Calculate the pH of 0.025 M NH3 with Kb = 1.8 × 10-5
Use this premium weak-base calculator to solve ammonia equilibrium accurately with either the quadratic method or the common square-root approximation. It computes hydroxide concentration, pOH, pH, percent ionization, and equilibrium concentrations.
Results
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How to calculate the pH of 0.025 M NH3 when Kb = 1.8 × 10-5
To calculate the pH of a 0.025 M ammonia solution, you treat ammonia, NH3, as a weak base in water. Unlike a strong base such as sodium hydroxide, ammonia does not ionize completely. Instead, it establishes an equilibrium with water:
NH3 + H2O ⇌ NH4+ + OH-
The base dissociation constant, Kb, tells you how far this equilibrium lies to the right. In this problem, Kb = 1.8 × 10-5, which means ammonia is a weak base, but still basic enough to produce a measurable concentration of hydroxide ions. Once the hydroxide concentration is known, the pOH is found from the negative logarithm, and the pH is then calculated from pH + pOH = 14.00 at 25°C.
Step 1: Write the ICE table
Start with the reaction:
NH3 + H2O ⇌ NH4+ + OH-
The water is a pure liquid and is omitted from the equilibrium expression. For an initial ammonia concentration of 0.025 M, the ICE table is:
- Initial: [NH3] = 0.025, [NH4+] = 0, [OH-] = 0
- Change: [NH3] = -x, [NH4+] = +x, [OH-] = +x
- Equilibrium: [NH3] = 0.025 – x, [NH4+] = x, [OH-] = x
Step 2: Apply the Kb expression
The equilibrium expression for ammonia is:
Kb = [NH4+][OH-] / [NH3]
Substitute the ICE table values:
1.8 × 10-5 = x2 / (0.025 – x)
At this point, you can either use the approximation method or solve the quadratic exactly. Because Kb is small and the concentration is much larger than the expected ionization, the approximation usually works well. Still, the exact method is preferred for a premium calculator because it avoids unnecessary error.
Step 3: Solve for x, the hydroxide concentration
Rearranging gives:
x2 + (1.8 × 10-5)x – (1.8 × 10-5)(0.025) = 0
Solving this quadratic yields:
x = [OH-] ≈ 6.62 × 10-4 M
That means the equilibrium hydroxide ion concentration in the ammonia solution is about 0.000662 M.
Step 4: Calculate pOH
pOH = -log[OH-]
pOH = -log(6.62 × 10-4) ≈ 3.18
Step 5: Convert pOH to pH
At 25°C:
pH = 14.00 – 3.18 = 10.82
Therefore, the pH of 0.025 M NH3 with Kb = 1.8 × 10-5 is approximately 10.82.
Why ammonia does not behave like a strong base
This question is common in general chemistry because students often compare ammonia to compounds such as NaOH or KOH. Strong bases dissociate essentially 100% in dilute water. If 0.025 M NaOH were dissolved, the hydroxide concentration would also be 0.025 M, giving a pOH of about 1.60 and a pH of about 12.40. Ammonia is very different. Since NH3 is a weak base, only a small fraction reacts with water to form NH4+ and OH-. That is why the pH is only 10.82 instead of above 12.
The exact magnitude of the difference depends on Kb. For ammonia, the Kb value of 1.8 × 10-5 indicates moderate basicity among weak bases, but still far weaker than complete hydroxide release from strong bases.
