Calculate The Ph Of A 0.25 M Nh4Br Solution

Use this premium acid-base calculator to determine the pH of ammonium bromide solution with either the exact quadratic method or the common weak-acid approximation. NH4Br is an acidic salt because NH4+ donates protons in water, while Br- is essentially neutral.

How to calculate the pH of a 0.25 M NH4Br solution

To calculate the pH of a 0.25 M NH4Br solution, you first identify what kind of salt ammonium bromide is. NH4Br is formed from NH3, a weak base, and HBr, a strong acid. In water, Br- does not appreciably react with water, so it is effectively a spectator ion. The chemically important species is NH4+, the conjugate acid of ammonia. Because NH4+ can donate a proton to water, an NH4Br solution is acidic.

The hydrolysis reaction is: NH4+ + H2O ⇌ NH3 + H3O+

Once you recognize NH4+ as a weak acid, the pH calculation becomes a standard weak-acid equilibrium problem. You need the acid dissociation constant of NH4+, but this is usually found indirectly from the base dissociation constant of NH3 using the relationship: Ka x Kb = Kw

At 25 C, a common textbook value for ammonia is Kb = 1.8 x 10^-5. Therefore: Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10

For a 0.25 M solution, set the initial weak-acid concentration equal to 0.25 M. If x represents the amount of NH4+ that dissociates, then:

• [NH4+] initial = 0.25
• [NH3] initial = 0
• [H3O+] initial is negligible compared with the acid contribution

At equilibrium:

• [NH4+] = 0.25 – x
• [NH3] = x
• [H3O+] = x

Insert these values into the weak-acid expression: Ka = x^2 / (0.25 – x)

Because Ka is very small, many courses allow the approximation 0.25 – x ≈ 0.25. Then: x ≈ sqrt(Ka x C) = sqrt((5.56 x 10^-10)(0.25)) = 1.18 x 10^-5 M

Since x = [H3O+], the pH is: pH = -log(1.18 x 10^-5) = 4.93

The exact quadratic solution gives essentially the same answer to the displayed precision. So the pH of a 0.25 M NH4Br solution is about 4.93.

Why NH4Br is acidic in water

Students often memorize categories of salts without understanding the reason behind them. A better approach is to ask what each ion does in water. Bromide is the conjugate base of the strong acid HBr, so it has almost no tendency to accept a proton from water. Ammonium, however, is the conjugate acid of ammonia, and ammonia is a weak base. Conjugate acids of weak bases have measurable acidity. That is why NH4Br lowers pH below 7.

This is also why NH4Br behaves very differently from salts such as NaBr or KBr. Sodium and potassium ions are spectators from strong bases, and bromide is the conjugate base of a strong acid. When all ions come from strong acid and strong base parents, the solution is approximately neutral. NH4Br is different because one ion, NH4+, is chemically active as a weak acid.

Quick decision rule for salt pH problems

1. Break the salt into ions.
2. Identify whether each ion comes from a strong or weak parent acid or base.
3. Ignore ions from strong parents unless told otherwise.
4. Write the hydrolysis reaction for any weakly acidic or weakly basic ion.
5. Use Ka or Kb, plus an ICE table, to solve for the equilibrium concentration.
6. Convert [H3O+] or [OH-] into pH.

Step-by-step worked example for 0.25 M NH4Br

1. Dissociate the salt

NH4Br(aq) → NH4+(aq) + Br-(aq)

The bromide ion is neutral in this context, so focus on ammonium.

2. Write the acid reaction

NH4+ + H2O ⇌ NH3 + H3O+

3. Convert Kb of NH3 into Ka of NH4+

If Kb(NH3) = 1.8 x 10^-5 and Kw = 1.0 x 10^-14:

Ka(NH4+) = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10

4. Set up the equilibrium expression

Let x = [H3O+] formed. Then:

Ka = x^2 / (0.25 – x)

5. Solve for x

Approximation method: x ≈ sqrt((5.56 x 10^-10)(0.25)) = 1.18 x 10^-5

Exact quadratic method: x = (-Ka + sqrt(Ka^2 + 4KaC)) / 2

With C = 0.25 and Ka = 5.56 x 10^-10, the value of x is again about 1.18 x 10^-5 M.

