Calculate The Ph Of A 0.50 M Solution Of Nano2

This interactive calculator estimates the pH of sodium nitrite solution by treating nitrite as the conjugate base of nitrous acid and solving the hydrolysis equilibrium. Default values are set for a 0.50 m NaNO2 solution at 25 degrees Celsius.

NaNO2 pH Calculator

Enter the concentration format, nitrous acid dissociation constant, and solution density if you want to convert from molality to molarity. The calculator then finds Kb, hydroxide concentration, pOH, and pH.

Expert Guide: How to Calculate the pH of a 0.50 m Solution of NaNO2

To calculate the pH of a 0.50 m solution of NaNO2, you need to recognize what sodium nitrite does in water. Sodium nitrite is a soluble ionic compound, so it dissociates essentially completely into Na+ and NO2-. The sodium ion is a spectator ion for acid-base purposes, but the nitrite ion is the conjugate base of nitrous acid, HNO2. Because HNO2 is a weak acid, its conjugate base is basic enough to react with water and produce hydroxide ions. That means the solution is basic, not neutral.

The heart of the problem is the hydrolysis equilibrium:

NO2- + H2O ⇌ HNO2 + OH-

This reaction tells you that nitrite ions accept a proton from water, making nitrous acid and hydroxide. The hydroxide produced determines the pOH and therefore the pH. The only equilibrium constant you usually look up directly is the acid dissociation constant for nitrous acid, Ka. Once you know Ka, you can calculate the base dissociation constant for nitrite using:

Kb = Kw / Ka

At 25 degrees Celsius, Kw is 1.0 × 10^-14. A common textbook value for Ka of HNO2 is about 4.5 × 10^-4, which corresponds to a pKa of about 3.35. Using those values gives:

Kb = (1.0 × 10^-14) / (4.5 × 10^-4) = 2.22 × 10^-11

Once Kb is known, the calculation follows the same pattern used for weak bases. If the prompt is written as 0.50 m, the lowercase m technically means molality, not molarity. In many classroom problems, students still treat it approximately like 0.50 M when the density is not given. A more careful treatment converts molality to molarity using the molar mass of NaNO2 and the density of the final solution. This calculator lets you do that. If you leave the density at 1.00 g/mL, the result is very close to the simpler classroom approximation.

Step-by-Step Chemistry Setup

1. Write the dissociation of sodium nitrite: NaNO2 → Na+ + NO2-
2. Identify NO2- as the conjugate base of the weak acid HNO2.
3. Write the base hydrolysis reaction: NO2- + H2O ⇌ HNO2 + OH-
4. Calculate Kb = Kw / Ka.
5. Use an ICE table with initial nitrite concentration C.
6. Solve for x = [OH-] either by the exact quadratic or the weak-base approximation.
7. Calculate pOH = -log[OH-] and then pH = 14.00 – pOH at 25 degrees Celsius.

Exact Calculation for a 0.50 m NaNO2 Solution

Let the initial nitrite concentration be approximately 0.50. For a weak base:

Kb = x^2 / (C – x)

where x = [OH-] formed at equilibrium.

Substitute values:

2.22 × 10^-11 = x^2 / (0.50 – x)

Because Kb is very small, x is much smaller than 0.50, so the approximation is excellent:

x ≈ √(KbC) = √((2.22 × 10^-11)(0.50)) = √(1.11 × 10^-11) ≈ 3.33 × 10^-6 M

Then:

pOH = -log(3.33 × 10^-6) ≈ 5.48

pH = 14.00 – 5.48 = 8.52

So the pH of a 0.50 m solution of NaNO2 is approximately 8.52 at 25 degrees Celsius when you use Ka = 4.5 × 10^-4 and assume the solution behaves close to 0.50 M. If you perform the exact quadratic, the answer changes only negligibly because x is tiny compared with the initial concentration.

Quantity	Symbol	Value Used	Meaning in the Calculation
Molality of sodium nitrite	m	0.50 m	Amount of NaNO2 per kilogram of solvent
Acid dissociation constant of HNO2	Ka	4.5 × 10^-4	Measures the strength of nitrous acid
Ion-product constant of water	Kw	1.0 × 10^-14	Used to convert Ka to Kb at 25 degrees Celsius
Base dissociation constant of NO2-	Kb	2.22 × 10^-11	Measures how strongly nitrite generates OH- in water
Hydroxide concentration	[OH-]	3.33 × 10^-6 M	Computed from the weak-base equilibrium
Final pH	pH	8.52	Basic solution because nitrite hydrolyzes water

Why NaNO2 Gives a Basic Solution

This is one of the most important conceptual points. Not every salt is neutral in water. A salt made from a strong base and a weak acid will usually make a basic solution. Sodium nitrite fits that pattern:

• NaOH is a strong base, so Na+ does not affect pH.
• HNO2 is a weak acid, so its conjugate base NO2- is basic.
• As NO2- reacts with water, it forms OH-, increasing pH above 7.

