Calculate the pH of a Solution of 1.5×10-5 M NH4OH
Use this interactive calculator to determine the pH, pOH, and hydroxide concentration for a dilute ammonium hydroxide solution. The tool applies the weak-base equilibrium expression using an exact quadratic solution for better accuracy at low concentration.
Calculator
Default values represent 1.5×10-5 M NH4OH with Kb = 1.8×10-5 at 25°C.
Results
Click Calculate pH to compute the pH of the solution.
Expert Guide: How to Calculate the pH of a Solution of 1.5×10-5 M NH4OH
When chemistry students are asked to calculate the pH of a solution of 1.5×10-5 M NH4OH, they are really solving a weak-base equilibrium problem. Ammonium hydroxide is commonly used in introductory chemistry problems as a representation of aqueous ammonia, a weak base that only partially ionizes in water. Because it does not dissociate completely, you cannot treat it the same way as a strong base such as NaOH or KOH. Instead, you must use the base dissociation constant, Kb, together with the initial concentration.
For this problem, the concentration is very small, only 1.5×10-5 M, which means the ionization is not negligible relative to the starting concentration. That is exactly why using an exact equilibrium method is better than blindly applying a shortcut. In this guide, you will learn the chemistry, the math, the assumptions, and the best way to interpret the final answer.
What reaction should be used?
The weak-base equilibrium for aqueous ammonia is typically written as:
NH3 + H2O ⇌ NH4+ + OH–
Many textbook and homework problems write the base as NH4OH, although modern chemistry courses often describe the species more accurately as dissolved NH3. In practice, when a problem gives “NH4OH” and asks for pH, it expects you to use the weak-base constant for ammonia in water. At 25°C, a commonly used value is:
Kb = 1.8×10-5
Step-by-step calculation for 1.5×10-5 M NH4OH
- Write the equilibrium expression:
Kb = [NH4+][OH–] / [NH3] - Let the initial concentration of the weak base be C = 1.5×10-5 M.
- Let x be the amount ionized. Then:
- [NH4+] = x
- [OH–] = x
- [NH3] = C – x
- Substitute into the Kb expression:
1.8×10-5 = x2 / (1.5×10-5 – x) - Rearrange into quadratic form:
x2 + Kbx – KbC = 0 - Insert values:
x2 + 1.8×10-5x – 2.7×10-10 = 0 - Solve with the quadratic formula:
x = [-Kb + √(Kb2 + 4KbC)] / 2 - This gives:
x ≈ 9.73×10-6 M - Since x = [OH–], calculate pOH:
pOH = -log(9.73×10-6) ≈ 5.01 - Then calculate pH:
pH = 14.00 – 5.01 = 8.99
Why the weak-base shortcut is not ideal here
A common approximation for weak bases is:
x = √(KbC)
This works best when x is much smaller than the initial concentration, usually when ionization is under about 5 percent. Here:
√(1.8×10-5 × 1.5×10-5) = √(2.7×10-10) ≈ 1.64×10-5
That result is actually larger than the initial concentration of the base, which is physically impossible. This tells you immediately that the approximation fails. The exact quadratic method is the right approach because the concentration is low and the base is not weak enough relative to the concentration to justify neglecting x in the denominator.
Comparison table: approximation versus exact solution
| Method | Formula Used | Calculated [OH-] | pOH | pH | Comment |
|---|---|---|---|---|---|
| Exact quadratic | x = [-Kb + √(Kb2 + 4KbC)] / 2 | 9.73×10-6 M | 5.01 | 8.99 | Correct and chemically valid |
| Shortcut approximation | x = √(KbC) | 1.64×10-5 M | 4.78 | 9.22 | Invalid because x exceeds the initial concentration |
What does the pH value mean?
