Calculate the pH of 0.29 m NH4Br Solution
Use this premium calculator to find the pH of ammonium bromide solution by treating NH4+ as a weak acid. The default example is 0.29 m, which is commonly approximated as 0.29 M in general chemistry pH problems.
Calculator Inputs
NH4+ + H2O ⇌ NH3 + H3O+
Ka(NH4+) = Kw / Kb(NH3)
Results
Click Calculate pH to compute the exact pH, hydronium concentration, hydroxide concentration, and ammonium dissociation for the selected input values.
How to calculate the pH of 0.29 m NH4Br solution
To calculate the pH of a 0.29 m NH4Br solution, the key idea is that ammonium bromide is a salt made from a weak base and a strong acid. The weak base is ammonia, NH3, and the strong acid is hydrobromic acid, HBr. Once NH4Br dissolves in water, it separates almost completely into NH4+ and Br-. The bromide ion is essentially neutral in water because it is the conjugate base of a strong acid. The ammonium ion, however, acts as a weak acid and donates protons to water to form hydronium ions.
That means the pH does not come from bromide at all. It comes from the weak acid equilibrium of NH4+. This is why chemistry students often solve this exact problem by finding the acid dissociation constant of NH4+ from the base dissociation constant of NH3. At 25 C, a frequently used value is Kb for NH3 = 1.8 × 10^-5. Since Kw = 1.0 × 10^-14, we can compute Ka for NH4+ = Kw / Kb = 5.56 × 10^-10.
NH4+ + H2O ⇌ NH3 + H3O+
Step 2: Find Ka
Ka = Kw / Kb = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10
Step 3: Let x = [H3O+]
Ka = x^2 / (0.29 – x)
Because Ka is very small and the concentration is much larger than the amount dissociated, we usually make the weak acid approximation 0.29 – x ≈ 0.29. Then:
x ≈ √((5.56 × 10^-10)(0.29))
x ≈ 1.27 × 10^-5 M
pH = -log(1.27 × 10^-5)
pH ≈ 4.90
If you use the exact quadratic method instead of the approximation, the answer changes negligibly for this case. The pH remains about 4.895 to 4.896 depending on rounding. So the default calculator result is the same answer most instructors expect in a standard aqueous equilibrium problem.
Why NH4Br gives an acidic solution
Many learners first see NH4Br and wonder whether it should be neutral because it is a salt. The important correction is that not all salts are neutral. Salt behavior depends on the acid and base that formed them:
- Strong acid + strong base generally gives a neutral salt.
- Strong acid + weak base gives an acidic salt.
- Weak acid + strong base gives a basic salt.
- Weak acid + weak base requires comparing Ka and Kb.
NH4Br falls into the strong acid + weak base category. HBr is strong and NH3 is weak. Therefore, NH4+ hydrolyzes in water and lowers the pH below 7. This is exactly why a 0.29 m NH4Br solution is acidic rather than neutral.
Does the lowercase m matter?
Strictly speaking, lowercase m usually means molality, not molarity. In many introductory chemistry exercises, especially quick pH practice sets, people often write concentration casually and then solve as if it were molarity. For a moderately dilute aqueous solution, using 0.29 m as approximately 0.29 M generally gives a very similar classroom answer. If you need highly rigorous thermodynamic precision, you would account for density and activity corrections. For most instructional purposes, this problem is solved with concentration treated as 0.29 M in the equilibrium expression.
Exact chemistry steps for the 0.29 NH4Br example
- Assume NH4Br dissociates fully: NH4Br → NH4+ + Br-.
- Set the initial ammonium concentration equal to the listed salt concentration, 0.29.
- Use the conjugate relationship between NH4+ and NH3.
- Calculate Ka of NH4+ from Kw / Kb.
- Set up an ICE table for NH4+ + H2O ⇌ NH3 + H3O+.
- Solve for x, where x is the hydronium concentration generated by hydrolysis.
- Convert x to pH using pH = -log[H3O+].
Here is the equilibrium setup in compact form:
Change: -x, +x, +x
Equilibrium: [NH4+] = 0.29 – x, [NH3] = x, [H3O+] = x
Ka = x^2 / (0.29 – x)
Using the approximation:
pH = -log(1.27 × 10^-5) = 4.895
This answer passes the 5 percent rule comfortably because x is tiny compared with 0.29. In fact, the percent dissociation is only around 0.0044 percent, showing that the weak acid approximation is fully justified here.
