Calculate the pH of a 0.20 M NH3 Solution
Use this calculator to determine the pH, pOH, hydroxide concentration, ammonium concentration, and percent ionization for aqueous ammonia. By default, the tool uses the accepted 25 C base dissociation constant for NH3, Kb = 1.8 × 10-5.
Equilibrium expression: Kb = [NH4+][OH–] / [NH3]
Exact setup: Kb = x2 / (C – x)
Equilibrium Snapshot
The chart compares the starting ammonia concentration with the calculated hydroxide concentration, ammonium concentration, and remaining ammonia at equilibrium.
How to calculate the pH of 0.20 M NH3 solution
Calculating the pH of a 0.20 M NH3 solution is a classic weak base equilibrium problem from general chemistry. Ammonia, NH3, is not a strong base, so it does not react completely with water. Instead, only a small fraction of dissolved NH3 accepts a proton from water to form NH4+ and OH-. Because the hydroxide concentration is generated by equilibrium rather than full dissociation, the solution must be solved using the base dissociation constant Kb.
At 25 C, ammonia has a commonly used Kb value of 1.8 × 10-5. The relevant chemical equation is:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)
If the initial ammonia concentration is 0.20 M, we usually start with an ICE table:
- Initial: [NH3] = 0.20, [NH4+] = 0, [OH-] = 0
- Change: [NH3] decreases by x, [NH4+] increases by x, [OH-] increases by x
- Equilibrium: [NH3] = 0.20 – x, [NH4+] = x, [OH-] = x
The equilibrium expression becomes:
Kb = x2 / (0.20 – x) = 1.8 × 10-5
For a weak base like NH3, the value of x is much smaller than 0.20, so many textbooks permit the approximation 0.20 – x ≈ 0.20. That gives:
x2 / 0.20 = 1.8 × 10-5
Solving for x:
x2 = 3.6 × 10-6
x = 1.897 × 10-3 M
Since x represents [OH-], the hydroxide concentration is approximately 1.90 × 10-3 M. Next:
pOH = -log(1.897 × 10-3) ≈ 2.72
Then use:
pH = 14.00 – 2.72 = 11.28
Therefore, the pH of a 0.20 M NH3 solution at 25 C is approximately 11.28. If you solve the quadratic expression exactly, you get essentially the same answer, confirming that the approximation is valid for this concentration and Kb value.
Step by step expert method
1. Recognize that NH3 is a weak base
This is the most important first step. If NH3 were a strong base, you would use direct stoichiometry to find [OH-]. But ammonia only partially ionizes in water, so equilibrium must be considered. This immediately tells you to use Kb, not a simple one line dissociation assumption.
2. Write the equilibrium reaction correctly
Ammonia accepts a proton from water:
- NH3 is the base
- H2O acts as the acid
- NH4+ is the conjugate acid
- OH- is the conjugate base formed in solution
Students often confuse this with NH3 producing OH- directly the way NaOH would. That is not the correct chemical picture. OH- forms only because NH3 establishes an equilibrium with water.
3. Build an ICE table
The ICE method keeps the algebra organized and prevents sign errors. For 0.20 M NH3:
- Initial NH3 concentration is 0.20 M
- Initial NH4+ and OH- are taken as approximately 0 from the weak base itself
- At equilibrium, NH3 decreases by x while NH4+ and OH- each increase by x
This leads to the equilibrium concentrations:
- [NH3] = 0.20 – x
- [NH4+] = x
- [OH-] = x
4. Substitute into the Kb expression
For ammonia:
Kb = [NH4+][OH-] / [NH3]
Substituting the ICE expressions:
1.8 × 10-5 = x2 / (0.20 – x)
You can solve this by approximation or by the exact quadratic formula. The calculator above lets you choose either method.
5. Check whether the approximation is valid
The common 5 percent rule says the approximation is acceptable if x is less than 5 percent of the initial concentration. Here, x ≈ 0.00190 M.
Percent ionization = (0.00190 / 0.20) × 100 ≈ 0.95%
Since 0.95 percent is well below 5 percent, the approximation is excellent.
Exact numerical result for 0.20 M NH3
Using the exact quadratic form:
x = [-Kb + √(Kb2 + 4KbC)] / 2
With C = 0.20 and Kb = 1.8 × 10-5:
- [OH-] = x ≈ 0.001888 M
- [NH4+] ≈ 0.001888 M
- [NH3] remaining ≈ 0.198112 M
- pOH ≈ 2.724
- pH ≈ 11.276
- Percent ionization ≈ 0.944%
Rounded appropriately for most classroom and lab contexts, the final answer is pH = 11.28.