6. Convert to pH

pH = -log(1.18 x 10^-5) = 4.93

Final answer: the pH of a 0.25 M NH4Br solution at 25 C is approximately 4.93.

Comparison table: expected pH at different NH4Br concentrations

The acidity of ammonium bromide depends on concentration. Higher concentration generally means lower pH, though the change is not linear because pH is logarithmic and weak-acid equilibria follow square-root behavior over common classroom ranges.

NH4Br concentration (M)	Ka of NH4+	Approximate [H3O+] (M)	Approximate pH
0.010	5.56 x 10^-10	2.36 x 10^-6	5.63
0.050	5.56 x 10^-10	5.27 x 10^-6	5.28
0.100	5.56 x 10^-10	7.45 x 10^-6	5.13
0.250	5.56 x 10^-10	1.18 x 10^-5	4.93
0.500	5.56 x 10^-10	1.67 x 10^-5	4.78
1.000	5.56 x 10^-10	2.36 x 10^-5	4.63

Data table: acid-base constants relevant to NH4Br calculations

The numbers below are the core data used in introductory and general chemistry calculations involving ammonium salts. Small variations may appear across textbooks because reported constants can differ slightly with ionic strength, temperature, and rounding convention. The values below match common classroom use at 25 C.

Species	Chemical role	Typical constant at 25 C	Interpretation
NH3	Weak base	Kb = 1.8 x 10^-5	Ammonia partially reacts with water to form NH4+ and OH-.
NH4+	Weak acid	Ka = 5.56 x 10^-10	Ammonium partially donates a proton to water.
HBr	Strong acid parent of Br-	Essentially complete dissociation	Bromide is negligibly basic in water.
Kw	Ion product of water	1.0 x 10^-14	Connects Ka and Kb for conjugate acid-base pairs.

Common mistakes when solving NH4Br pH problems

• Treating NH4Br as neutral. It is not neutral because NH4+ is acidic.
• Using Kb directly for NH4+. NH4+ is an acid, so convert Kb of NH3 into Ka of NH4+ first.
• Forgetting that Br- is a spectator ion. Bromide does not significantly affect pH in this problem.
• Using the strong acid formula. NH4+ is a weak acid, so [H3O+] is not simply 0.25 M.
• Ignoring units and temperature assumptions. Most general chemistry values assume 25 C and mol/L.

Approximation vs exact solution

For weak-acid problems like NH4+, the approximation x << C usually works very well. You can check it by comparing x to the initial concentration. Here, x is about 1.18 x 10^-5 while C is 0.25. The fraction dissociated is only about 0.0047%, which is far below the 5% guideline commonly used in chemistry courses. That means the approximation is excellent.

Even so, the exact quadratic method is useful if you want full rigor or if your concentration becomes very small. This calculator includes both methods so you can compare them instantly and see why they converge so closely for 0.25 M NH4Br.

Interpretation of the result

A pH of 4.93 means the solution is mildly acidic, not strongly acidic. This makes sense because NH4+ is only a weak acid. It raises hydronium concentration above pure water, but nowhere near the level of a strong acid at the same formal concentration. In practical terms, NH4Br solutions can acidify aqueous media and may matter in equilibrium, analytical, and biological contexts, but they do not behave like concentrated mineral acids.

Authoritative references for acid-base data and pH concepts

• U.S. Environmental Protection Agency: pH overview and interpretation
• National Institute of Standards and Technology: pH standards and measurement resources
• University of Wisconsin chemistry resource on weak acids and equilibria

Bottom line

To calculate the pH of a 0.25 M NH4Br solution, recognize that ammonium bromide is an acidic salt. Convert the known Kb of ammonia to Ka for ammonium, set up the weak-acid equilibrium, solve for hydronium concentration, and then compute pH. Using Kb(NH3) = 1.8 x 10^-5 at 25 C gives Ka(NH4+) = 5.56 x 10^-10. For a 0.25 M NH4Br solution, the hydronium concentration is about 1.18 x 10^-5 M, giving a final pH of 4.93.