Calculate the pH of 0.32 M NH4Br Solution
Use this premium calculator to find the pH of ammonium bromide solutions by applying salt hydrolysis, acid-base equilibrium, and the relationship between Kb of NH3 and Ka of NH4+.
NH4Br pH Calculator
Ammonium bromide is a salt of a weak base and a strong acid, so its aqueous solution is acidic because NH4+ donates protons to water. Enter your values below.
How to calculate the pH of 0.32 M NH4Br solution
To calculate the pH of 0.32 M NH4Br solution, you need to recognize what kind of salt ammonium bromide is. NH4Br is formed from ammonium hydroxide or, more accurately, ammonia as the weak base component, and hydrobromic acid as the strong acid component. In water, the bromide ion behaves as a spectator ion because it is the conjugate base of a strong acid. The ammonium ion, however, is the conjugate acid of the weak base NH3. That means the chemistry of the solution is controlled by NH4+, which hydrolyzes according to the equilibrium NH4+ + H2O ⇌ NH3 + H3O+.
This is the key reason a 0.32 M NH4Br solution is acidic. Many students mistakenly expect a salt solution to be neutral, but salts from strong acids and weak bases do not produce a pH of 7. Instead, the conjugate acid cation reacts with water and generates hydronium. Once you identify NH4+ as a weak acid, the calculation becomes a standard weak acid equilibrium problem. The central steps are to find Ka for NH4+, solve for the hydronium concentration, and then convert that concentration into pH.
Step 1: Identify the species that controls pH
When NH4Br dissolves, it dissociates essentially completely:
NH4Br → NH4+ + Br-
The bromide ion has negligible effect on pH because it is the conjugate base of HBr, a strong acid. The ammonium ion does affect pH because it can act as a weak acid. Therefore, for a 0.32 M NH4Br solution, the initial concentration of NH4+ is 0.32 M. This concentration becomes the starting acid concentration in the equilibrium setup.
Step 2: Relate Ka of NH4+ to Kb of NH3
In many textbook and exam problems, you are given the base dissociation constant of ammonia rather than the acid dissociation constant of ammonium. At 25 C, a commonly used value is:
- Kb(NH3) = 1.8 × 10^-5
- Kw = 1.0 × 10^-14
Use the conjugate acid-base relationship:
Ka(NH4+) = Kw / Kb(NH3)
Substituting the standard values gives:
Ka = (1.0 × 10^-14) / (1.8 × 10^-5) = 5.56 × 10^-10
This very small Ka tells you ammonium is a weak acid, but because the concentration is fairly high at 0.32 M, the resulting solution is still measurably acidic.
Step 3: Set up the weak acid equilibrium
Now write the acid ionization expression for ammonium:
NH4+ + H2O ⇌ NH3 + H3O+
Initial concentrations:
- [NH4+] = 0.32 M
- [NH3] = 0
- [H3O+] = 0 from the salt, ignoring pure water initially
Let x be the amount of NH4+ that ionizes:
- [NH4+] = 0.32 – x
- [NH3] = x
- [H3O+] = x
The equilibrium expression is:
Ka = x^2 / (0.32 – x)
Substitute Ka = 5.56 × 10^-10:
5.56 × 10^-10 = x^2 / (0.32 – x)
Step 4: Solve for hydronium concentration
Because Ka is very small compared with the starting concentration, the weak acid approximation works extremely well. Under that approximation, 0.32 – x is treated as approximately 0.32. Then:
x^2 = (5.56 × 10^-10)(0.32)
x^2 = 1.779 × 10^-10
x = 1.33 × 10^-5 M
Since x = [H3O+], the hydronium concentration is approximately 1.33 × 10^-5 M. Now convert to pH:
pH = -log10(1.33 × 10^-5) = 4.88
So the pH of 0.32 M NH4Br solution is approximately 4.88 at 25 C.
Exact solution versus approximation
If you solve the quadratic equation exactly, the value changes only in the fifth or sixth decimal place because x is tiny relative to 0.32. That is why chemistry instructors often accept the approximation method for this problem. The exact form is:
x = (-Ka + √(Ka^2 + 4KaC)) / 2
With C = 0.32 M and Ka = 5.56 × 10^-10, the exact value of x remains essentially 1.33 × 10^-5 M, leading to the same practical pH of 4.88.
