Calculate the pH of a 0.10 M Solution of N₂H₄
Use this premium weak-base calculator to find the pH, pOH, hydroxide concentration, percent ionization, and equilibrium values for hydrazine in water. The default setup is preloaded for a 0.10 M N₂H₄ solution.
Hydrazine pH Calculator
Hydrazine, N₂H₄, is a weak base. This tool uses the base dissociation expression and solves the equilibrium with the quadratic formula for high accuracy.
Expert Guide: How to Calculate the pH of a 0.10 M Solution of N₂H₄
To calculate the pH of a 0.10 M solution of N₂H₄, you treat hydrazine as a weak Brønsted base in water. That means it does not react completely the way NaOH would. Instead, only a small fraction of dissolved hydrazine molecules accept a proton from water, producing hydrazinium ions and hydroxide ions. Because pH depends directly on the hydroxide concentration generated at equilibrium, the entire problem is an equilibrium problem rather than a simple stoichiometry problem.
The reaction is:
N₂H₄ + H₂O ⇌ N₂H₅+ + OH–
For this equilibrium, the weak-base dissociation constant is commonly taken as Kb = 1.3 × 10-6 at 25°C in many general chemistry settings. Since the starting concentration is 0.10 M, the solution is basic but not strongly basic. The pH ends up a little above 10.5, not close to 13 or 14. That is the key conceptual point students often miss: hydrazine is a weak base, so it generates far less OH– than a strong base at the same formal concentration.
Step 1: Write the equilibrium expression
Let the initial concentration of hydrazine be C = 0.10 M. Let x be the amount of OH– produced at equilibrium. Then the ICE setup is:
- Initial: [N₂H₄] = 0.10, [N₂H₅+] = 0, [OH–] = 0
- Change: [N₂H₄] = -x, [N₂H₅+] = +x, [OH–] = +x
- Equilibrium: [N₂H₄] = 0.10 – x, [N₂H₅+] = x, [OH–] = x
Substitute into the base equilibrium expression:
Kb = x² / (0.10 – x)
Using Kb = 1.3 × 10-6:
1.3 × 10-6 = x² / (0.10 – x)
Step 2: Solve for x, the hydroxide concentration
You can solve this two ways. In many classrooms, the approximation method is used first because x is much smaller than 0.10. If x is small enough, then 0.10 – x is approximately 0.10. That gives:
x ≈ √(Kb × C)
x ≈ √((1.3 × 10-6)(0.10))
x ≈ √(1.3 × 10-7) ≈ 3.61 × 10-4 M
Since x represents [OH–], you now know the hydroxide concentration. For a more exact answer, solve the quadratic:
x² + (1.3 × 10-6)x – (1.3 × 10-7) = 0
The positive root is approximately:
x = 3.60 × 10-4 M
The approximation and the exact result are extremely close, which tells you the weak-base approximation is valid here.
Step 3: Convert hydroxide concentration to pOH
Use the definition of pOH:
pOH = -log[OH–]
pOH = -log(3.60 × 10-4) ≈ 3.44
Step 4: Convert pOH to pH
At 25°C, use:
pH + pOH = 14.00
So:
pH = 14.00 – 3.44 = 10.56
Why the answer is not extremely high
Students often see a 0.10 M base and expect a pH near 13. That would be true only for a strong base such as NaOH, where nearly every formula unit contributes hydroxide directly. Hydrazine does not behave that way. It is a weak base, which means equilibrium strongly favors the reactant side. Only a small percentage of N₂H₄ molecules convert into N₂H₅+ and OH–. In this example, the percent ionization is only about 0.36%.
Percent ionization of hydrazine at 0.10 M
Percent ionization is a useful check on reasonableness:
% ionization = (x / C) × 100
% ionization = (3.60 × 10-4 / 0.10) × 100 ≈ 0.36%
Because this is far below 5%, the approximation 0.10 – x ≈ 0.10 is justified. This is one of the standard criteria for deciding whether the square-root shortcut is acceptable.
Exact calculation workflow in ordered steps
- Write the weak-base equilibrium reaction for N₂H₄ in water.
- Set up an ICE table with initial concentration 0.10 M and change variable x.
- Substitute equilibrium concentrations into Kb = [N₂H₅+][OH–] / [N₂H₄].
