Calculate The Ph Of 0.1 M Honh2

Use this interactive weak-base calculator to estimate the pH of hydroxylamine solution, NH2OH or HONH2. Enter concentration, base dissociation constant, and calculation method to get pH, pOH, hydroxide concentration, and percent ionization with a live chart.

Hydroxylamine pH Calculator

Default values are set for a 0.1 M HONH2 solution at 25 C using a typical literature Kb value near 9.1 × 10^-9.

Formula used for a weak base: Kb = [BH⁺][OH^–] / [B]. For HONH2 in water, solve for x = [OH^–], then compute pOH = -log10(x) and pH = 14.00 – pOH at 25 C.

How to calculate the pH of 0.1 M HONH2 correctly

To calculate the pH of 0.1 M HONH2, you treat hydroxylamine as a weak base in water. HONH2 is commonly written as NH2OH, and in aqueous solution it accepts a proton from water to form its conjugate acid, NH3OH⁺, while producing OH^–. Because hydroxylamine does not fully ionize, the pH cannot be found by assuming complete dissociation the way you would for a strong base such as NaOH. Instead, you use the weak-base equilibrium expression and either solve exactly with a quadratic equation or use a weak-base approximation when the ionization is small.

The key reaction is:

HONH2 + H2O ⇌ HONH3⁺ + OH^–

If the initial concentration of HONH2 is 0.1 M and the base dissociation constant Kb is approximately 9.1 × 10^-9 at 25 C, then the equilibrium setup is straightforward. Let x represent the amount of HONH2 that reacts. At equilibrium, the concentration of OH^– is x, the concentration of HONH3⁺ is also x, and the remaining concentration of HONH2 is 0.1 – x.

The equilibrium expression becomes:

Kb = x² / (0.1 – x)

Substitute the Kb value:

9.1 × 10^-9 = x² / (0.1 – x)

Since Kb is very small, x will be much smaller than 0.1, so the approximation method gives:

x ≈ √(Kb × C) = √(9.1 × 10^-9 × 0.1) = √(9.1 × 10^-10) ≈ 3.02 × 10^-5 M

This means:

• [OH^–] ≈ 3.02 × 10^-5 M
• pOH = -log10(3.02 × 10^-5) ≈ 4.52
• pH = 14.00 – 4.52 = 9.48

So the pH of 0.1 M HONH2 is about 9.48 under standard 25 C conditions when using a typical Kb value near 9.1 × 10^-9. If your chemistry text or laboratory manual uses a slightly different Kb, such as 8.7 × 10^-9 or 1.1 × 10^-8, your final pH may vary by a few hundredths of a pH unit. That is normal and reflects source-to-source differences in tabulated equilibrium constants.

Why HONH2 gives a basic solution

Hydroxylamine is a weak Brønsted-Lowry base because the nitrogen atom can accept a proton. In water, that proton transfer is incomplete, so only a small fraction of dissolved HONH2 molecules produce hydroxide ions. Even though the ionization is limited, the OH^– that forms is enough to shift the solution above neutral pH. This is why a 0.1 M solution of hydroxylamine has a pH above 7 but nowhere near the pH of a 0.1 M strong base.

A useful intuition check is to compare weak and strong bases at the same concentration. If 0.1 M NaOH were present, the hydroxide concentration would be about 0.1 M and the pOH would be 1, giving a pH near 13. In contrast, 0.1 M HONH2 only generates hydroxide on the order of 10^-5 M, so the pH is much lower, around 9.5. That huge difference comes from incomplete ionization.

Solution	Concentration	Approximate [OH-]	Approximate pOH	Approximate pH at 25 C
HONH2, weak base	0.1 M	3.02 × 10^-5 M	4.52	9.48
NaOH, strong base	0.1 M	1.00 × 10^-1 M	1.00	13.00
Neutral water	Pure water	1.00 × 10^-7 M	7.00	7.00

Exact method versus approximation method

Students are often taught the square-root shortcut for weak acids and weak bases, but it is worth knowing when that method is valid. For weak bases, the approximation assumes x is so small relative to the initial concentration C that C – x is effectively just C. A common validation rule is the 5 percent rule. After calculating x, divide it by the initial concentration and convert to a percent. If the ionization is less than 5 percent, the approximation is usually considered acceptable.

