Calculate The Ph Of 0.20 M Aniline A Weak Base

Use this interactive chemistry calculator to solve the pH of aniline solution from concentration and base strength. The tool handles either Kb or pKb input, shows equilibrium steps, and visualizes the concentration changes with a responsive chart.

Expert Guide: How to Calculate the pH of 0.20 M Aniline, a Weak Base

Aniline, with the formula C₆H₅NH₂, is a classic example of a weak base in aqueous chemistry. If you are asked to calculate the pH of 0.20 M aniline, you are solving a weak-base equilibrium problem rather than a strong-base stoichiometry problem. That distinction matters because aniline does not fully react with water. Instead, only a very small fraction of dissolved aniline molecules accept a proton from water to form the anilinium ion and hydroxide.

The relevant equilibrium is:

C₆H₅NH₂ + H₂O ⇌ C₆H₅NH₃⁺ + OH^–

Because hydroxide is produced, the solution becomes basic, so the pH will be above 7. However, because aniline is weak, the pH will not be nearly as high as a 0.20 M solution of sodium hydroxide. The entire calculation depends on the base dissociation constant, Kb, or equivalently on pKb. A commonly used room-temperature value for aniline is approximately Kb = 4.3 × 10^-10, corresponding to pKb ≈ 9.37.

Core Setup for the 0.20 M Aniline Problem

Start by identifying the known quantities:

• Initial aniline concentration, C = 0.20 M
• Base dissociation constant, Kb = 4.3 × 10^-10
• Water temperature assumed to be 25°C so pH + pOH = 14.00

Now construct an ICE table:

• Initial: [C₆H₅NH₂] = 0.20, [C₆H₅NH₃⁺] = 0, [OH^–] = 0
• Change: -x, +x, +x
• Equilibrium: 0.20 – x, x, x

Substitute into the Kb expression:

Kb = ([C₆H₅NH₃⁺][OH^–]) / ([C₆H₅NH₂])

which becomes:

4.3 × 10^-10 = x² / (0.20 – x)

Since Kb is very small, x will be tiny compared with 0.20, so the common weak-base approximation is valid:

0.20 – x ≈ 0.20

This simplifies the expression to:

x² = (4.3 × 10^-10)(0.20) = 8.6 × 10^-11

Taking the square root:

x = [OH^–] ≈ 9.27 × 10^-6 M

Then calculate pOH:

pOH = -log(9.27 × 10^-6) ≈ 5.03

Finally:

pH = 14.00 – 5.03 = 8.97

Final answer for a 0.20 M aniline solution at 25°C using Kb = 4.3 × 10^-10: pH ≈ 8.97.

Why Aniline Is Such a Weak Base

Students often wonder why aniline is much less basic than ammonia or simple alkylamines. The reason lies in resonance and electron delocalization. The nitrogen lone pair in aniline is partially delocalized into the benzene ring. When that lone pair participates in resonance, it becomes less available to accept a proton from water. Lower proton affinity means lower Kb and therefore a lower pH than one might expect for a 0.20 M nitrogen-containing base.

Compare that with ammonia, whose lone pair remains more localized and more available for protonation. Ammonia therefore exhibits a significantly larger Kb and yields much greater hydroxide concentration at the same formal concentration.

Exact Method Versus Approximation

For this specific problem, the square-root approximation works very well, but an exact quadratic solution can also be used. The full equation is:

x² + Kb x – Kb C = 0

Substituting values:

x² + (4.3 × 10^-10)x – (8.6 × 10^-11) = 0

The positive root gives essentially the same hydroxide concentration, confirming that the approximation is excellent. In practice, a useful validation rule is the 5% rule. If x/C is less than 5%, the approximation is acceptable. Here, x/C is roughly:

(9.27 × 10^-6) / 0.20 × 100 ≈ 0.0046%

That is far below 5%, so the approximation is completely justified.

Comparison Table: Weak Base Strengths at 25°C

The following table helps place aniline in context. The values below are representative textbook-scale values commonly used in general chemistry and show why aniline behaves as a much weaker base than ammonia or methylamine.

Base	Approximate Kb	Approximate pKb	Relative Basicity Insight
Aniline, C6H5NH2	4.3 × 10^-10	9.37	Very weak aromatic amine because the lone pair is resonance-delocalized into the ring.
Ammonia, NH3	1.8 × 10^-5	4.74	Much stronger base than aniline because the lone pair is more available.
Methylamine, CH3NH2	4.4 × 10^-4	3.36	Even stronger than ammonia due to electron-donating alkyl substitution.

