Calculate the pH of a 0.20 M Solution of NH4Br
Use this premium calculator to determine the pH of ammonium bromide solution by converting the base dissociation constant of ammonia into the acid dissociation constant of NH4+, then solving for hydrogen ion concentration. The default setup is for a 0.20 M NH4Br solution, but you can adjust the inputs for study, homework checking, and lab review.
Enter molarity of ammonium bromide.
For aqueous solutions, M and mol/L are equivalent.
Default value commonly used for ammonia at 25°C.
Default ionic product of water at 25°C.
Quadratic is more rigorous. Approximation is useful for teaching and quick checks.
How to calculate the pH of a 0.20 M solution of NH4Br
If you need to calculate the pH of a 0.20 M solution of NH4Br, the key idea is that ammonium bromide is not a neutral salt. It is made from NH4+, the conjugate acid of the weak base ammonia, and Br-, the conjugate base of the strong acid HBr. Because bromide is the conjugate base of a strong acid, it is essentially neutral in water. The chemistry that controls the pH comes from the ammonium ion, which behaves as a weak acid.
That means a 0.20 M NH4Br solution will be acidic, with a pH below 7 at 25°C. To find the exact pH, you first convert the base dissociation constant of ammonia, Kb, into the acid dissociation constant Ka for NH4+. Then you solve the weak acid equilibrium. This calculator performs that work instantly, but it is important to understand the reasoning behind the answer because this type of problem appears often in general chemistry, AP Chemistry, introductory analytical chemistry, and lab calculations.
Step 1: Identify the acid and spectator ion
In water, ammonium bromide dissociates completely:
NH4Br(aq) → NH4+(aq) + Br-(aq)
The bromide ion does not significantly react with water, so it does not affect pH in a meaningful way. The ammonium ion does react:
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
This equilibrium produces hydronium ions, so the solution becomes acidic. Whenever you see a salt derived from a weak base and a strong acid, expect an acidic solution.
Step 2: Convert Kb of NH3 into Ka of NH4+
Many textbooks give the base dissociation constant for ammonia rather than the acid dissociation constant for ammonium. The relationship between them at 25°C is:
Ka × Kb = Kw
Using the common values:
- Kb for NH3 = 1.8 × 10-5
- Kw = 1.0 × 10-14
So:
Ka = Kw / Kb = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
This Ka value tells you that NH4+ is a weak acid, but it is still acidic enough to lower the pH of a 0.20 M solution well below neutral.
Step 3: Set up the equilibrium expression
Let the initial concentration of NH4+ be 0.20 M. Let x represent the amount that dissociates:
- Initial: [NH4+] = 0.20, [NH3] = 0, [H3O+] = 0
- Change: -x, +x, +x
- Equilibrium: [NH4+] = 0.20 – x, [NH3] = x, [H3O+] = x
Substituting into the Ka expression:
Ka = [NH3][H3O+] / [NH4+] = x2 / (0.20 – x)
Since Ka is small, most classroom solutions use the weak acid approximation:
x2 / 0.20 ≈ 5.56 × 10-10
Therefore:
x = √(KaC) = √[(5.56 × 10-10)(0.20)] = 1.05 × 10-5 M
Because x = [H3O+], the pH is:
pH = -log(1.05 × 10-5) ≈ 4.98
Why the result is not extremely acidic
Students sometimes expect any salt containing a positive ion from an acid-base pair to behave like a strong acid, but NH4+ is only a weak acid. Its Ka is on the order of 10-10, which means only a tiny fraction of the ammonium ions donate a proton to water. That is why the pH is acidic but still near 5 rather than near 1 or 2.
The hydrogen ion concentration is approximately 1.05 × 10-5 M, which is only a small amount compared with the 0.20 M starting concentration of NH4+. This also validates the common approximation that 0.20 – x is very close to 0.20.
When to use the quadratic equation instead of the approximation
For this problem, the approximation works very well because the percent ionization is tiny. However, a more exact treatment solves:
x2 + Ka x – KaC = 0
giving:
x = [-Ka + √(Ka2 + 4KaC)] / 2
Plugging in the values yields almost the same hydrogen ion concentration and therefore nearly the same pH. In chemistry classes, instructors often ask students to verify that the approximation is valid by checking the 5% rule:
- Percent ionization = (x / C) × 100
- For NH4Br at 0.20 M, percent ionization is about 0.0053%
Because that value is far below 5%, the approximation is fully justified.
