Calculate the pH of 0.25 m Solution of Aniline
Use this premium weak-base calculator to determine the pH, pOH, hydroxide concentration, and degree of ionization for an aqueous aniline solution. By default, the tool uses the accepted 25 degrees Celsius base dissociation constant for aniline and applies the exact quadratic solution for equilibrium.
Aniline pH Calculator
How the calculator works
Aniline, C6H5NH2, is a weak base. In water it reacts as:
The equilibrium expression is:
For an initial concentration C and equilibrium hydroxide concentration x:
The exact quadratic solution used here is:
With the default values C = 0.25 and Kb = 4.3 × 10-10, the expected pH is about 9.02 at 25 degrees Celsius, assuming ideal dilute behavior.
Expert Guide: How to Calculate the pH of a 0.25 m Solution of Aniline
If you need to calculate the pH of a 0.25 m solution of aniline, the key chemical idea is that aniline is a weak base, not a strong one. That means it does not fully react with water. Instead, only a small fraction of the dissolved aniline molecules accept a proton from water to generate hydroxide ions. The pH therefore depends on an equilibrium calculation rather than simple complete dissociation.
Aniline has the formula C6H5NH2. Structurally, it is an aromatic amine. The nitrogen atom contains a lone pair of electrons, which allows aniline to act as a Brønsted-Lowry base by accepting a proton. However, because the lone pair is partially delocalized into the benzene ring, aniline is less basic than many aliphatic amines. That reduced basicity is exactly why the pH of a 0.25 solution is only mildly basic rather than strongly alkaline.
Step 1: Write the base ionization reaction
The equilibrium between aniline and water is:
This tells us that every hydroxide ion produced is accompanied by one anilinium ion, C6H5NH3+.
Step 2: Use the base dissociation constant
At 25 degrees Celsius, a commonly used value for the base dissociation constant of aniline is:
Equivalent data may also be reported as pKb:
Because Kb is small, aniline ionizes only slightly in water. That is the central reason the resulting pH is near 9 rather than 12 or 13.
Step 3: Interpret the meaning of 0.25 m
In many classroom and general chemistry contexts, users type “0.25 m” while really intending “0.25 M.” Strictly speaking, lowercase m denotes molality, while uppercase M denotes molarity. For dilute aqueous solutions, the numerical values can be close enough that introductory calculations often treat them similarly. This calculator lets you keep the entered value as an effective analytical concentration or use the common approximation that dilute aqueous density is near 1.00 kg/L.
For the standard textbook-style answer, we take the analytical concentration as 0.25. The exact unit distinction has little effect on the final pH at this concentration unless highly precise thermodynamic corrections are required.
Step 4: Set up the ICE table
Let the initial concentration of aniline be 0.25 and let x be the amount that reacts:
| Species | Initial | Change | Equilibrium |
|---|---|---|---|
| C6H5NH2 | 0.25 | -x | 0.25 – x |
| C6H5NH3+ | 0 | +x | x |
| OH– | 0 | +x | x |
Now substitute into the equilibrium expression:
Step 5: Solve for hydroxide concentration
Insert Kb = 4.3 × 10-10:
Since aniline is a weak base, x will be much smaller than 0.25. A quick approximation would be:
This x value is the hydroxide concentration:
Because the approximation is extremely good here, the exact quadratic calculation gives nearly the same answer. In fact, the percent ionization is tiny, so the 5 percent rule is easily satisfied.
Step 6: Convert to pOH and pH
Once you know the hydroxide concentration, pOH is straightforward:
At 25 degrees Celsius, use:
Therefore:
Why aniline is only a weak base
Students often expect an amine to produce a strongly basic solution, but aromatic amines behave differently from many simple alkyl amines. The nitrogen lone pair in aniline is partially shared with the aromatic ring by resonance. This lowers its availability to bind a proton. As a result, aniline has a much smaller Kb than bases such as methylamine or ethylamine.
| Base | Typical Kb at 25 degrees Celsius | Typical pKb | Relative basicity comment |
|---|---|---|---|
| Aniline | 4.3 × 10-10 | 9.37 | Weak aromatic amine due to resonance delocalization |
| Ammonia | 1.8 × 10-5 | 4.74 | Much stronger base than aniline in water |
| Methylamine | 4.4 × 10-4 | 3.36 | Far stronger than aniline because lone pair is more available |
This comparison helps explain why a 0.25 solution of aniline has a pH just above 9, while similarly concentrated solutions of stronger weak bases can be significantly more alkaline.
Exact vs approximate solution
For weak acid and weak base problems, instructors often teach a shortcut:
That works when x is small compared with the initial concentration. For aniline at 0.25, the estimate gives:
The exact quadratic equation is:
Plugging in the same values yields practically the same hydroxide concentration. In a classroom setting, both methods are acceptable if the approximation is justified. In a digital calculator, the exact approach is preferred because it removes uncertainty and works over a wider range of inputs.
Percent ionization of aniline at 0.25 concentration
Another useful quantity is percent ionization:
Using x ≈ 1.04 × 10-5 and C = 0.25:
This extremely low ionization confirms that aniline remains mostly in its unprotonated molecular form in solution.
Common mistakes when calculating the pH of aniline
- Assuming aniline is a strong base and setting [OH–] equal to 0.25 directly.
- Confusing Ka and Kb values.
- Using pH = -log[OH–] instead of first calculating pOH.
- Ignoring whether the problem states molarity or molality, especially in more advanced courses.
- Forgetting that aromatic amines are much less basic than many aliphatic amines.
How concentration changes the pH
As the concentration of aniline increases, the pH rises, but not dramatically. Because aniline is weak, the hydroxide concentration depends on the square root of Kb times concentration under the weak-base approximation. That means doubling concentration does not double the pH shift. The change is gradual.
| Aniline concentration | Approximate [OH–] | Approximate pOH | Approximate pH |
|---|---|---|---|
| 0.010 | 2.07 × 10-6 | 5.68 | 8.32 |
| 0.050 | 4.64 × 10-6 | 5.33 | 8.67 |
| 0.100 | 6.56 × 10-6 | 5.18 | 8.82 |
| 0.250 | 1.04 × 10-5 | 4.98 | 9.02 |
| 0.500 | 1.47 × 10-5 | 4.83 | 9.17 |
The table shows a useful trend: even large changes in analytical concentration create modest changes in pH for a weak base like aniline. This is an important intuition for acid-base chemistry.
What if the problem gives pKa instead of Kb?
Sometimes you are given information about the conjugate acid, anilinium ion, rather than the base itself. In that case, use:
Or, in logarithmic form:
If you know pKa of anilinium, you can find pKb of aniline and then proceed with the same calculation. This relationship is especially helpful in analytical chemistry and buffer problems.
Real-world considerations
In highly accurate work, a chemist may consider activity coefficients, temperature dependence, solvent composition, and the exact conversion between molality and molarity. Those effects matter more in concentrated, nonideal, or mixed-solvent systems. For standard educational problems, however, the equilibrium method shown above is the accepted route. The result of about pH 9.02 is the right practical answer.
Authoritative chemistry references
Final takeaway
- Recognize aniline as a weak base.
- Use the equilibrium reaction with water.
- Apply Kb ≈ 4.3 × 10-10.
- Solve for [OH–] using the weak-base equation or exact quadratic form.
- Convert [OH–] to pOH, then to pH.
Following these steps gives a pH of approximately 9.02 for a 0.25 solution of aniline at 25 degrees Celsius. If you want a fast answer, use the calculator above. If you want mastery, use the guide to understand every chemical assumption behind the result.