Calculate the pH of a 0.60 m Solution of Aniline
Use this premium calculator to determine the pH, pOH, hydroxide concentration, and percent ionization for aniline in water. By default, it uses a concentration of 0.60 m and the commonly cited base dissociation constant for aniline at 25 degrees Celsius.
Expert Guide: How to Calculate the pH of a 0.60 m Solution of Aniline
Aniline, with formula C6H5NH2, is an aromatic amine and a classic example of a weak base in aqueous chemistry. If you are asked to calculate the pH of a 0.60 m solution of aniline, the key idea is that aniline does not fully react with water. Instead, it establishes an equilibrium in which only a small fraction of molecules accept protons from water to form the anilinium ion and hydroxide ion. That small amount of hydroxide is what makes the solution basic.
In many introductory and intermediate chemistry problems, the notation 0.60 m is interpreted practically as a concentration close to 0.60 M when the density of the solution is not supplied. Strictly speaking, molality and molarity are different units. Molality is moles of solute per kilogram of solvent, while molarity is moles of solute per liter of solution. Because acid base equilibrium constants are usually applied using molarity, most textbook solutions either state an approximation or provide density data if a more exact conversion is needed. This calculator lets you do either.
The chemical equilibrium for aniline in water
The relevant weak base reaction is:
For a weak base, the base dissociation constant is written as:
A commonly used value for aniline at 25 degrees Celsius is Kb = 4.3 × 10-10, which corresponds to pKb ≈ 9.37. Because the Kb is small, aniline is a much weaker base than ammonia. The aromatic ring delocalizes electron density and makes the nitrogen lone pair less available for protonation.
Step by step solution for a 0.60 m aniline solution
- Assume the given 0.60 m behaves approximately as 0.60 M unless density data are supplied.
- Set up an ICE table for the equilibrium:
- Initial: [C6H5NH2] = 0.60, [C6H5NH3+] = 0, [OH–] = 0
- Change: -x, +x, +x
- Equilibrium: 0.60 – x, x, x
- Substitute into the Kb expression:
4.3 × 10^-10 = x^2 / (0.60 – x)
- Since Kb is very small, x will be tiny compared with 0.60, so the approximation 0.60 – x ≈ 0.60 is excellent:
x^2 = (4.3 × 10^-10)(0.60) = 2.58 × 10^-10x = [OH-] = √(2.58 × 10^-10) ≈ 1.61 × 10^-5 M
- Calculate pOH:
pOH = -log(1.61 × 10^-5) ≈ 4.79
- Convert to pH:
pH = 14.00 – 4.79 = 9.21
So the pH of a 0.60 m solution of aniline is approximately 9.21 under the usual 25 degree Celsius assumption. If you solve the equilibrium exactly with the quadratic formula, you get essentially the same answer because the extent of ionization is extremely small relative to the initial concentration.
Why the approximation works so well
Students often wonder whether it is safe to ignore x in the denominator. Here, the percentage ionization is tiny:
That is far below the common 5 percent guideline. Because of that, the exact and approximate solutions are nearly identical. For weak bases like aniline, this kind of simplification is not just convenient, it is chemically justified.
Molality versus molarity in this problem
The wording 0.60 m can trigger an important conceptual discussion. Molality is not the same as molarity. If a problem gives only molality and no density, many instructors accept the approximation that 0.60 m is close enough to 0.60 M for a hand calculation. If density is known, you can convert more precisely using:
For aniline, the molar mass is about 93.13 g/mol. If the density were 1.00 g/mL, then a 0.60 m solution would correspond to a molarity slightly lower than 0.60 M. The pH would still remain close to 9.2 because the base is weak and the square root relationship softens the effect of moderate concentration changes.
Comparison table: aniline versus other weak bases
The table below shows why aniline gives a mildly basic pH rather than a strongly basic one. Its Kb is much smaller than that of ammonia and somewhat smaller than that of pyridine.
| Base | Representative formula | Kb at 25 degrees Celsius | pKb | Relative basic strength |
|---|---|---|---|---|
| Aniline | C6H5NH2 | 4.3 × 10^-10 | 9.37 | Weakest of the three listed |
| Pyridine | C5H5N | 1.7 × 10^-9 | 8.77 | About 4 times stronger than aniline |
| Ammonia | NH3 | 1.8 × 10^-5 | 4.74 | Much stronger weak base |
This comparison is chemically useful. In aniline, the nitrogen lone pair can interact with the benzene ring, reducing its availability to bind a proton. In ammonia, there is no aromatic ring withdrawing availability from the lone pair, so the base is much stronger.
How concentration affects the pH of aniline solutions
Because aniline is a weak base, increasing the concentration does increase pH, but not dramatically. The hydroxide concentration depends roughly on the square root of concentration when the weak base approximation is valid. That means a tenfold increase in concentration raises pH by only about 0.5 units.
| Aniline concentration treated as M | Calculated [OH-] (M) | pOH | pH at 25 degrees Celsius | Percent ionization |
|---|---|---|---|---|
| 0.010 | 2.07 × 10^-6 | 5.68 | 8.32 | 0.0207% |
| 0.10 | 6.56 × 10^-6 | 5.18 | 8.82 | 0.0066% |
| 0.60 | 1.61 × 10^-5 | 4.79 | 9.21 | 0.0027% |
| 1.00 | 2.07 × 10^-5 | 4.68 | 9.32 | 0.0021% |
Common mistakes when solving this problem
- Using Ka instead of Kb. Aniline is acting as a base here, so the correct equilibrium constant is Kb.
- Forgetting to convert from pOH to pH. The equilibrium gives hydroxide, so pOH is found first and then converted by subtracting from 14.00 at 25 degrees Celsius.
- Assuming complete ionization. Aniline is a weak base, so it ionizes only slightly.
- Ignoring the meaning of 0.60 m. If your instructor expects a distinction between molality and molarity, mention the approximation or perform a density conversion.
- Incorrect logarithm handling. Make sure the negative sign is included when calculating pOH = -log[OH–].
Exact quadratic solution
If you want the exact answer rather than the approximation, solve:
Rearrange to:
Using C = 0.60 and Kb = 4.3 × 10-10:
This gives x ≈ 1.606 × 10-5 M, which still leads to pH ≈ 9.21. The exact method confirms that the approximation is valid.
Practical interpretation of the result
A pH around 9.21 means the solution is definitely basic, but not strongly caustic like a concentrated sodium hydroxide solution. In real laboratory practice, aromatic amines such as aniline are handled with attention to both acid base behavior and toxicological safety. The pH tells you about hydroxide concentration, but safe handling also depends on volatility, skin absorption, and occupational exposure concerns. For that reason, chemical data should always be paired with authoritative safety references.
Authoritative references for deeper study
- PubChem: Aniline chemical data and properties
- NIST Chemistry WebBook: Aniline reference data
- U.S. EPA: Aniline technical fact sheet
Final takeaway
To calculate the pH of a 0.60 m solution of aniline, you treat aniline as a weak base, write the equilibrium with water, insert the Kb expression, and solve for the hydroxide concentration. Using Kb = 4.3 × 10-10 and the standard classroom approximation that 0.60 m is approximately 0.60 M, you obtain [OH–] ≈ 1.61 × 10-5 M, pOH ≈ 4.79, and pH ≈ 9.21. That answer is chemically sensible, mathematically justified, and consistent with aniline being a weak aromatic base.
If you need a more exact answer for research, process chemistry, or highly concentrated systems, use density to convert molality to molarity and solve with the exact quadratic equation. For most educational contexts, however, the reported pH of approximately 9.2 is the accepted result.