Calculate the pH of a 0.050m C2H5 2NH Solution
Use this premium weak-base calculator to estimate the pH, pOH, hydroxide concentration, conjugate acid concentration, and percent ionization for a dilute aqueous diethylamine solution. The default setup models 0.050 mol/kg as approximately 0.050 M, which is a standard classroom assumption for dilute water solutions.
Weak Base pH Calculator
Equilibrium Visualization
The chart compares the initial base concentration with equilibrium concentrations of unreacted base, hydroxide ion, and conjugate acid. For a weak base such as diethylamine, only a small fraction ionizes, so the pH rises above 7 without reaching the extremely high values characteristic of strong bases.
Reaction modeled: (C2H5)2NH + H2O ⇌ (C2H5)2NH2+ + OH-
How to calculate the pH of a 0.050m C2H5 2NH solution
To calculate the pH of a 0.050m C2H5 2NH solution, you treat the solute as a weak Brønsted base in water. The formula written as C2H5 2NH is commonly understood as diethylamine, more precisely written as (C2H5)2NH. In aqueous solution, diethylamine accepts a proton from water to produce its conjugate acid, (C2H5)2NH2+, and hydroxide ion, OH-. Because hydroxide is generated, the solution is basic and the pH is greater than 7. The key to the calculation is the base dissociation constant, Kb, which measures how strongly the amine reacts with water.
For dilute aqueous problems in general chemistry, a value near Kb = 1.3 × 10-3 at 25 C is often used for diethylamine. If the problem states 0.050m, that formally means molality, or 0.050 moles of solute per kilogram of solvent. In many classroom and introductory problem settings, especially for dilute water solutions, molality and molarity are treated as numerically close, so 0.050m is approximated as 0.050 M. That is exactly the assumption this calculator uses by default.
Bottom line: using Kb = 1.3 × 10-3 and C = 0.050, the hydroxide concentration comes out to about 7.43 × 10-3 M, giving pOH ≈ 2.13 and pH ≈ 11.87.
1. Write the equilibrium reaction
The first step is to write the weak-base equilibrium with water:
(C2H5)2NH + H2O ⇌ (C2H5)2NH2+ + OH-
This tells you that every mole of base that reacts forms one mole of hydroxide and one mole of conjugate acid. That one-to-one stoichiometry is essential when setting up the ICE table.
2. Set up the ICE table
Let the initial concentration of diethylamine be 0.050 M. Let x be the amount that reacts.
- Initial: [(C2H5)2NH] = 0.050, [(C2H5)2NH2+] = 0, [OH-] = 0
- Change: [-x, +x, +x]
- Equilibrium: [0.050 – x, x, x]
Now substitute these equilibrium concentrations into the Kb expression:
Kb = [(C2H5)2NH2+][OH-] / [(C2H5)2NH]
1.3 × 10-3 = x2 / (0.050 – x)
3. Solve for x
There are two common ways to proceed. The first is the weak-base approximation, which assumes x is small compared with 0.050. The second is the exact quadratic solution, which is more rigorous and is what the calculator uses by default.
Approximation method
If x is small, then 0.050 – x is approximated as 0.050:
1.3 × 10-3 ≈ x2 / 0.050
x2 ≈ 6.5 × 10-5
x ≈ 8.06 × 10-3 M
This gives pOH = -log(8.06 × 10-3) ≈ 2.09, so pH ≈ 11.91. That is close, but not exact.
Exact quadratic method
Starting from:
1.3 × 10-3 = x2 / (0.050 – x)
Rearrange to standard quadratic form:
x2 + (1.3 × 10-3)x – (1.3 × 10-3)(0.050) = 0
Solving gives the positive root:
x ≈ 7.43 × 10-3 M
That means:
- [OH-] ≈ 7.43 × 10-3 M
- [(C2H5)2NH2+] ≈ 7.43 × 10-3 M
- [(C2H5)2NH] remaining ≈ 0.04257 M
Then:
- pOH = -log(7.43 × 10-3) ≈ 2.13
- pH = 14.00 – 2.13 ≈ 11.87
4. Check the percent ionization
Percent ionization tells you how much of the base actually reacted:
Percent ionization = (x / 0.050) × 100
Percent ionization ≈ (0.00743 / 0.050) × 100 ≈ 14.9%
This is a useful quality check. Because the ionization is much larger than 5%, the small-x approximation is not ideal here. That is why the exact quadratic method is the better choice for a 0.050 M diethylamine solution.
Why diethylamine gives a basic solution
Diethylamine is an amine, and amines are weak bases because the nitrogen atom has a lone pair that can accept a proton. However, they are weaker than strong bases like sodium hydroxide because the proton transfer with water is not complete. Instead, the system reaches equilibrium. The pH depends on both the initial concentration and the Kb value. A larger Kb means stronger proton acceptance from water and therefore more hydroxide production. Diethylamine is considerably more basic than aniline, for example, because the ethyl groups donate electron density to nitrogen, making proton acceptance more favorable.
