Calculate the pH of 0.050 M CH3NH3Br
Use this interactive weak-acid salt calculator to find the pH of methylammonium bromide solutions, review the equilibrium steps, and visualize how concentration affects acidity.
How to calculate the pH of 0.050 M CH3NH3Br
To calculate the pH of 0.050 M CH3NH3Br, you first recognize that methylammonium bromide is a salt made from a weak base, methylamine CH3NH2, and a strong acid, hydrobromic acid HBr. Because bromide is the conjugate base of a strong acid, it does not significantly affect pH in water. The chemistry comes almost entirely from the methylammonium ion CH3NH3+, which behaves as a weak acid.
That means the solution is acidic, not neutral. Students often look at a salt and assume neutrality, but that only applies when the cation and anion both come from strong parent species. Here, the cation is the conjugate acid of a weak base, so hydrolysis occurs:
CH3NH3+ + H2O ⇌ CH3NH2 + H3O+
The hydronium produced in this equilibrium lowers the pH. For a standard textbook calculation at 25°C, a commonly used value is Kb(CH3NH2) = 4.4 × 10^-4. Since the acid constant of the conjugate acid and the base constant of the weak base are related through water ionization, you use:
Ka = Kw / Kb
With Kw = 1.0 × 10^-14, the resulting acid dissociation constant is:
Ka = (1.0 × 10^-14) / (4.4 × 10^-4) = 2.27 × 10^-11
Step-by-step setup
- Write the acid hydrolysis equation for CH3NH3+.
- Determine Ka from the known Kb of methylamine.
- Set the initial concentration of CH3NH3+ equal to the salt concentration, 0.050 M.
- Use an ICE table to express the equilibrium concentrations.
- Solve for [H3O+] and calculate pH from pH = -log[H3O+].
ICE table for 0.050 M CH3NH3Br
Because CH3NH3Br dissolves essentially completely, the initial methylammonium concentration is 0.050 M. Let x represent the amount that dissociates:
- Initial: [CH3NH3+] = 0.050, [CH3NH2] = 0, [H3O+] = 0
- Change: -x, +x, +x
- Equilibrium: 0.050 – x, x, x
Substituting into the acid equilibrium expression gives:
Ka = x² / (0.050 – x)
Since Ka is very small, the amount dissociated is tiny compared with 0.050 M. The common approximation becomes:
x ≈ √(KaC) = √((2.27 × 10^-11)(0.050))
Solving gives:
x ≈ 1.07 × 10^-6 M
Therefore:
pH = -log(1.07 × 10^-6) ≈ 5.97
So the pH of 0.050 M CH3NH3Br is about 5.97 at 25°C when using Kb = 4.4 × 10^-4 for methylamine.
Why CH3NH3Br is acidic in water
The best way to understand this result is to focus on conjugate acid-base behavior. Methylamine, CH3NH2, is a weak base because the nitrogen lone pair can accept a proton, but it does not do so as aggressively as a strong base. Once methylamine gains a proton, it becomes CH3NH3+. That species can donate a proton back to water. In other words, it is a weak acid.
Bromide, by contrast, is the conjugate base of the strong acid HBr. Conjugate bases of strong acids are negligibly basic in water. They are spectators for most pH calculations. That is why nearly all of the pH effect comes from methylammonium hydrolysis.
This principle generalizes to many ammonium-type salts. If you are given a salt of the form RNH3+X- or NH4+X-, and the parent amine is a weak base, the solution will usually be acidic. If you are instead given a salt of a weak acid and a strong base, the solution will usually be basic.