For 0.1 M HONH2:

percent ionization = (3.02 × 10^-5 / 0.1) × 100 ≈ 0.030 percent

That is far below 5 percent, so the approximation is excellent here. The exact quadratic method still gives nearly the same answer, making it a good confirmation tool. In this calculator, you can toggle between the exact solution and the approximation to see how closely they agree.

1. Write the equilibrium reaction for HONH2 in water.
2. Construct an ICE table with initial, change, and equilibrium concentrations.
3. Use Kb = x² / (C – x).
4. Solve for x exactly or use the square-root approximation.
5. Calculate pOH from x.
6. Convert pOH to pH using pH = 14.00 – pOH at 25 C.

Common equilibrium data used for hydroxylamine

One reason chemistry problems involving hydroxylamine can seem confusing is that sources may list either the Kb of HONH2 or the Ka of its conjugate acid, HONH3⁺. These are related by the water ion product:

Ka × Kb = Kw = 1.0 × 10^-14 at 25 C

If you are given Ka for the conjugate acid instead of Kb for the base, convert it using:

Kb = Kw / Ka

Parameter	Typical value at 25 C	Meaning	How it affects pH
Kb for HONH2	About 8.7 × 10^-9 to 1.1 × 10^-8	Base strength of hydroxylamine	Higher Kb gives a slightly higher pH
Ka for HONH3^+	About 9.1 × 10^-7 to 1.15 × 10^-6	Acid strength of the conjugate acid	Higher Ka gives a slightly lower Kb
Kw	1.0 × 10^-14	Water autoionization constant	Connects pH, pOH, Ka, and Kb
pKw	14.00	Used to convert pOH to pH at 25 C	pH = 14.00 – pOH

Detailed worked example for 0.1 M HONH2

Let us work through the full setup one more time in a clean problem-solving format.

• Given: C = 0.1 M
• Given: Kb = 9.1 × 10^-9
• Find: pH

Step 1: Write the equilibrium.
HONH2 + H2O ⇌ HONH3⁺ + OH^–

Step 2: Build the equilibrium concentrations.
Initial: [HONH2] = 0.1, [HONH3⁺] = 0, [OH^–] = 0
Change: -x, +x, +x
Equilibrium: 0.1 – x, x, x

Step 3: Apply Kb.
9.1 × 10^-9 = x² / (0.1 – x)

Step 4: Solve for x.
Approximation gives x ≈ 3.02 × 10^-5 M

Step 5: Convert to pOH.
pOH = -log10(3.02 × 10^-5) ≈ 4.52

Step 6: Convert to pH.
pH = 14.00 – 4.52 = 9.48

This is the answer most instructors expect unless a different Kb is specified. If your assignment says to calculate the pH of 0.1 M HONH2 without giving Kb, check your textbook table or instructor notes. The pH is still in the same neighborhood, usually around 9.47 to 9.52 depending on the constant used.

Frequent mistakes to avoid

• Do not treat HONH2 like a strong base. It does not dissociate completely.
• Do not set [OH^–] equal to 0.1 M. That would massively overestimate the pH.
• Make sure you use Kb, not Ka, unless you convert properly.
• Remember that x represents OH^–, so calculate pOH first and then convert to pH.
• At 25 C, use pH + pOH = 14.00. If your course uses another temperature, pKw can change.

Why source data can differ slightly

Equilibrium constants can vary a bit between references because of rounding, ionic strength assumptions, and the exact way data were measured and reported. In introductory chemistry, this usually leads to very small differences in the final pH. For a problem like 0.1 M HONH2, that means your answer might be 9.47, 9.48, or 9.50 depending on the reference table. These are all chemically consistent when based on valid data.

Reliable chemistry references and educational sources

For general acid-base concepts, equilibrium calculations, and water chemistry, consult authoritative educational and government resources such as:

• LibreTexts Chemistry for educational explanations and worked equilibrium methods
• U.S. Environmental Protection Agency on pH for fundamental pH concepts and definitions
• Michigan State University acid-base tutorial for equilibrium background and pH relationships

If you need reference values for constants, always prioritize the equilibrium data table used by your course or laboratory. Then apply the same calculation framework shown above. With that approach, calculating the pH of 0.1 M HONH2 becomes a clean weak-base problem instead of a memorization task.

Final takeaway: for a typical 25 C problem using Kb = 9.1 × 10^-9, the pH of 0.1 M HONH2 is approximately 9.48.