Calculate the pH of 0.40 M NH3 with Kb = 1.8 × 10-5
Use this premium weak-base calculator to find the equilibrium hydroxide concentration, pOH, pH, percent ionization, and equilibrium species concentrations for aqueous ammonia. The default values are set to the classic chemistry problem: 0.40 M NH3 and Kb = 1.8 × 10-5.
Kb = [NH4+][OH-] / [NH3]
Enter the starting molarity of ammonia.
For 1.8 × 10^-5, enter 1.8 here.
For 1.8 × 10^-5, enter -5 here.
Use the exact method for the most reliable answer.
How to calculate the pH of 0.40 M NH3 when Kb = 1.8 × 10-5
To calculate the pH of a 0.40 M ammonia solution, you need to recognize that NH3 is a weak base, not a strong base. That single idea drives the entire setup. Because ammonia only partially reacts with water, you cannot simply assume the hydroxide concentration equals the starting concentration of NH3. Instead, you use an equilibrium expression based on the base dissociation constant, Kb. For ammonia, a common textbook value is Kb = 1.8 × 10-5, which tells you that only a small fraction of NH3 molecules accept a proton from water.
The reaction is:
NH3 + H2O ⇌ NH4+ + OH-
This means every mole of NH3 that reacts forms one mole of NH4+ and one mole of OH-. Since pH is directly related to hydronium concentration, and pOH is directly related to hydroxide concentration, the practical path is to solve for [OH-], then calculate pOH, and finally convert to pH using the relationship pH + pOH = 14.00 at 25 degrees Celsius.
Step 1: Set up the ICE table
An ICE table is one of the cleanest ways to solve weak acid and weak base equilibrium problems. ICE stands for Initial, Change, and Equilibrium.
- Initial: [NH3] = 0.40 M, [NH4+] = 0, [OH-] = 0
- Change: [NH3] decreases by x, [NH4+] increases by x, [OH-] increases by x
- Equilibrium: [NH3] = 0.40 – x, [NH4+] = x, [OH-] = x
Now substitute the equilibrium concentrations into the Kb expression:
Kb = [NH4+][OH-] / [NH3] = x2 / (0.40 – x)
Since Kb = 1.8 × 10-5, you get:
1.8 × 10-5 = x2 / (0.40 – x)
Step 2: Solve for x, which equals [OH-]
You now have two common options. In a classroom, the approximation method is often taught first because it is fast. The exact method uses the quadratic equation and is slightly more accurate. Since ammonia ionizes only a little, both methods are close here.
- Approximation method: If x is small relative to 0.40, then 0.40 – x is approximately 0.40.
- Exact method: Solve the full quadratic equation x2 + Kb x – KbC = 0.
Using the approximation:
x = √(Kb × C) = √[(1.8 × 10-5)(0.40)] = √(7.2 × 10-6) ≈ 2.68 × 10-3 M
Using the exact quadratic equation:
x = [-Kb + √(Kb2 + 4KbC)] / 2
Substituting Kb = 1.8 × 10-5 and C = 0.40 gives:
x ≈ 2.67 × 10-3 M
This x value is the equilibrium hydroxide concentration, [OH-].
Step 3: Convert [OH-] to pOH and then to pH
Now apply the standard logarithmic relationship:
- pOH = -log[OH-]
- pH = 14.00 – pOH
With [OH-] ≈ 2.67 × 10-3 M:
pOH = -log(2.67 × 10-3) ≈ 2.57
pH = 14.00 – 2.57 ≈ 11.43
That is the standard answer chemistry students should expect for this problem.
Why this works for ammonia
Ammonia is a weak Brønsted-Lowry base. It accepts a proton from water, but not completely. A strong base like NaOH dissociates essentially 100%, so you can directly use stoichiometry. A weak base like NH3 needs equilibrium analysis because the extent of reaction is limited by its Kb value. The magnitude of Kb matters a great deal: the larger the Kb, the more OH- is generated and the higher the pH will be at the same initial concentration.
For ammonia, Kb = 1.8 × 10-5 is small enough to ensure partial ionization, but large enough that the resulting solution is still noticeably basic. That is why a 0.40 M NH3 solution has a pH over 11, even though it is not a strong base.
The 5% rule check
Whenever you use the small-x approximation, check whether it was valid. The 5% rule says that if x is less than 5% of the initial concentration, the approximation is acceptable.
Here, using the exact value x ≈ 0.00267 M:
% ionization = (0.00267 / 0.40) × 100 ≈ 0.67%
Since 0.67% is far below 5%, the approximation is excellent. That is why both the exact and approximate methods give nearly the same pH.
| Method | Calculated [OH-] (M) | pOH | pH | Difference from Exact |
|---|---|---|---|---|
| Exact quadratic | 0.002674 | 2.573 | 11.427 | 0.000 |
| Approximation, x = √(KbC) | 0.002683 | 2.571 | 11.429 | About 0.002 pH units |
| Percent ionization | 0.667% | Not applicable | Not applicable | Supports small-x validity |
Common mistakes when solving weak base pH problems
Students often lose points on weak-base questions for the same predictable reasons. Avoiding these errors can save time and improve accuracy.
