Calculate the pH of a 0.050 M C2H5NH2 Solution
Use this premium weak base calculator to find the pH, pOH, hydroxide concentration, percent ionization, and equilibrium concentrations for ethylamine, C2H5NH2, in water. The calculator uses the exact quadratic solution and visualizes the equilibrium species with Chart.js.
Weak Base pH Calculator
How to Calculate the pH of a 0.050 M C2H5NH2 Solution
To calculate the pH of a 0.050 M C2H5NH2 solution, you treat ethylamine as a weak Brønsted base that partially reacts with water to form its conjugate acid and hydroxide ions. The chemistry is straightforward, but the details matter because weak bases do not dissociate completely. That means you cannot simply assume the hydroxide concentration equals the starting concentration. Instead, you write the equilibrium expression, solve for the amount ionized, calculate pOH, and then convert pOH to pH.
Ethylamine, written as C2H5NH2, belongs to the family of weak amine bases. In water, it accepts a proton from H2O and forms C2H5NH3+ while producing OH-. Since pH depends on the concentration of hydrogen ions and hydroxide ions in solution, the generation of OH- drives the solution to a pH above 7. A 0.050 M solution is concentrated enough that the pH is clearly basic, but because the base is weak, the final pH is still much lower than what you would expect from a strong base of the same formal concentration.
Step 1: Write the balanced equilibrium
The first step is the base ionization reaction:
Water is a pure liquid, so it is not included in the equilibrium constant expression. The relevant constant is the base dissociation constant, Kb. For ethylamine at 25 degrees C, a commonly used textbook value is approximately 5.6 × 10-4.
Step 2: Set up an ICE table
An ICE table tracks the Initial, Change, and Equilibrium concentrations:
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| C2H5NH2 | 0.050 | -x | 0.050 – x |
| C2H5NH3+ | 0 | +x | x |
| OH- | 0 | +x | x |
Here, x represents the amount of ethylamine that reacts with water. At equilibrium, that same amount appears as both C2H5NH3+ and OH- because the reaction stoichiometry is 1:1:1.
Step 3: Write the Kb expression
Using the ICE table, the equilibrium expression becomes:
Substitute Kb = 5.6 × 10-4:
Step 4: Solve for x exactly
Although many chemistry classes first try the weak base shortcut x ≈ √(KbC), the exact quadratic solution is more robust. Rearranging the equation gives:
With Kb = 5.6 × 10-4 and C = 0.050:
Apply the quadratic formula and keep the positive root:
This gives:
So the equilibrium hydroxide concentration is about 0.00502 M.
Step 5: Convert [OH-] to pOH and pH
Now calculate pOH:
At 25 degrees C, use pH + pOH = 14.00:
Final answer: the pH of a 0.050 M C2H5NH2 solution is approximately 11.70.
Interpretation of the Result
A pH of 11.70 means the solution is moderately basic. It is nowhere near as basic as a 0.050 M strong base such as NaOH, which would have [OH-] = 0.050 M and pH around 12.70. That full pH unit difference is significant because the pH scale is logarithmic. Ethylamine only partially ionizes, so its hydroxide concentration is much lower than its formal concentration.
This is one of the most important takeaways for acid-base chemistry: concentration alone does not determine pH. Strength matters. A weak base at 0.050 M and a strong base at 0.050 M do not produce the same pH because the fraction that reacts with water is very different.
Percent ionization
You can also quantify how much ethylamine ionizes:
That means only about one tenth of the ethylamine molecules are protonated at equilibrium under these conditions. This is a useful confirmation that the compound behaves as a weak base.