Exact vs approximate solution
In introductory chemistry, the approximation x = √(KaC) is used because it saves time and is usually very accurate for weak acids at moderate concentrations. For CH3NH3Br at 0.050 M, this shortcut works very well. However, in a more exact treatment, you solve the quadratic equation:
x² + Kax – KaC = 0
Then:
x = (-Ka + √(Ka² + 4KaC)) / 2
Since Ka is tiny compared with C, the exact and approximate answers differ by a negligible amount for this specific case. The calculator above lets you choose either method so you can compare them directly.
| Parameter | Value used | Meaning | Impact on pH |
|---|---|---|---|
| Salt concentration | 0.050 M | Initial [CH3NH3+] | Higher concentration generally lowers pH |
| Kb of CH3NH2 | 4.4 × 10-4 | Base strength of methylamine | Larger Kb means smaller Ka for CH3NH3+ |
| Kw | 1.0 × 10-14 | Water ionization constant at 25°C | Used to convert Kb to Ka |
| Calculated Ka | 2.27 × 10-11 | Acid strength of CH3NH3+ | Controls hydronium formation |
| [H3O+] | 1.07 × 10-6 M | Equilibrium hydronium concentration | Determines the final pH |
| Final pH | 5.97 | Acidic solution | Below neutral pH 7 |
Concentration dependence and real numerical trends
One useful insight is that weak-acid salt solutions do not become dramatically more acidic in a linear way as concentration rises. Because hydronium is approximately proportional to the square root of concentration for a weak acid, pH changes more gradually than many learners expect. That means a tenfold increase in concentration only changes pH by about 0.5 units for a simple weak acid approximation.
Using the same Ka = 2.27 × 10^-11, the table below shows approximate pH values at several concentrations. These values are realistic and consistent with weak acid equilibrium behavior at 25°C.
| CH3NH3Br concentration (M) | Approximate [H3O+] (M) | Approximate pH | Interpretation |
|---|---|---|---|
| 0.001 | 1.51 × 10-7 | 6.82 | Only slightly acidic |
| 0.010 | 4.76 × 10-7 | 6.32 | Weakly acidic |
| 0.050 | 1.07 × 10-6 | 5.97 | Standard textbook example |
| 0.100 | 1.51 × 10-6 | 5.82 | Moderately more acidic |
| 0.500 | 3.37 × 10-6 | 5.47 | Acidic, but still a weak acid system |
Common mistakes when solving this problem
- Treating CH3NH3Br as a strong acid. It is not. The acidic species is CH3NH3+, which is a weak acid.
- Using Kb directly to find pH. Kb belongs to CH3NH2, not CH3NH3+. You must convert it to Ka first.
- Forgetting that Br- is a spectator ion. Bromide does not appreciably raise pH.
- Assuming pH = 7 because it is a salt. Many salts are not neutral. Salt pH depends on the strengths of the parent acid and base.
- Ignoring units and temperature. Equilibrium constants depend on temperature, and concentration must be in molarity.
How this problem fits into acid-base chemistry
The pH of 0.050 M CH3NH3Br is a classic conjugate-acid problem. It sits between simple strong-acid calculations and more advanced buffer or polyprotic systems. If you can identify the parent weak base, convert Kb to Ka, build the ICE table, and solve for hydronium, you understand an important category of equilibrium chemistry.
This same method works for ammonium chloride, anilinium salts, substituted ammonium salts, and many organic amine hydrochlorides or hydrobromides. In pharmaceutical, analytical, and environmental contexts, understanding the pH of protonated amine salts matters because pH can affect solubility, extraction behavior, protonation state, and reaction kinetics.
Quick summary process
- Identify the pH-active ion: CH3NH3+.
- Find the parent base: CH3NH2.
- Convert Kb to Ka using Ka = Kw / Kb.
- Use Ka = x² / (C – x).
- Solve for x = [H3O+].
- Compute pH = -log[H3O+].
Authoritative chemistry and pH references
For broader background on pH, equilibrium constants, and conjugate acid-base reasoning, these authoritative resources are useful:
Bottom line
If you are asked to calculate the pH of 0.050 M CH3NH3Br, the chemically correct approach is to treat CH3NH3+ as a weak acid. Using a standard methylamine base constant of 4.4 × 10^-4, you obtain Ka = 2.27 × 10^-11, then solve for hydronium concentration in a 0.050 M solution. The result is a pH of approximately 5.97.
Use the calculator above to verify the answer, test different concentrations, and compare exact versus approximate methods. It is a fast way to understand not only this single homework-style question, but the wider pattern of weak-base conjugate-acid salts in water.