Calculate the pH of 0.15 M NH4Br Solution
Use this premium calculator to determine the pH of ammonium bromide solution from concentration and ammonia Kb, then review the chemistry behind every step.
NH4Br pH Calculator
Ammonium bromide is a salt of a weak base (NH3) and a strong acid (HBr). The bromide ion is neutral in water, while NH4+ acts as a weak acid.
Expert Guide: How to Calculate the pH of 0.15 M NH4Br Solution
When chemistry students are asked to calculate the pH of a 0.15 M NH4Br solution, the most important skill is recognizing what kind of salt is present. NH4Br, or ammonium bromide, is formed from ammonium ion, NH4+, and bromide ion, Br-. Bromide comes from hydrobromic acid, a strong acid, so it does not significantly hydrolyze in water. Ammonium, however, is the conjugate acid of ammonia, NH3, which is a weak base. That means the solution is acidic, not neutral.
This distinction matters because many salt pH problems can look similar at first glance. Some salts are neutral, some are acidic, and others are basic. The pH of NH4Br is controlled almost entirely by the weak acidity of NH4+. Once you identify that, the rest of the process follows a familiar weak acid equilibrium approach.
Why NH4Br Makes an Acidic Solution
In water, NH4Br dissociates essentially completely:
NH4Br → NH4+ + Br-
The bromide ion is the conjugate base of a strong acid, so it has negligible basicity in water. The ammonium ion is different. It donates a proton to water according to:
NH4+ + H2O ⇌ NH3 + H3O+
This reaction generates hydronium ion, H3O+, and lowers the pH below 7. Because NH4+ is a weak acid, the equilibrium lies mostly to the left, but even a small amount of proton donation is enough to make the solution measurably acidic.
Step 1: Start with the Known Base Constant of Ammonia
The standard tabulated base dissociation constant for ammonia at 25 degrees C is commonly taken as:
- Kb(NH3) = 1.8 × 10-5
Since NH4+ is the conjugate acid of NH3, its acid dissociation constant is found using the conjugate relationship:
Ka × Kb = Kw
At 25 degrees C, the ionic product of water is:
- Kw = 1.0 × 10-14
So:
Ka(NH4+) = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
Step 2: Set Up the Weak Acid Equilibrium
If the NH4Br concentration is 0.15 M, then the initial ammonium ion concentration is also 0.15 M because the salt dissociates completely.
Let x be the concentration of H3O+ formed at equilibrium. Then the ICE setup is:
- Initial: [NH4+] = 0.15, [NH3] = 0, [H3O+] = 0
- Change: [NH4+] = -x, [NH3] = +x, [H3O+] = +x
- Equilibrium: [NH4+] = 0.15 – x, [NH3] = x, [H3O+] = x
Substitute into the acid expression:
Ka = [NH3][H3O+] / [NH4+] = x2 / (0.15 – x)
Step 3: Solve for x
Because Ka is small and the concentration is relatively large, the weak acid approximation is valid. That means 0.15 – x is essentially 0.15 for routine classroom work. Then:
x2 / 0.15 = 5.56 × 10-10
x2 = 8.34 × 10-11
x = 9.13 × 10-6 M
Since x = [H3O+], the pH becomes:
pH = -log(9.13 × 10-6) = 5.04
Exact Method Versus Approximation
For stronger weak acids or much lower concentrations, it is sometimes better to solve the full quadratic equation. Here the approximation is excellent because the ionization is tiny compared with the initial concentration. The percent ionization is:
(9.13 × 10-6 / 0.15) × 100 = 0.0061%
That is far below 5%, so the approximation clearly holds.
| Property | Value Used | Meaning | Effect on pH |
|---|---|---|---|
| NH4Br concentration | 0.15 M | Initial concentration of NH4+ | Higher concentration generally lowers pH slightly for this acidic salt |
| Kb of NH3 | 1.8 × 10-5 | Base strength of ammonia | Larger Kb means smaller Ka for NH4+, which raises pH |
| Kw | 1.0 × 10-14 | Water autoionization constant at 25 degrees C | Needed to convert Kb to Ka |
| Ka of NH4+ | 5.56 × 10-10 | Acid strength of ammonium | Directly determines how much H3O+ forms |
| Calculated [H3O+] | 9.13 × 10-6 M | Equilibrium hydronium concentration | Converts directly to pH 5.04 |
How NH4Br Compares with Other Salt Solutions
A very common source of mistakes is treating all ionic compounds as if they give neutral solutions. In reality, the pH depends on whether the ions are conjugates of strong or weak acids and bases. The table below compares ammonium bromide with other familiar salts.