- Treating NH3 like a strong base. You cannot set [OH-] = 0.40 M. That would produce an impossibly high pH for a weak base.
- Using Ka instead of Kb. The problem gives Kb because ammonia is acting as a base.
- Forgetting to calculate pOH first. Since the equilibrium gives [OH-], the next step is pOH, then pH.
- Skipping the equilibrium setup. Weak bases require an ICE table or an equivalent equilibrium framework.
- Rounding too early. Log calculations are sensitive to rounding. Carry extra digits until the final answer.
What if the concentration changed?
The pH of ammonia depends on both the starting concentration and the Kb value. If concentration increases, the hydroxide concentration increases, but not linearly. Because weak-base equilibrium usually leads to a square-root relationship under the approximation, a large concentration increase produces a smaller than expected shift in pH.
| Weak Base | Typical Kb at 25 C | Approximate pH at 0.40 M | Relative Basic Strength |
|---|---|---|---|
| Ammonia, NH3 | 1.8 × 10^-5 | 11.43 | Moderate weak base |
| Pyridine, C5H5N | 1.7 × 10^-9 | 9.92 | Much weaker than NH3 |
| Methylamine, CH3NH2 | 4.4 × 10^-4 | 12.12 | Stronger weak base than NH3 |
This comparison helps place ammonia in context. Ammonia is not among the strongest weak bases, but it is significantly more basic than species like pyridine. At the same 0.40 M concentration, its pH falls in a distinctly basic range because its Kb is several orders of magnitude larger.
Detailed worked solution for 0.40 M NH3
1. Write the balanced reaction
NH3 + H2O ⇌ NH4+ + OH-
2. Define the equilibrium variable
Let x = [OH-] formed at equilibrium. Then [NH4+] = x and [NH3] = 0.40 – x.
3. Substitute into the Kb expression
Kb = x2 / (0.40 – x) = 1.8 × 10-5
4. Solve the equation
Multiply both sides:
x2 = (1.8 × 10-5)(0.40 – x)
x2 + 1.8 × 10-5x – 7.2 × 10-6 = 0
Apply the quadratic formula:
x = [-1.8 × 10-5 + √((1.8 × 10-5)2 + 4(7.2 × 10-6))] / 2
The physically meaningful positive root is approximately:
x = 2.674 × 10-3 M
5. Find pOH
pOH = -log(2.674 × 10-3) = 2.573
6. Find pH
pH = 14.000 – 2.573 = 11.427
Rounded appropriately, the answer is pH = 11.43.
Interpreting the chemistry behind the answer
A pH of 11.43 tells you the solution is basic, but it also tells you something more subtle: ammonia is only partially protonated. The overwhelming majority of dissolved NH3 molecules remain as NH3 at equilibrium, while only a small fraction becomes NH4+. This is exactly what you should expect from a weak base with a relatively small Kb. The measured basicity comes from the OH- formed, yet that OH- concentration is still only a few thousandths of a molar, much smaller than the initial 0.40 M NH3 concentration.
This also explains why percent ionization remains under 1%. Strong bases achieve their basicity through nearly complete dissociation. Weak bases achieve it through an equilibrium that strongly favors the unreacted form. In ammonia’s case, that balance still generates enough OH- to raise the pH substantially.
When to use Kb versus Ka
If the species is acting as a base and you are given Kb, use the base reaction with water and solve directly for OH-. If instead you are given Ka for the conjugate acid NH4+, you could convert between the constants using:
Ka × Kb = Kw = 1.0 × 10-14
For ammonia and ammonium, this relationship is especially useful in buffer problems. However, for this standalone NH3 solution, working directly with Kb is the most efficient route.
Authoritative chemistry references
If you want to verify ammonia properties, review acid-base theory, or consult broader chemical data, these sources are useful starting points:
- NIST Chemistry WebBook on ammonia
- U.S. Environmental Protection Agency ammonia resources
- CDC NIOSH ammonia information
Final answer summary
For the problem “calculate the pH of 0.40 M NH3, Kb = 1.8 × 10-5,” the correct setup is a weak-base equilibrium calculation. Using the ammonia reaction with water, an ICE table, and the Kb expression, the equilibrium hydroxide concentration is about 2.67 × 10-3 M. That gives a pOH of about 2.57 and a final pH of about 11.43.
This problem is a model example of weak-base chemistry because it reinforces several essential concepts at once: partial ionization, equilibrium expressions, logarithmic pH calculations, and approximation validation with the 5% rule. If you can solve this ammonia question confidently, you are well prepared for many similar equilibrium problems involving weak bases in general chemistry.