Calculate the pH of 0.900 M Anilinium C6H5NH3Cl
Use this premium acid-base calculator to determine the pH of an anilinium chloride solution, inspect the equilibrium math, and visualize how concentration affects acidity.
Press Calculate to compute the pH of the anilinium chloride solution and generate the concentration vs pH chart.
How to calculate the pH of 0.900 M anilinium C6H5NH3Cl
Anilinium chloride, written as C6H5NH3Cl, is the salt formed from the weak base aniline and the strong acid hydrochloric acid. In water, chloride behaves as a spectator ion, while the anilinium ion, C6H5NH3+, acts as a weak acid. That means the pH of a 0.900 M solution is controlled by the acid dissociation of the conjugate acid of aniline rather than by complete proton release as in a strong acid. This distinction is the key to solving the problem correctly.
In many classroom and laboratory settings, a pKa near 4.60 to 4.63 is used for the anilinium ion at 25 degrees C. Since pKa and Ka are related by Ka = 10-pKa, that gives a Ka on the order of 2.3 to 2.5 × 10-5. Because the solution concentration is fairly high at 0.900 M, the hydrogen ion concentration generated by dissociation is still much smaller than the initial acid concentration, so the weak acid approximation works well. However, this calculator also offers the exact quadratic solution, which is the best practice for precision.
What species are present in solution?
When anilinium chloride dissolves, it separates into ions:
The chloride ion does not affect pH significantly because it is the conjugate base of a strong acid. The acid-base equilibrium that matters is:
That equilibrium tells you the solution is acidic. The stronger the conjugate acid, the lower the pH. Because aniline is only a weak base, its conjugate acid is a weak acid, not a strong one. Therefore, you cannot set [H+] equal to 0.900 M. Doing so would overestimate the acidity dramatically.
Step by step calculation
- Identify the acid species: C6H5NH3+.
- Use the acid dissociation constant of the anilinium ion.
- Set up an ICE table for the weak acid equilibrium.
- Solve for x = [H3O+].
- Compute pH = -log[H3O+].
Let the initial concentration of anilinium ion be 0.900 M. If x dissociates, then at equilibrium:
- [C6H5NH3+] = 0.900 – x
- [C6H5NH2] = x
- [H3O+] = x
The equilibrium expression is:
Using pKa = 4.60 gives:
Substitute into the expression:
With the weak acid approximation, 0.900 – x ≈ 0.900, so:
Then:
If you solve the exact quadratic equation instead of using the approximation, you get nearly the same answer, typically around pH 2.32. This is the most appropriate final answer when pKa = 4.60 is assumed. If your textbook uses pKa = 4.63, the result shifts slightly upward by only a few hundredths of a pH unit.
Why the concentration matters
The higher the concentration of the weak acid, the more hydronium forms, although the relationship is not linear because of equilibrium. Doubling the concentration does not halve the pH. Instead, weak acid systems often follow an approximate square root dependence, which means pH changes more gradually than many students expect. A 0.900 M solution is much more acidic than a 0.0900 M solution, but not by a full pH unit.
| Initial anilinium concentration (M) | Assumed pKa | Ka | Approximate [H3O+] (M) | Calculated pH |
|---|---|---|---|---|
| 0.010 | 4.60 | 2.51 × 10^-5 | 5.01 × 10^-4 | 3.30 |
| 0.050 | 4.60 | 2.51 × 10^-5 | 1.12 × 10^-3 | 2.95 |
| 0.100 | 4.60 | 2.51 × 10^-5 | 1.58 × 10^-3 | 2.80 |
| 0.500 | 4.60 | 2.51 × 10^-5 | 3.54 × 10^-3 | 2.45 |
| 0.900 | 4.60 | 2.51 × 10^-5 | 4.75 × 10^-3 | 2.32 |
| 1.000 | 4.60 | 2.51 × 10^-5 | 5.01 × 10^-3 | 2.30 |
Common mistakes to avoid
- Treating anilinium chloride as a strong acid. It is a salt of a weak base and a strong acid, so the cation is only partially dissociated as an acid.
- Using Kb of aniline directly without converting. If you start with the base constant of aniline, convert using KaKb = Kw.
- Ignoring units or logarithm signs. pH must come from the negative base-10 logarithm of hydronium concentration.
- Dropping x from the denominator without checking. It works here, but exact solving is safer and is included in the calculator above.
Relationship between pKa, pKb, and acidity
Aniline is a significantly weaker base than aliphatic amines because the nitrogen lone pair is partially delocalized into the aromatic ring. That reduced basicity makes its conjugate acid, the anilinium ion, more acidic than the conjugate acids of many common amines. In water at 25 degrees C:
| Species | Acid or base descriptor | Typical pKa or pKb | Interpretation |
|---|---|---|---|
| Aniline, C6H5NH2 | Weak base | pKb ≈ 9.37 to 9.40 | Accepts protons only weakly in water |
| Anilinium ion, C6H5NH3+ | Weak acid | pKa ≈ 4.60 to 4.63 | Partially donates protons to water |
| Ammonium ion, NH4+ | Weak acid | pKa ≈ 9.25 | Less acidic than anilinium |
| Hydrochloric acid, HCl | Strong acid | pKa well below 0 | Nearly complete dissociation in water |
This comparison helps explain why anilinium chloride solutions are acidic but not as strongly acidic as strong acid solutions of the same formal concentration. A 0.900 M HCl solution would have a pH close to 0.05, while 0.900 M anilinium chloride has a pH around 2.32. That difference is enormous chemically, corresponding to hydronium concentrations that differ by more than two orders of magnitude.
Exact solution versus approximation
The approximation method uses x = √(KaC), which is derived by assuming x is much smaller than the initial concentration. For 0.900 M anilinium chloride, x is roughly 0.00475 M, which is only about 0.53% of 0.900 M. Since this is well below the common 5% threshold, the approximation is excellent. Still, exact solving is preferable when you want cleaner reporting or are working near the limits of approximation validity.
The quadratic equation for the equilibrium is:
Solving gives:
Only the positive root has physical meaning. Once x is known, pH follows directly. In this calculator, both approaches are available so you can compare them instantly.
When would activity corrections matter?
At concentrations approaching 1.0 M, ideal-solution assumptions become less reliable. Introductory chemistry courses usually ignore activity coefficients and use concentration directly in equilibrium expressions. That is appropriate for most educational problems, including this one. In more advanced physical chemistry or analytical chemistry contexts, ionic strength can alter effective acidity, and pH meter readings may differ slightly from a simple equilibrium calculation. For a homework, exam, or standard reference problem asking for the pH of 0.900 M anilinium chloride, the textbook concentration-based answer near 2.32 is normally expected.
Authoritative chemistry references
For foundational acid-base theory, equilibrium behavior, and water chemistry, the following sources are useful:
- National Institute of Standards and Technology (NIST)
- Chemistry LibreTexts educational resource
- U.S. Environmental Protection Agency chemistry and water resources
Final answer summary
If you assume pKa = 4.60 for the anilinium ion at 25 degrees C, then the hydronium concentration produced by a 0.900 M solution of anilinium chloride is about 4.75 × 10-3 M, giving a pH of approximately 2.32. If a slightly different literature pKa is used, such as 4.63, the result changes only slightly. The chemistry behind the answer is that C6H5NH3+ is a weak acid, so the solution is acidic, but not nearly as acidic as a strong acid at the same concentration.
Use the interactive tool above to test different concentrations, compare exact and approximate methods, and visualize how the pH trend responds to changes in the initial amount of anilinium ion.