| Salt | Parent Acid | Parent Base | Expected Solution Type | Typical pH Behavior |
|---|---|---|---|---|
| NH4Br | HBr, strong acid | NH3, weak base | Acidic | Below 7 because NH4+ hydrolyzes |
| NaCl | HCl, strong acid | NaOH, strong base | Neutral | Near 7 under standard conditions |
| CH3COONa | CH3COOH, weak acid | NaOH, strong base | Basic | Above 7 because acetate hydrolyzes |
| NH4Cl | HCl, strong acid | NH3, weak base | Acidic | Very similar logic to NH4Br |
| Na2CO3 | H2CO3, weak acid | NaOH, strong base | Basic | Often significantly above 7 |
Step-by-Step Shortcut for Exams
- Identify NH4Br as a salt from a weak base and a strong acid.
- Conclude the solution is acidic because NH4+ acts as a weak acid.
- Use Kb of NH3 to find Ka of NH4+ through Ka = Kw / Kb.
- Set initial NH4+ concentration equal to the salt molarity, 0.15 M.
- Apply the weak acid formula x = √(KaC) when justified.
- Compute pH = -log[H3O+].
Numerical Check at Different NH4Br Concentrations
The pH of ammonium bromide changes with concentration, but not in a linear way. Because the hydronium concentration depends on the square root of Ka times concentration for the weak acid approximation, doubling the concentration does not halve the pH. The trend is more gradual.
| NH4Br Concentration | Approximate [H3O+] | Approximate pH | Interpretation |
|---|---|---|---|
| 0.010 M | 2.36 × 10-6 M | 5.63 | Dilute solution, still acidic |
| 0.050 M | 5.27 × 10-6 M | 5.28 | Moderately acidic |
| 0.150 M | 9.13 × 10-6 M | 5.04 | Standard problem value |
| 0.500 M | 1.67 × 10-5 M | 4.78 | More concentrated, lower pH |
| 1.000 M | 2.36 × 10-5 M | 4.63 | Significantly acidic for an ammonium salt |
Common Mistakes Students Make
- Assuming NH4Br is neutral. It is not neutral because NH4+ is acidic.
- Using Kb directly for NH4+. NH4+ is an acid, so you need Ka, not Kb.
- Forgetting that Br- is a spectator ion. Bromide does not change the pH appreciably.
- Mixing up NH3 and NH4+. NH3 accepts a proton; NH4+ donates a proton.
- Skipping the 5% rule check. In this case the approximation is valid, but it should still be justified.
Why the pH Is Not Extremely Low
Even though the solution is acidic, the pH is only around 5.04 rather than 1 or 2. That is because NH4+ is a weak acid. Its Ka is only 5.56 × 10-10, which means only a tiny fraction of ammonium ions actually donate a proton to water. By contrast, a strong acid would ionize nearly completely and produce far more hydronium ion.
Real-World Relevance of This Calculation
Weak acid and weak base salt calculations matter in analytical chemistry, environmental chemistry, biochemistry, and chemical process work. Ammonium salts appear in buffer systems, fertilizer chemistry, atmospheric deposition studies, and laboratory reagent preparation. Understanding the pH of ammonium-containing solutions helps predict corrosion, reaction pathways, ammonia-ammonium speciation, and biological compatibility.
If you want authoritative background reading on pH and aqueous equilibria, the following resources are useful:
- USGS: pH and Water
- Purdue University: Weak Acid Equilibrium Concepts
- University of Wisconsin: Acid-Base Equilibrium Tutorial
When Temperature Changes the Answer
The values of Kw and equilibrium constants change with temperature. In general chemistry problems, 25 degrees C is usually assumed unless the problem states otherwise. If temperature changes, you should use the corresponding Kw and, if available, the updated Kb or Ka values. The calculator above allows you to change both constants so you can model nonstandard conditions more accurately.
Summary
To calculate the pH of 0.15 M NH4Br solution, identify NH4Br as a salt of a weak base and strong acid. Treat NH4+ as a weak acid, compute its Ka from the known Kb of ammonia, and solve the equilibrium. Using Kb = 1.8 × 10-5 and Kw = 1.0 × 10-14, the resulting Ka is 5.56 × 10-10. For a 0.15 M solution, the hydronium concentration is approximately 9.13 × 10-6 M, giving a pH of about 5.04. This is the standard textbook result and a great example of how conjugate acid-base relationships control the pH of salt solutions.