Why NH4Br is not neutral
This topic often causes confusion because the compound contains both a cation and an anion, and students may think they cancel. In acid-base chemistry, the origin of each ion matters. Bromide comes from HBr, a strong acid, and therefore has negligible basicity in water. Ammonium comes from ammonia, a weak base, and is acidic. Since one ion is acid-active and the other is essentially neutral, the solution becomes acidic overall. This is a classic example of salt hydrolysis.
| Quantity | Typical value at 25 C | Why it matters |
|---|---|---|
| Kb of NH3 | 1.8 × 10^-5 | Used to derive Ka of NH4+ |
| Kw of water | 1.0 × 10^-14 | Connects Ka and Kb for conjugate pairs |
| Ka of NH4+ | 5.56 × 10^-10 | Determines how much NH4+ releases H3O+ |
| Initial [NH4+] | 0.32 M | Sets the weak acid concentration |
| [H3O+] in solution | 1.33 × 10^-5 M | Directly converted into pH |
| Final pH | 4.88 | Shows the solution is acidic |
How concentration changes the pH of NH4Br solutions
The pH of NH4Br depends strongly on concentration. A more concentrated ammonium bromide solution contains more NH4+, which pushes the equilibrium toward greater hydronium production. However, because NH4+ is a weak acid, the pH does not change linearly with concentration. It changes logarithmically through the acid equilibrium relationship. This is why doubling the concentration does not halve the pH, and why a high-concentration salt solution may still have only moderate acidity compared with a strong acid.
| NH4Br concentration (M) | Approximate [H3O+] (M) | Approximate pH at 25 C | Percent ionization of NH4+ |
|---|---|---|---|
| 0.010 | 2.36 × 10^-6 | 5.63 | 0.0236% |
| 0.050 | 5.27 × 10^-6 | 5.28 | 0.0105% |
| 0.100 | 7.45 × 10^-6 | 5.13 | 0.0075% |
| 0.320 | 1.33 × 10^-5 | 4.88 | 0.0042% |
| 0.500 | 1.67 × 10^-5 | 4.78 | 0.0033% |
| 1.000 | 2.36 × 10^-5 | 4.63 | 0.0024% |
Common mistakes when solving this problem
- Treating NH4Br as a neutral salt. It is not neutral because NH4+ is acidic.
- Using the bromide ion in the pH calculation. Br- is essentially a spectator ion for acid-base purposes.
- Using Kb directly instead of converting to Ka. For pH of the salt solution, you need the acid behavior of NH4+.
- Forgetting the negative sign in pH. pH is the negative logarithm of hydronium concentration.
- Mixing strong acid formulas with weak acid formulas. NH4+ is weakly acidic, so use equilibrium, not complete dissociation.
When the approximation is valid
The 5% rule is a quick check. If x divided by the initial concentration is less than 5%, the approximation is acceptable. Here, x is about 1.33 × 10^-5 and the initial concentration is 0.32 M. The fraction is roughly 4.16 × 10^-5, or 0.0042%, which is far below 5%. That means the approximation is excellent. In real coursework, either the quadratic or the approximation leads to the same reported pH for this system.
Temperature and equilibrium effects
At temperatures other than 25 C, the value of Kw changes, which slightly changes Ka for NH4+ when you derive it from Kb. In introductory chemistry, problems are usually assumed to be at 25 C unless otherwise stated. If your instructor provides a different value of Kw or Kb, use those values. The calculator above lets you compare a few common temperatures to show how pH shifts with changing equilibrium constants.
Practical interpretation of the result
A pH of 4.88 means the solution is acidic but not strongly acidic. That makes sense chemically. NH4+ is a weak acid, so it does not dissociate extensively, yet at 0.32 M there are enough ammonium ions present to produce a measurable amount of hydronium. This kind of solution is common in laboratory discussions of buffer chemistry, salt hydrolysis, and acid-base conjugate pairs. Understanding why NH4Br gives an acidic pH also helps you classify other salts quickly. For example, NaCl is neutral, NH4Cl is acidic, and CH3COONa is basic.
Short answer summary
If you need the direct result only, here it is:
- NH4Br dissociates into NH4+ and Br-
- NH4+ is a weak acid, Br- is neutral
- Ka(NH4+) = 1.0 × 10^-14 / 1.8 × 10^-5 = 5.56 × 10^-10
- [H3O+] ≈ √(Ka × 0.32) = 1.33 × 10^-5 M
- pH ≈ 4.88
Authoritative references for pH, ammonia chemistry, and acid-base background
For deeper study, review these authoritative educational and government resources: