Calculate the pH of 0.38 M NH4Br Solution
Use this premium calculator to determine the acidity of ammonium bromide solution by treating NH4Br as a salt of a weak base and a strong acid. The bromide ion is neutral, while NH4+ acts as a weak acid in water.
How to calculate the pH of 0.38 M NH4Br solution
To calculate the pH of a 0.38 M NH4Br solution, you first need to identify what kind of salt ammonium bromide is. NH4Br is formed from NH3, a weak base, and HBr, a strong acid. When this salt dissolves in water, it separates completely into NH4+ and Br-. The bromide ion is the conjugate base of a strong acid, so it has negligible basic behavior in water. The ammonium ion, however, is the conjugate acid of the weak base ammonia, so it donates protons weakly and makes the solution acidic.
This means the pH is not found by assuming the solution is neutral and it is not found by treating NH4Br as a strong acid either. Instead, the correct approach is to use the acid dissociation behavior of NH4+:
NH4+ + H2O ⇌ NH3 + H3O+
The key constant for this equilibrium is the acid dissociation constant of NH4+, written as Ka. Because many textbooks provide the base dissociation constant of NH3 rather than Ka for NH4+, the standard conversion is:
Ka(NH4+) = Kw / Kb(NH3)
At 25°C, Kw = 1.0 × 10^-14 and a common value for Kb of NH3 is 1.8 × 10^-5. Therefore:
Ka = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10
Now let the initial concentration of NH4+ be 0.38 M. Because NH4Br dissociates completely, the starting concentration of NH4+ is the same as the formal concentration of the salt. Set up the equilibrium table:
- Initial: [NH4+] = 0.38, [NH3] = 0, [H3O+] ≈ 0
- Change: -x, +x, +x
- Equilibrium: [NH4+] = 0.38 – x, [NH3] = x, [H3O+] = x
Substitute into the Ka expression:
Ka = x² / (0.38 – x)
Using the weak acid approximation because Ka is small relative to concentration, 0.38 – x is approximately 0.38:
x² / 0.38 = 5.56 × 10^-10
x² = 2.11 × 10^-10
x = [H3O+] = 1.45 × 10^-5 M
Then:
pH = -log(1.45 × 10^-5) = 4.84
So the pH of a 0.38 M NH4Br solution at 25°C is approximately 4.84. This is the value most chemistry instructors expect when the standard Kb of ammonia is used. The exact quadratic method gives essentially the same value because x is much smaller than 0.38 M, so the approximation is excellent.
Why NH4Br solution is acidic
Students often memorize rules about salts but do not always see why they work. NH4Br is a textbook example of a salt whose pH depends on the parent acid and base. HBr is a strong acid, so Br- is an extremely weak base and does not significantly react with water. NH3 is a weak base, so its conjugate acid, NH4+, does react with water and produces hydronium ions. The resulting solution is acidic.
A useful rule is this:
- Strong acid + strong base salt → approximately neutral
- Strong acid + weak base salt → acidic
- Weak acid + strong base salt → basic
- Weak acid + weak base salt → depends on relative Ka and Kb
Ammonium bromide falls squarely into the strong acid plus weak base category. That is why the correct pH is less than 7, but not extremely low, because NH4+ is only a weak acid.
Common constants used in the calculation
Exact values can vary slightly by source and temperature, but introductory chemistry courses usually use the values below at 25°C. That is why pH answers may differ in the second decimal place depending on the textbook table provided by the instructor.
| Quantity | Typical value at 25°C | Why it matters |
|---|---|---|
| Kb of NH3 | 1.8 × 10^-5 | Used to convert to Ka for NH4+ |
| Kw of water | 1.0 × 10^-14 | Relates conjugate acid and base constants |
| Ka of NH4+ | 5.56 × 10^-10 | Determines hydronium formation from NH4+ |
| Formal concentration of NH4Br | 0.38 M | Initial NH4+ concentration after dissolution |
| Calculated [H3O+] | 1.45 × 10^-5 M | Directly converted into pH |
| Calculated pH | 4.84 | Final answer for the solution |
Step by step method used by chemistry students and instructors
- Write the dissociation of the salt: NH4Br → NH4+ + Br-
- Identify spectator behavior: Br- does not significantly hydrolyze because it comes from the strong acid HBr.
- Write the acid equilibrium: NH4+ + H2O ⇌ NH3 + H3O+
- Find Ka for NH4+: Ka = Kw / Kb
- Set up an ICE table: use 0.38 M as the initial NH4+ concentration.
- Solve for x: x is the equilibrium hydronium concentration.
- Compute pH: pH = -log[H3O+]
- Check the approximation: x/0.38 should be well below 5 percent.
For this problem, the approximation check is excellent:
(1.45 × 10^-5) / 0.38 × 100 ≈ 0.0038%
That is far below 5%, so the approximation is completely justified.