That pattern is reliable in general chemistry. If you can identify the parent acid and parent base, you can often predict whether a salt solution will be acidic, basic, or nearly neutral before doing any algebra.

Molality vs Molarity: Why the Prompt Matters

Students often overlook the notation. Lowercase m means molality, while uppercase M means molarity. Molality is defined as moles of solute per kilogram of solvent. Molarity is moles of solute per liter of solution. They are not exactly the same, although they can be numerically close in dilute aqueous solutions.

For a 0.50 m NaNO2 solution, if you take 1.000 kg of water, you have 0.50 mol NaNO2. Since the molar mass of sodium nitrite is about 69.00 g/mol, the solute mass is about 34.50 g. The total solution mass is then about 1034.50 g. If the solution density is roughly 1.00 g/mL, the volume is roughly 1034.50 mL or 1.0345 L. The corresponding molarity is:

M ≈ 0.50 / 1.0345 = 0.483 M

Using 0.483 M instead of 0.500 M changes the pH only slightly because the base is weak and pH depends on the square root of concentration. That is why many introductory solutions simply use 0.50 as the concentration for the hydrolysis step unless the problem explicitly gives density data.

Assumption	Concentration Used in Equilibrium	Estimated [OH-]	Estimated pH	Comment
Treat 0.50 m as approximately 0.50 M	0.500 M	3.33 × 10^-6 M	8.52	Standard classroom shortcut
Convert 0.50 m using density = 1.00 g/mL	0.483 M	3.27 × 10^-6 M	8.51	More careful but very close
If concentration were diluted to 0.10 M	0.100 M	1.49 × 10^-6 M	8.17	Shows concentration dependence
If concentration were increased to 1.00 M	1.000 M	4.71 × 10^-6 M	8.67	Even at higher concentration, the solution stays mildly basic

Common Mistakes to Avoid

• Using Ka directly in the ICE table for nitrite. Because NO2- is a base, you need Kb, not Ka, for the hydrolysis expression.
• Forgetting that Na+ is a spectator ion. Sodium does not make the solution acidic or basic here.
• Confusing 0.50 m with 0.50 M. They are close only if the solution density is near 1.00 g/mL and the solute amount is modest.
• Solving for pH directly from x. The hydrolysis gives hydroxide, so you must find pOH first, then pH.
• Ignoring temperature dependence. Ka and Kw are temperature sensitive, so a 25 degrees Celsius answer may shift at other temperatures.

When Is the Approximation Valid?

The weak-base approximation assumes that x is very small compared with the initial concentration C. A useful guideline is that if the percent ionization is under about 5 percent, the approximation is considered good. For sodium nitrite at 0.50 concentration units, the hydroxide formed is only about 3.33 × 10^-6 M. Compared with 0.50, that is vanishingly small. The percent hydrolysis is about:

(3.33 × 10^-6 / 0.50) × 100 ≈ 0.00067%

That is far below 5 percent, so the approximation is exceptionally safe. In fact, the exact quadratic and the shortcut produce essentially the same pH to the digits usually reported in general chemistry.

How This Calculator Works

The calculator above follows a rigorous sequence. First, it reads your concentration, unit choice, density, Ka, and preferred solution method. If you select molality, it converts that molality to molarity using the molar mass of NaNO2, which is approximately 69.00 g/mol, and the density you entered. Next, it computes Kb from Ka and Kw. Then it solves for equilibrium hydroxide either with the exact quadratic formula or with the square-root approximation. Finally, it reports the effective molarity, Kb, hydroxide concentration, pOH, and pH.

The chart adds a practical visualization by plotting pH across a range of NaNO2 concentrations. This helps you see that sodium nitrite remains only mildly basic over a wide concentration range because nitrite is a weak base. The pH rises with concentration, but not dramatically, since the dependence is tied to equilibrium rather than complete hydroxide release as with a strong base like NaOH.

Reference Concepts from Authoritative Sources

If you want to strengthen your understanding of pH, weak acid-base equilibria, and water chemistry, these authoritative resources are helpful:

• USGS: pH and Water
• MIT OpenCourseWare: Acids and Bases
• Purdue University Chemistry: Acid-Base Equilibria Review

Final Answer Summary

For a 0.50 m solution of NaNO2, using Ka(HNO2) = 4.5 × 10^-4 at 25 degrees Celsius, the nitrite ion behaves as a weak base with Kb = 2.22 × 10^-11. Solving the hydrolysis equilibrium gives [OH-] ≈ 3.33 × 10^-6 M, which corresponds to pOH ≈ 5.48 and pH ≈ 8.52. If you account more carefully for the fact that the problem gives molality rather than molarity, the result stays extremely close, around pH 8.51 to 8.52 depending on the density assumption.

In most general chemistry settings, the accepted answer is that a 0.50 m NaNO2 solution is mildly basic with a pH of about 8.5.