Exact Method vs Approximation Method
Many students are taught to estimate x using the shortcut x ≈ √(KbC). For this problem:
This leads to a pOH of about 2.28 and a pH of about 11.72. That estimate is very close to the exact answer of 11.70, but not identical. The difference is small here, which is why the approximation is often acceptable in introductory chemistry. However, the exact solution removes ambiguity and is especially useful when your instructor expects a polished answer or when percent ionization is not trivially small.
| Method | [OH-] (M) | pOH | pH | Comment |
|---|---|---|---|---|
| Exact quadratic | 0.00502 | 2.30 | 11.70 | Best formal answer |
| Square root approximation | 0.00529 | 2.28 | 11.72 | Fast estimate, slightly high |
| Strong base assumption | 0.0500 | 1.30 | 12.70 | Incorrect for ethylamine |
Comparison with Other Bases
Ethylamine is a stronger weak base than ammonia. That means, at the same concentration, ethylamine generally produces a slightly higher pH than NH3. This trend reflects the electron donating effect of the ethyl group, which helps the nitrogen atom accept a proton more readily. In practical acid-base calculations, that stronger Kb translates into higher [OH-] and a lower pOH.
| Base | Typical Kb at 25 degrees C | Approximate pH at 0.050 M | Relative basicity |
|---|---|---|---|
| Ammonia, NH3 | 1.8 × 10^-5 | About 10.98 | Weaker |
| Ethylamine, C2H5NH2 | 5.6 × 10^-4 | About 11.70 | Stronger weak base |
| Sodium hydroxide, NaOH | Strong base, effectively complete dissociation | About 12.70 | Much stronger in water |
Common Mistakes When Solving This Problem
- Treating ethylamine like a strong base. If you set [OH-] = 0.050 M directly, you overestimate the pH by about 1 full unit.
- Using Ka instead of Kb. Ethylamine is a base, so you should use Kb unless the problem explicitly gives Ka for the conjugate acid.
- Forgetting the relationship between pOH and pH. At 25 degrees C, pH = 14.00 – pOH.
- Dropping the x term too quickly. The approximation often works, but it should be justified. The exact quadratic method always works.
- Using the wrong formula for percent ionization. For a base, percent ionization = [OH-] / initial concentration × 100 when the stoichiometry is 1:1.
Short Worked Example
If you needed to present a compact solution on a homework assignment, you could write it like this:
- C2H5NH2 + H2O ⇌ C2H5NH3+ + OH-
- ICE table gives Kb = x² / (0.050 – x)
- Using Kb = 5.6 × 10-4, solve x² + 5.6 × 10-4x – 2.8 × 10-5 = 0
- x = [OH-] = 0.00502 M
- pOH = -log(0.00502) = 2.30
- pH = 14.00 – 2.30 = 11.70
That format is concise, chemically correct, and easy to grade.
Why the Kb Value Matters
Published equilibrium constants can vary slightly depending on the source, ionic strength, and temperature. If your textbook uses a different Kb value for ethylamine, your pH may differ by a few hundredths. For example, using 6.4 × 10-4 instead of 5.6 × 10-4 would produce a slightly larger hydroxide concentration and a slightly higher pH. That is why it is smart to show your Kb value in your setup. It tells the reader exactly why your final number came out as it did.
Authoritative Reference Links
For broader background on pH, equilibrium, and acid-base chemistry, these sources are useful:
- USGS: pH and Water
- LibreTexts Chemistry (.edu host mirror content available through many universities)
- NIST Chemistry WebBook
Practical Takeaway
If you remember only one thing, remember this: to calculate the pH of a 0.050 M C2H5NH2 solution, you must use weak base equilibrium, not full dissociation. Start with the reaction of ethylamine with water, set up the ICE table, solve Kb = x² / (C – x), determine [OH-], and then convert to pH. Using Kb = 5.6 × 10-4 at 25 degrees C, the final answer is pH ≈ 11.70.
This value makes chemical sense. It is clearly basic, it is stronger than ammonia at the same concentration, and it is noticeably less basic than a strong base such as sodium hydroxide. In classroom work, exam settings, or laboratory calculations, this combination of chemistry insight and careful equilibrium math is exactly what instructors expect to see.