A pH of about 8.99 means the solution is basic, but not strongly basic. Because NH4OH is a weak base and the concentration is very dilute, the hydroxide concentration produced is modest. This is an important concept in acid-base chemistry: the identity of the solute matters, but so does concentration. A weak base at low concentration may produce a pH only slightly above neutral.
Students sometimes assume any base must have a pH near 11 or 12, but that is only true for stronger bases or more concentrated solutions. In this case, the solution is just mildly basic.
Percent ionization for this solution
Percent ionization helps you understand how much of the base actually reacts with water. It is calculated as:
Percent ionization = (x / C) × 100
For this problem:
(9.73×10-6 / 1.5×10-5) × 100 ≈ 64.9%
That is extremely high for a weak-base textbook approximation, and it confirms why the “small x” assumption is not appropriate. High percent ionization often appears when the initial concentration is very low.
Data table: pH trend for NH4OH at different concentrations
| Initial NH4OH Concentration | Kb Used | Calculated [OH-] (Exact) | pOH | pH at 25°C |
|---|---|---|---|---|
| 1.0×10-2 M | 1.8×10-5 | 4.15×10-4 M | 3.38 | 10.62 |
| 1.0×10-3 M | 1.8×10-5 | 1.25×10-4 M | 3.90 | 10.10 |
| 1.0×10-4 M | 1.8×10-5 | 3.46×10-5 M | 4.46 | 9.54 |
| 1.5×10-5 M | 1.8×10-5 | 9.73×10-6 M | 5.01 | 8.99 |
Common mistakes students make
- Treating NH4OH as a strong base. It is not fully dissociated, so using [OH–] = 1.5×10-5 directly is incorrect.
- Using the shortcut without checking validity. If x is not much smaller than C, the approximation can fail badly.
- Confusing pH and pOH. You must calculate pOH first from hydroxide concentration, then convert to pH.
- Ignoring significant figures. Since the concentration and Kb each have about two significant figures, the final pH is usually reported as 8.99 or 9.0 depending on your class rules.
- Using the wrong equilibrium constant. Be sure you use Kb for ammonia, not Ka for ammonium.
Does autoionization of water matter here?
Pure water contributes 1.0×10-7 M OH– at 25°C. In this problem, the calculated hydroxide concentration from the base is around 9.73×10-6 M, which is much larger than 1.0×10-7 M. So the contribution from water is relatively small and can usually be neglected in a standard general chemistry calculation. However, for extremely dilute acids and bases, water autoionization can become more important and may require a more advanced treatment.
How this problem fits into equilibrium chemistry
This calculation is a classic example of the relationship between equilibrium constants and solution composition. The larger the Kb, the more a base reacts with water to form OH–. The larger the initial concentration, the more total OH– can be generated. But when concentrations become very small, the degree of ionization rises and approximation methods can break down. That is why equilibrium chemistry often requires both conceptual understanding and careful algebra.
The NH4OH problem also reinforces an important pattern:
- Strong base problems are usually stoichiometric first and simple.
- Weak base problems require equilibrium setup.
- Dilute weak base problems often require an exact quadratic solution.
Authoritative references for acid-base constants and water chemistry
For trusted chemistry data and educational references, consult:
National Institute of Standards and Technology (NIST)
LibreTexts Chemistry, hosted by academic institutions
U.S. Environmental Protection Agency (EPA)
Practical conclusion
If you need to calculate the pH of a solution of 1.5×10-5 M NH4OH, the best method is to use the weak-base equilibrium expression and solve the quadratic exactly. With Kb = 1.8×10-5, the hydroxide concentration is about 9.73×10-6 M, the pOH is about 5.01, and the pH is about 8.99. This answer makes chemical sense: the solution is basic, but only mildly so because the concentration is extremely dilute.
As a rule, whenever the weak-base concentration is close in magnitude to the base dissociation constant, pause before using the shortcut formula. Check the assumptions, solve carefully, and make sure the result is physically reasonable. That habit will improve both your exam accuracy and your understanding of acid-base equilibrium.