium calculator should also support the exact solution, especially if concentrations become lower or if the base constant changes. Starting from:
Kb = x² / (C – x)
Rearrange to standard quadratic form:
x² + Kb x – Kb C = 0
Then solve using:
x = [-Kb + √(Kb² + 4KbC)] / 2
Only the positive root is chemically meaningful. For 0.150 M aniline, the exact and approximate answers are essentially identical because the percent ionization is very small.
| Method | [OH⁻] Produced | pOH | pH | Percent Ionization |
|---|---|---|---|---|
| Approximation | 8.03 × 10-6 M | 5.095 | 8.905 | 0.00535% |
| Exact quadratic | 8.03 × 10-6 M | 5.095 | 8.905 | 0.00535% |
The percent ionization is far below 5%, so the common approximation is justified. In general chemistry, that is the checkpoint many instructors expect you to verify after solving.
Common Mistakes When Solving the pH of Aniline
- Treating aniline like a strong base. If you assume full dissociation, the pH becomes wildly incorrect.
- Using Ka instead of Kb. Aniline is a base, so the natural equilibrium constant is Kb. If you use the conjugate acid instead, convert carefully.
- Confusing pH and pOH. Weak base calculations usually give [OH⁻] first, which means pOH comes before pH.
- Dropping significant digits too early. Carry enough digits through the square root and logarithm steps, then round at the end.
- Ignoring temperature dependence. Published equilibrium constants can vary slightly by source or experimental condition.
What the Chemistry Means in Real Terms
A pH of about 8.9 indicates a mildly basic solution, not a strongly corrosive one. This matters in laboratory planning. Aromatic amines like aniline have chemical behavior that is shaped not only by acid-base chemistry but also by organic structure, solubility, and safety concerns. If you are preparing or handling aniline solutions, pH is only one property to consider.
For identification and safety context, authoritative references include the NIST Chemistry WebBook entry for aniline, the U.S. EPA fact sheet on aniline, and toxicology information from the National Library of Medicine resources hosted by NIH. While these sources are not all acid-base tutorials, they are valuable for chemical identity, properties, and health context.
How to Recognize When Water Autoionization Is Negligible
Another advanced point is whether you must account for water itself producing OH⁻. At 25 degrees Celsius, pure water has 1.0 × 10-7 M each of H⁺ and OH⁻. In this problem, the aniline-generated hydroxide concentration is approximately 8.03 × 10-6 M, which is about 80 times larger than the contribution from pure water. That means water autoionization is negligible here.
If the base concentration were much lower, or if the base were even weaker, then a more rigorous treatment might be needed. This is one reason exact calculators are useful even for seemingly simple chemistry questions.
Interpreting Percent Ionization
Percent ionization is often underused as a teaching tool. In this example:
Percent ionization = ([OH⁻] / initial base concentration) × 100
= (8.03 × 10-6 / 0.150) × 100 ≈ 0.00535%
That tiny value confirms two things at once. First, aniline is a weak base. Second, the approximation made in the ICE table is excellent. If less than a hundredth of a percent reacts, then subtracting x from 0.150 changes almost nothing numerically.
Exam Strategy for Problems Like This
- Write the correct equilibrium equation first.
- Identify whether the substance is a weak acid or weak base.
- Use an ICE table to organize concentration changes.
- Substitute into the equilibrium expression.
- Choose approximation or quadratic method.
- Convert [OH⁻] to pOH, then to pH.
- Check whether the final pH makes chemical sense.
That final check matters. Since aniline is a weak base, the pH should be above 7 but not dramatically high. A result like 13 would signal a setup error. A result below 7 would indicate either a wrong constant, a wrong sign, or confusion between aniline and anilinium.
How This Calculator Helps
The calculator above is built for both speed and clarity. It reads the concentration and Kb value, lets you choose an exact or approximate method, and then reports hydroxide concentration, pOH, pH, and percent ionization. It also renders a visual chart so you can immediately see how tiny the ionized fraction is relative to the starting concentration.
That visual perspective is useful for students, tutors, and chemistry content creators. Weak equilibrium problems can feel abstract because the meaningful chemistry hides in very small numbers. A chart bridges that gap by making the scale visible.
Final Takeaway
To calculate the pH of 0.150 M aniline, treat aniline as a weak base and use its Kb value. With Kb = 4.3 × 10-10, the hydroxide concentration is about 8.03 × 10-6 M, the pOH is about 5.095, and the final pH is about 8.905. The solution is basic, but only mildly so, because aniline is much less basic than ammonia and simple alkyl amines.
If you need to solve similar questions, remember the big idea: pH depends on both concentration and intrinsic acid-base strength. For aniline, the weak-base constant is the decisive factor.