Comparison table: weak base strengths of selected nitrogen bases
| Base | Representative Kb at 25 C | pKb | Relative basicity in water |
|---|---|---|---|
| Ammonia, NH3 | 1.8 × 10-5 | 4.74 | Weak base |
| Methylamine, CH3NH2 | 4.4 × 10-4 | 3.36 | Stronger than ammonia |
| Diethylamine, (C2H5)2NH | 1.3 × 10-3 | 2.89 | Quite basic for a simple amine |
| Aniline, C6H5NH2 | 4.3 × 10-10 | 9.37 | Very weak base in water |
This comparison helps explain why the pH of a diethylamine solution is noticeably basic. Its Kb is much larger than the Kb of ammonia, so at the same formal concentration it generates much more OH-. At the same time, it is still a weak base, so the reaction does not go to completion the way NaOH would.
Comparison table: expected pH values for 0.050 M base solutions
| Base | Assumed Kb | Approximate [OH-] | Estimated pH at 0.050 M |
|---|---|---|---|
| Ammonia | 1.8 × 10-5 | 9.5 × 10-4 M | 10.98 |
| Methylamine | 4.4 × 10-4 | 4.5 × 10-3 M | 11.65 |
| Diethylamine | 1.3 × 10-3 | 7.4 × 10-3 M | 11.87 |
| Sodium hydroxide | Strong base | 5.0 × 10-2 M | 12.70 |
The table highlights an important conceptual point. A 0.050 M solution of a weak base like diethylamine has a high pH, but it still does not behave like a strong base at the same concentration. The equilibrium nature of the weak base limits hydroxide formation.
Common mistakes when solving this problem
- Using Ka instead of Kb. Diethylamine is a base, so the equilibrium constant you need is Kb, unless the problem gives you the Ka of the conjugate acid and asks you to convert using KaKb = Kw.
- Confusing molality with molarity. The symbol m means molality, while M means molarity. In dilute water solutions they can be close, but they are not formally identical.
- Forgetting to calculate pOH first. Since the base generates OH-, solve for hydroxide, find pOH, then convert to pH.
- Using the small-x approximation when it is not valid. For this problem, the percent ionization is around 15%, so the approximation introduces noticeable error.
- Assuming complete dissociation. Weak bases do not dissociate fully in water.
When can you treat 0.050m as 0.050 M?
Molality is moles of solute per kilogram of solvent, while molarity is moles of solute per liter of solution. In rigorous physical chemistry, you should distinguish them carefully because solution density matters. However, for dilute aqueous solutions near room temperature, the density is often close enough to 1.00 g/mL that the numeric difference becomes small. This is why many educational pH calculations use the same numerical value. If you were doing high precision analytical work, concentrated solutions, or temperature-sensitive work, you would need density data and activity corrections.
Real chemistry context for amine basicity
The basicity of amines depends on electronic effects, solvation, and steric factors. Ethyl groups donate electron density to nitrogen through inductive effects, making the lone pair more available for protonation than in ammonia. That is one reason diethylamine is a stronger base than ammonia in water. However, water also stabilizes charged species through solvation, and bulky groups can interfere with that stabilization. So while alkyl substitution often increases electron donation, observed aqueous basicity is always a balance between electron effects and solvation effects.
Step by step summary for students
- Identify the solute as a weak base: diethylamine, (C2H5)2NH.
- Write the equilibrium with water to produce OH-.
- Set up an ICE table with initial concentration 0.050.
- Use Kb = x2 / (0.050 – x).
- Solve for x exactly using the quadratic formula.
- Interpret x as [OH-].
- Compute pOH = -log[OH-].
- Compute pH = 14 – pOH.
Final answer
Under the standard dilute-solution assumption that 0.050m ≈ 0.050 M and using Kb = 1.3 × 10-3 for diethylamine at 25 C, the solution has:
- [OH-] ≈ 7.43 × 10-3 M
- pOH ≈ 2.13
- pH ≈ 11.87
That makes the solution clearly basic, but still consistent with the behavior of a weak base rather than a strong base.
Authoritative references for acid-base chemistry
- Chemistry LibreTexts educational resource
- U.S. Environmental Protection Agency acid-base and water chemistry resources
- MIT Department of Chemistry educational content
For classroom problem solving, those resources can help you review equilibrium expressions, logarithms in pH calculations, and the distinction between strong and weak bases. If your instructor provides a different Kb value, use that exact value because the final pH will shift slightly with the equilibrium constant used.