Important constants and benchmark values
The following table summarizes the numerical values commonly used in general chemistry for this calculation. These are the values behind the pH estimate of approximately 4.90.
| Quantity | Typical value at 25 C | Meaning for the problem |
|---|---|---|
| Kb of NH3 | 1.8 × 10^-5 | Base strength of ammonia |
| Kw of water | 1.0 × 10^-14 | Used to convert Kb to Ka |
| Ka of NH4+ | 5.56 × 10^-10 | Weak acid strength of ammonium ion |
| pKa of NH4+ | 9.25 | Shows NH4+ is a weak acid |
| Input concentration | 0.29 | Initial NH4+ concentration from NH4Br |
| Calculated [H3O+] | 1.27 × 10^-5 M | Hydronium generated by hydrolysis |
| Calculated pH | 4.895 | Final result |
Comparison table: how concentration affects pH of NH4Br
One useful way to understand the calculation is to compare several NH4Br concentrations using the same acid-base constant set. Since NH4+ is a weak acid, increasing concentration lowers pH, but not in a one-to-one linear way. The hydronium concentration depends on the square root of Ka × C when the weak acid approximation applies.
| NH4Br concentration | Approximate [H3O+] | Approximate pH | Interpretation |
|---|---|---|---|
| 0.010 | 2.36 × 10^-6 M | 5.63 | Mildly acidic |
| 0.050 | 5.27 × 10^-6 M | 5.28 | Acidic but still weakly hydrolyzed |
| 0.100 | 7.45 × 10^-6 M | 5.13 | Common textbook benchmark |
| 0.290 | 1.27 × 10^-5 M | 4.90 | The target example in this page |
| 0.500 | 1.67 × 10^-5 M | 4.78 | More acidic due to higher NH4+ |
| 1.000 | 2.36 × 10^-5 M | 4.63 | Still weakly acidic, not strongly acidic |
Common mistakes when solving NH4Br pH problems
1. Treating NH4Br as a strong acid
NH4Br is not a strong acid itself. It is a salt whose cation, NH4+, acts as a weak acid. If someone assumes full proton donation like HCl or HBr, they will massively overestimate the hydronium concentration and get a pH that is far too low.
2. Using Kb directly instead of converting to Ka
The species in solution that matters is NH4+, not NH3. Therefore, use Ka for NH4+, not Kb for NH3, unless you are explicitly converting with Ka = Kw / Kb.
3. Forgetting that Br- is neutral
Bromide is the conjugate base of HBr, a strong acid. Conjugate bases of strong acids are negligibly basic in water. This means Br- does not appreciably raise the pH.
4. Confusing molality and molarity
If a problem states 0.29 m, the most rigorous interpretation is molality. However, many pH exercises are written informally and solved using molarity. For dilute or moderate aqueous solutions in an educational setting, the difference is often small enough that instructors accept the conventional approximation. If your course emphasizes solution density and activity, ask whether a molality-to-molarity conversion is expected.
5. Ignoring the approximation check
Even when the weak acid shortcut looks reasonable, it is good chemistry practice to verify that x is much smaller than the initial concentration. In this example, it is. Since x / 0.29 × 100 ≈ 0.0044 percent, the approximation is excellent.
What the result means in practical terms
A pH of about 4.90 means the solution is definitely acidic, but not strongly acidic. It sits far above the pH of a strong acid solution of similar formal concentration. This result reflects the weak acidity of NH4+, which hydrolyzes only slightly. The equilibrium produces enough hydronium to lower the pH below 7, yet the vast majority of the ammonium remains undissociated with respect to its acid reaction.
That is an important conceptual distinction. The salt dissociates completely into ions, but the acidic reaction of NH4+ with water occurs only to a very small extent. In other words, NH4Br dissolves thoroughly, but NH4+ only partially reacts as an acid.
When you should use the exact quadratic method
For this problem, the approximation works beautifully. Still, a strong chemistry workflow includes the exact method when:
- The concentration is very low.
- The acid is not especially weak.
- Your instructor or lab report requires exact numerical treatment.
- You need to compare several methods for error analysis.
The exact equation comes from rearranging:
which becomes
x^2 + Ka x – Ka C = 0
x = [-Ka + √(Ka^2 + 4KaC)] / 2
Substituting Ka = 5.56 × 10^-10 and C = 0.29 gives nearly the same x value as the square-root shortcut. That is why this calculator lets you choose either method.
Authoritative references for acid-base constants and pH concepts
For deeper study, consult authoritative references such as the USGS explanation of pH and water, the NIST Chemistry WebBook, and Purdue University’s chemistry help pages on weak acid equilibrium calculations. These sources are useful for confirming equilibrium ideas, reviewing pH fundamentals, and checking accepted chemistry constants and methods.
Final answer
If you are solving the standard textbook problem calculate the pH of 0.29 m NH4Br solution, the expected answer at 25 C is:
This result comes from recognizing NH4+ as a weak acid, calculating its Ka from the Kb of NH3, and solving the weak acid equilibrium for hydronium concentration. If you want to explore how the answer changes with concentration, Kb values, or calculation method, use the calculator above.