What the Equilibrium Numbers Mean

One of the best ways to understand this calculation is to interpret the equilibrium concentrations physically. In a 0.20 M aniline solution, the overwhelming majority of aniline molecules remain unprotonated. Only about 9.27 × 10^-6 M converts into anilinium ion and hydroxide. That means the extent of reaction is extremely small. So while the solution is basic, it is only mildly basic. This is the hallmark of weak-base equilibria with very small Kb values.

If you imagine 1.00 liter of solution, the equilibrium amounts are approximately:

• Aniline remaining: 0.1999907 mol
• Anilinium formed: 9.27 × 10^-6 mol
• Hydroxide formed: 9.27 × 10^-6 mol

That dramatic imbalance is why the pH ends up below 9 even though the starting concentration appears fairly large.

Comparison Table: 0.20 M Base Solutions and Expected Basicity

0.20 M Solute	Type	Approximate [OH^–]	Approximate pH at 25°C
Aniline	Weak base	9.27 × 10^-6 M	8.97
Ammonia	Weak base	About 1.9 × 10^-3 M	11.28
Sodium hydroxide	Strong base	0.20 M	13.30

Step-by-Step Shortcut for Exams

If you are under time pressure on a quiz or exam, use this rapid method:

1. Write the base equilibrium: B + H₂O ⇌ BH⁺ + OH^–.
2. Set [OH^–] = x and use Kb = x²/(C – x).
3. If Kb is small, replace C – x with C.
4. Compute x = √(KbC).
5. Find pOH = -log x.
6. Find pH = 14 – pOH.

For aniline specifically:

1. x = √[(4.3 × 10^-10)(0.20)]
2. x = √(8.6 × 10^-11)
3. x = 9.27 × 10^-6
4. pOH = 5.03
5. pH = 8.97

Common Mistakes to Avoid

• Using strong-base logic: Do not set [OH^–] equal to 0.20 M. That would only apply to complete dissociation, such as NaOH.
• Confusing Ka and Kb: Aniline is a base, so use Kb unless the problem gives Ka for the conjugate acid and you convert properly.
• Forgetting pH from pOH: Weak base problems naturally lead to [OH^–], so pOH comes first and pH comes second.
• Incorrect square root handling: Scientific notation errors are common. Always verify your exponent arithmetic.
• Ignoring the 5% rule: Although the approximation is fine here, it is good practice to check.

When to Use pKb Instead of Kb

Some instructors or databases report the basicity of aniline as pKb instead of Kb. The conversion is simple:

Kb = 10^-pKb

If pKb = 9.37, then:

Kb = 10^-9.37 ≈ 4.27 × 10^-10

This leads to the same pH result within rounding. The calculator above accepts either entry mode so you can work from whichever form your source uses.

Why pH Can Depend Slightly on Data Source

You may see small variations in the final pH, such as 8.96, 8.97, or 8.98. These differences usually come from:

• Using Kb = 4.0 × 10^-10 versus 4.3 × 10^-10
• Rounding intermediate values too early
• Using pKb values rounded to one or two decimal places
• Temperature assumptions affecting Kw slightly

In a typical general chemistry setting, any answer very close to pH 8.97 is consistent with the standard data for 0.20 M aniline at room temperature.

Authoritative Chemistry References

For broader background on pH, aqueous equilibria, and weak acid-base principles, consult these reliable sources:

• U.S. Environmental Protection Agency: pH Basics and Water Chemistry
• Michigan State University: Acids, Bases, and Equilibrium Concepts
• Purdue University Chemistry: Acid-Base Equilibria Review

Bottom Line

To calculate the pH of 0.20 M aniline, treat aniline as a weak base, write the equilibrium with water, use its Kb value, solve for hydroxide concentration, then convert through pOH to pH. Using Kb = 4.3 × 10^-10, the hydroxide concentration is about 9.27 × 10^-6 M, the pOH is 5.03, and the final result is:

pH ≈ 8.97

This result highlights a central principle of chemical equilibrium: concentration alone does not determine solution pH. The intrinsic strength of the base matters just as much, and in the case of aniline, resonance stabilization makes that basicity quite weak.