- Solve x² / (0.10 – x) = 1.3 × 10-6 for x.
- Use x as [OH–] and compute pOH = -log x.
- Use pH = 14.00 – pOH at 25°C.
Comparison table: hydrazine versus stronger and weaker common bases
The table below helps put hydrazine into context. Kb values are standard reference-scale values used in chemistry instruction, and the pH values for 0.10 M solutions are estimated at 25°C.
| Base | Approximate Kb at 25°C | Estimated pH of 0.10 M solution | Interpretation |
|---|---|---|---|
| Hydrazine, N₂H₄ | 1.3 × 10-6 | 10.56 | Weak base that produces modest OH– |
| Ammonia, NH₃ | 1.8 × 10-5 | 11.13 | Stronger weak base than hydrazine |
| Aniline, C₆H₅NH₂ | 4.3 × 10-10 | 8.82 | Much weaker base with far less OH– |
| Sodium hydroxide, NaOH | Strong base | 13.00 | Nearly complete dissociation |
How pH changes as hydrazine concentration changes
Another helpful way to understand the calculation is to compare several starting concentrations while holding Kb fixed. The trend is clear: as concentration increases, pH rises, but because hydrazine is weak, the increase is gradual and not linear with concentration.
| Initial N₂H₄ concentration | Approximate [OH–] | Approximate pOH | Approximate pH at 25°C |
|---|---|---|---|
| 0.0010 M | 3.54 × 10-5 M | 4.45 | 9.55 |
| 0.010 M | 1.13 × 10-4 M | 3.95 | 10.05 |
| 0.10 M | 3.60 × 10-4 M | 3.44 | 10.56 |
| 1.00 M | 1.14 × 10-3 M | 2.94 | 11.06 |
Common mistakes when solving this problem
- Treating N₂H₄ as a strong base. If you simply assume [OH–] = 0.10 M, you would get pH = 13, which is far too high.
- Using Ka instead of Kb. Since hydrazine acts as a base here, Kb is the relevant equilibrium constant.
- Forgetting to convert from pOH to pH. Once you find [OH–], the first logarithmic quantity is pOH, not pH.
- Ignoring temperature assumptions. At temperatures other than 25°C, pKw is not exactly 14.00.
- Dropping the quadratic without checking percent ionization. The approximation is fine here, but you should know why it is fine.
When should you use the quadratic formula?
For many weak acid and weak base problems, the square-root approximation is accepted if percent ionization is under 5%. In this specific problem, the approximation works very well because x is tiny compared with 0.10. However, if the concentration were much smaller, or if the equilibrium constant were larger, the approximation could become poor. The quadratic formula is always the more rigorous route and is preferred in calculators, software, and high-precision work.
Why molarity matters more than the label alone
The phrase “0.10 M hydrazine” carries the whole concentration term in the equilibrium expression. A different concentration changes the denominator and therefore changes the amount of ionization. This is why pH cannot be discussed for hydrazine without its concentration. A 0.0010 M hydrazine solution has a much lower pH than a 1.00 M hydrazine solution, even though both contain the same chemical base.
Safety and reference context
Hydrazine is not just a classroom molecule. It is an industrial and aerospace chemical with significant safety concerns, and it should be handled only under proper laboratory or regulated industrial conditions. For toxicological and safety information, consult official government resources rather than informal summaries. For chemical identity and compound data, official chemistry databases and educational chemistry departments are the most reliable places to verify nomenclature, formula, and broader context.
Useful authoritative references include:
- PubChem, National Institutes of Health: Hydrazine compound record
- CDC NIOSH: Hydrazine safety and occupational information
- University of Wisconsin chemistry tutorial on acid-base concepts
Bottom line
If you are asked to calculate the pH of a 0.10 M solution of N₂H₄, the correct chemistry approach is to use weak-base equilibrium. Start with the hydrazine hydrolysis reaction, write the Kb expression, solve for hydroxide concentration, calculate pOH, and then convert to pH. Using a typical textbook value of Kb = 1.3 × 10-6, the final answer is:
pH ≈ 10.56
That result is chemically reasonable, mathematically consistent, and aligned with the expected behavior of a weak base at moderate concentration.