Comparison table: NH4Br versus related nitrogen salts
| Salt | Parent acid/base origin | Acidic, basic, or neutral? | Main ion controlling pH | Typical pH trend at 0.20 M |
|---|---|---|---|---|
| NH4Br | Weak base NH3 + strong acid HBr | Acidic | NH4+ | About 5.0 |
| NH4Cl | Weak base NH3 + strong acid HCl | Acidic | NH4+ | About 5.0 |
| NaBr | Strong base NaOH + strong acid HBr | Neutral | None significantly | Near 7.0 |
| NH4CH3COO | Weak base NH3 + weak acid CH3COOH | Depends on Ka vs Kb | NH4+ and CH3COO- | Near neutral to slightly acidic/basic |
Numerical summary for the 0.20 M NH4Br problem
| Quantity | Value | Meaning |
|---|---|---|
| Initial [NH4+] | 0.20 M | Concentration from complete NH4Br dissociation |
| Kb of NH3 | 1.8 × 10-5 | Standard textbook value at 25°C |
| Kw | 1.0 × 10-14 | Ionic product of water at 25°C |
| Ka of NH4+ | 5.56 × 10-10 | Calculated from Kw/Kb |
| [H3O+] | 1.05 × 10-5 M | Hydronium concentration at equilibrium |
| pH | 4.98 | Acidic solution |
| Percent ionization | 0.0053% | Shows weak acid approximation is valid |
Common mistakes students make
- Treating NH4Br as neutral. This is incorrect because NH4+ is the conjugate acid of a weak base and hydrolyzes in water.
- Using bromide in the equilibrium. Br- comes from a strong acid and does not significantly affect pH.
- Using Kb directly instead of converting to Ka. For pH from hydronium concentration, you need the acid equilibrium of NH4+.
- Forgetting that NH4Br dissociates completely before hydrolysis. The salt first gives 0.20 M NH4+ and 0.20 M Br-.
- Rounding too early. Carrying extra digits through the Ka and x calculations gives a more reliable final pH.
How concentration changes the pH
For weak acids like NH4+, higher concentration generally lowers pH because more acid is present, even though the percent ionization may decrease. If you were comparing 0.020 M, 0.20 M, and 2.0 M NH4Br solutions, the most concentrated solution would have the lowest pH. However, because ammonium is a weak acid, the change in pH is not as dramatic as it would be for a strong acid of the same formal concentration.
The chart in this calculator visualizes the current solution by comparing the formal NH4Br concentration with the much smaller equilibrium concentrations of H3O+ and OH-. This makes it easy to see why ammonium bromide is acidic but only mildly so.
Why NH4Br is important in chemistry education
Problems involving NH4Br are valuable because they combine several fundamental topics: salt hydrolysis, conjugate acid-base relationships, equilibrium constants, and logarithmic pH calculations. A student who can correctly solve the pH of NH4Br usually understands how to classify salts and how to connect Ka, Kb, and Kw.
This problem also reinforces a powerful chemistry habit: do not judge pH from a formula alone. Instead, ask where the ions came from. If a cation is the conjugate acid of a weak base, it often makes the solution acidic. If an anion is the conjugate base of a weak acid, it often makes the solution basic. When both ions come from strong partners, the solution is usually neutral.
Fast method you can remember for exams
- Write the salt ions: NH4+ and Br-
- Ignore Br- because it is from a strong acid
- Use Ka = Kw / Kb for NH4+
- Set [H3O+] ≈ √(KaC)
- Take the negative log to get pH
For the default case:
- Ka = 5.56 × 10-10
- [H3O+] = 1.05 × 10-5 M
- pH = 4.98
Authoritative references for acid-base constants and pH concepts
- U.S. Environmental Protection Agency: pH overview
- Purdue University Chemistry: acids, bases, and equilibrium review
- University of Wisconsin Chemistry: acid-base equilibrium tutorial
Bottom line
To calculate the pH of a 0.20 M solution of NH4Br, treat NH4+ as a weak acid and Br- as a spectator ion. Convert the ammonia Kb into the ammonium Ka, solve the equilibrium, and compute pH from the hydronium concentration. Using standard constants at 25°C, the pH comes out to approximately 4.98. That makes sense chemically because ammonium bromide is the salt of a weak base and a strong acid, so its aqueous solution must be acidic.