Comparison table: NH3 concentration vs pH at 25 C
The table below uses Kb = 1.8 × 10-5 and exact weak base equilibrium calculations at 25 C. These values show how pH rises as the starting ammonia concentration increases.
| Initial NH3 concentration (M) | Calculated [OH-] (M) | pOH | pH | Percent ionization |
|---|---|---|---|---|
| 0.010 | 4.15 × 10-4 | 3.38 | 10.62 | 4.15% |
| 0.050 | 9.40 × 10-4 | 3.03 | 10.97 | 1.88% |
| 0.100 | 1.33 × 10-3 | 2.88 | 11.12 | 1.33% |
| 0.200 | 1.89 × 10-3 | 2.72 | 11.28 | 0.94% |
| 0.500 | 2.99 × 10-3 | 2.52 | 11.48 | 0.60% |
Comparison table: NH3 and other common weak bases
Ammonia is an important benchmark weak base in chemistry because its Kb is moderate, well known, and used frequently in introductory equilibrium calculations. The values below are standard textbook scale comparisons often used in academic chemistry courses.
| Weak base | Representative Kb at 25 C | Relative basic strength | Typical note |
|---|---|---|---|
| Ammonia, NH3 | 1.8 × 10-5 | Moderate weak base | Standard teaching example for weak base pH |
| Methylamine, CH3NH2 | 4.4 × 10-4 | Stronger than NH3 | Organic amines often have higher Kb values than NH3 |
| Pyridine, C5H5N | 1.7 × 10-9 | Much weaker than NH3 | Aromatic stabilization reduces basicity |
Why the answer is not 13 or 14
A common mistake is to treat 0.20 M NH3 as if it were 0.20 M OH-. If that were true, the pOH would be 0.70 and the pH would be 13.30. But that reasoning applies to strong bases such as NaOH, KOH, or Ba(OH)2 with appropriate stoichiometry. Ammonia is weak, so only a small fraction generates OH-. In fact, less than 1 percent of the 0.20 M NH3 ionizes under these conditions. That is why the pH is about 11.28 instead of something close to 13.
Most common mistakes in NH3 pH calculations
- Using Ka instead of Kb for ammonia
- Assuming complete dissociation like a strong base
- Forgetting to calculate pOH first from [OH-]
- Using pH = -log[OH-] instead of pOH = -log[OH-]
- Neglecting the 14.00 relationship at 25 C
- Rounding too early, which can shift the final pH by a few hundredths
- Applying the x is small approximation without checking the 5 percent rule
How this applies in laboratory and industrial settings
Ammonia solutions are important in analytical chemistry, environmental chemistry, and industrial formulations. In the laboratory, NH3 and NH4+ make an important conjugate acid-base pair that appears in buffer systems and coordination chemistry. In water treatment and environmental monitoring, understanding the equilibrium behavior of dissolved ammonia matters because pH affects ammonia speciation, toxicity, and treatment outcomes. The 0.20 M example is more concentrated than many natural waters, but the same equilibrium principles apply.
In practice, chemists also consider temperature, ionic strength, and activity effects in more advanced calculations. However, for standard educational problems and many routine calculations, using Kb = 1.8 × 10-5 at 25 C with ideal solution assumptions gives a solid and accepted answer.
Quick summary of the full solution
- Write the equilibrium reaction for NH3 in water.
- Set up an ICE table with initial concentration 0.20 M.
- Use Kb = 1.8 × 10-5.
- Solve Kb = x2 / (0.20 – x).
- Find [OH-] = x ≈ 1.89 × 10-3 M.
- Compute pOH ≈ 2.72.
- Compute pH = 14.00 – 2.72 = 11.28.
Authoritative chemistry references
For further validation and deeper reading, review these authoritative sources:
- NIST Chemistry WebBook on ammonia properties
- U.S. Environmental Protection Agency overview of pH in water
- University of Wisconsin acid-base equilibrium instructional resource
Final answer
Using Kb = 1.8 × 10-5 at 25 C, the pH of a 0.20 M NH3 solution is approximately 11.28. The solution is basic because ammonia accepts protons from water and generates OH-, but the pH is far below that of a strong base of the same formal concentration because ammonia ionizes only partially.