| Base / Solution | Initial Concentration (M) | Base Strength Data | Approximate [OH-] (M) | Approximate pH at 25°C |
|---|---|---|---|---|
| NH3 | 0.025 | Kb = 1.8 × 10^-5 | 6.62 × 10^-4 | 10.82 |
| NaOH | 0.025 | Strong base, essentially complete dissociation | 2.50 × 10^-2 | 12.40 |
| NH3 | 0.100 | Kb = 1.8 × 10^-5 | 1.33 × 10^-3 | 11.12 |
| NH3 | 0.010 | Kb = 1.8 × 10^-5 | 4.15 × 10^-4 | 10.62 |
Approximation versus exact quadratic solution
In many textbook problems, you may see the approximation:
Kb = x2 / 0.025
which comes from assuming that x is small enough that 0.025 – x ≈ 0.025. Then:
x = √(Kb × C) = √((1.8 × 10-5)(0.025)) ≈ 6.71 × 10-4 M
This produces a pOH of about 3.17 and a pH of about 10.83. That is very close to the exact answer, because the percent ionization is only a few percent. The 5% rule is commonly used to judge whether the approximation is acceptable. Here, x / 0.025 × 100% is about 2.65%, so the approximation is valid.
| Method | [OH-] (M) | pOH | pH | Difference from Exact pH |
|---|---|---|---|---|
| Exact quadratic | 6.62 × 10^-4 | 3.18 | 10.82 | 0.00 |
| Square-root approximation | 6.71 × 10^-4 | 3.17 | 10.83 | About 0.01 pH unit |
Common mistakes students make on this problem
- Using Ka instead of Kb. Since NH3 is a base, the problem must be solved with Kb directly, not Ka. If you only had Ka for NH4+, then you could convert using KaKb = Kw.
- Assuming complete dissociation. Writing [OH-] = 0.025 M would treat ammonia as a strong base, which is incorrect.
- Forgetting to calculate pOH first. The species produced by the base reaction is OH-, so pOH is computed before pH.
- Dropping x without checking. The approximation is usually fine here, but in weaker dilution ranges or for larger Kb values, the exact quadratic can matter.
- Using the wrong logarithm sign. pOH is negative log of hydroxide concentration, not positive log.
What the result means chemically
A pH of 10.82 means the solution is distinctly basic, but not strongly caustic in the way a comparable concentration of sodium hydroxide would be. The equilibrium heavily favors unreacted NH3 over NH4+ and OH-, so most of the ammonia remains in molecular form. This is typical behavior for weak bases: only a small percentage ionizes, yet enough hydroxide is formed to move the pH well above neutral.
The percent ionization can be estimated from:
percent ionization = x / initial concentration × 100%
Using x ≈ 6.62 × 10-4 M and initial concentration 0.025 M:
percent ionization ≈ 2.65%
This means about 97.35% of the ammonia remains as NH3 at equilibrium. In other words, only a small portion accepts protons from water to form ammonium and hydroxide.
Reference data and authoritative chemistry resources
If you want to verify weak-base chemistry principles, acid-base equilibria, and standard water ion-product assumptions, these authoritative sources are useful:
- LibreTexts Chemistry for equilibrium, ICE tables, and weak base treatment
- National Institute of Standards and Technology (NIST) for scientific constants and measurement standards
- U.S. Environmental Protection Agency (EPA) for ammonia-related chemistry and water quality context
You can also consult university chemistry departments and open educational resources on weak acids and weak bases. Good examples include course notes from .edu domains and chemistry education libraries.
Fast exam shortcut for this exact problem
If you are under time pressure, the fastest route is:
- Write NH3 + H2O ⇌ NH4+ + OH-
- Use x = √(Kb × C)
- x = √((1.8 × 10^-5)(0.025)) ≈ 6.7 × 10^-4
- pOH = -log(6.7 × 10^-4) ≈ 3.17
- pH = 14.00 – 3.17 = 10.83
Then, if your instructor wants exactness, you can mention that the quadratic gives 10.82, and because the percent ionization is below 5%, the approximation is justified.
Final summary
To calculate the pH of 0.025 M NH3 with Kb = 1.8 × 10-5, set up the weak base equilibrium, define x as the hydroxide concentration, and solve the expression Kb = x2 / (0.025 – x). The exact quadratic result gives [OH-] ≈ 6.62 × 10-4 M, pOH ≈ 3.18, and pH ≈ 10.82. The square-root approximation also works well and produces nearly the same answer. This problem is a classic example of how weak bases differ from strong bases: the solution is clearly basic, but only a small fraction of the dissolved base ionizes.