Exact quadratic solution versus approximation
In many chemistry courses, instructors accept the weak acid approximation because NH4+ is only weakly acidic and the concentration is fairly large. However, advanced students or analytical chemistry workflows may prefer the exact quadratic expression:
x² + Ka x – Ka C = 0
Where:
- x = [H3O+]
- Ka = 5.56 × 10^-10
- C = 0.38 M
The physically meaningful root is:
x = (-Ka + √(Ka² + 4KaC)) / 2
When you evaluate that expression, you obtain essentially the same hydronium concentration as the approximation, so the pH remains about 4.84. This is a good example of a system where the simple method and exact method agree closely.
| Method | [H3O+] (M) | pH | Difference from exact pH |
|---|---|---|---|
| Weak acid approximation | 1.4535 × 10^-5 | 4.8376 | Very small |
| Exact quadratic solution | 1.4529 × 10^-5 | 4.8378 | Reference result |
| Percent ionization | 0.00382% | Not a pH value | Confirms approximation validity |
How concentration affects the pH of NH4Br
Because NH4+ is a weak acid, the pH of ammonium bromide changes with concentration. More concentrated solutions usually have a slightly lower pH because more NH4+ is present to produce hydronium ions. However, the pH does not decrease in a one-to-one way as it would with a strong acid of equal concentration.
The pattern for a weak acid like NH4+ follows the square-root relation approximately:
[H3O+] ≈ √(KaC)
That means if the concentration changes by a factor of 100, the hydronium concentration changes by about a factor of 10, and the pH changes by about 1 unit. This is why weak acid solutions do not become as acidic as equally concentrated strong acid solutions.
Illustrative concentration comparison for NH4Br at 25°C
| NH4Br concentration (M) | Approximate [H3O+] (M) | Approximate pH | Interpretation |
|---|---|---|---|
| 0.010 | 2.36 × 10^-6 | 5.63 | Mildly acidic |
| 0.050 | 5.27 × 10^-6 | 5.28 | Still weakly acidic |
| 0.100 | 7.45 × 10^-6 | 5.13 | Acidicity increasing |
| 0.380 | 1.45 × 10^-5 | 4.84 | Target problem value |
| 1.000 | 2.36 × 10^-5 | 4.63 | More acidic, but still weak acid behavior |
Frequent mistakes when solving this problem
1. Treating NH4Br as a neutral salt
This is probably the most common mistake. Students see a salt and assume pH 7. That only works for salts made from strong acids and strong bases such as NaCl or KNO3. NH4Br is not in that category.
2. Using HBr to determine pH directly
HBr is indeed a strong acid, but it is not present as undissociated HBr in the final salt solution. The solution contains NH4+ and Br-. The acidity comes from NH4+, not from a separate quantity of free HBr added to water.
3. Forgetting to convert Kb to Ka
Because NH4+ is the conjugate acid of NH3, you cannot use Kb directly in the acid equilibrium expression. You must convert it using Ka = Kw / Kb.
4. Using the wrong initial concentration
Since NH4Br dissociates completely, the initial concentration of NH4+ equals the concentration of NH4Br. In this case, that means 0.38 M NH4+ at the start of the hydrolysis calculation.
5. Ignoring temperature assumptions
If your instructor specifies a temperature other than 25°C, Kw and sometimes equilibrium constants may differ. Unless stated otherwise, chemistry problems of this type almost always assume 25°C standard conditions.
What the final answer means chemically
A pH of 4.84 means the solution is clearly acidic, yet far less acidic than a 0.38 M strong acid would be. For comparison, a 0.38 M solution of a fully dissociated strong acid would have a pH near 0.42 because pH = -log(0.38). That huge difference highlights the limited proton donation ability of NH4+, which hydrolyzes only slightly.
The percent ionization is tiny, under 0.004%, so almost all of the ammonium remains as NH4+ in solution. Nevertheless, even a very small amount of hydrolysis is enough to shift the pH below 7 and produce a measurable acidic environment.
Authoritative references for acid-base equilibrium and pH
If you want to verify constants or review weak acid and conjugate acid calculations, these sources are useful:
- U.S. Environmental Protection Agency: pH Basics
- Purdue University: Acid Dissociation Constants and Equilibria
- National Institute of Standards and Technology: Chemistry and Measurement Resources
Quick answer summary
If all you need is the final result, here is the streamlined solution:
- NH4Br dissociates into NH4+ and Br-.
- Br- is neutral because it is the conjugate base of strong acid HBr.
- NH4+ is a weak acid with Ka = Kw / Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10.
- For 0.38 M NH4+, [H3O+] ≈ √(KaC) = √[(5.56 × 10^-10)(0.38)] = 1.45 × 10^-5 M.
- pH = -log(1.45 × 10^-5) = 4.84.