Calculate The Ph Of A 0.010 M Nano2 Solution.

Chemistry Calculator

Use this interactive calculator to determine the pH, pOH, hydroxide concentration, and equilibrium behavior of sodium nitrite solutions from weak acid-base chemistry.

NaNO2 pH Calculator

Equilibrium Chart

This chart compares the initial nitrite concentration with the calculated hydroxide and hydronium concentrations after hydrolysis.

How to calculate the pH of a 0.010 M NaNO2 solution

Sodium nitrite, NaNO2, is an ionic salt that dissociates completely in water into Na+ and NO2-. The sodium ion is essentially a spectator ion in this context, but the nitrite ion matters a great deal because it is the conjugate base of nitrous acid, HNO2. Since HNO2 is a weak acid, its conjugate base, NO2-, reacts with water to produce a small amount of hydroxide ions. That hydrolysis makes the solution basic, so the pH of a 0.010 M NaNO2 solution is greater than 7.

At the core of this problem is the equilibrium:

NO2- + H2O ⇌ HNO2 + OH-

This tells us that nitrite accepts a proton from water, forming nitrous acid and hydroxide ion. To calculate the pH, you first need the base dissociation constant, Kb, for nitrite. In most chemistry courses, you are given Ka for HNO2 rather than Kb for NO2-. That means you use the conjugate acid-base relationship:

Kb = Kw / Ka

Using a common textbook value of Ka(HNO2) = 4.5 × 10^-4 at about 25 degrees C, and Kw = 1.0 × 10^-14, the nitrite base constant is:

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

Now let the initial concentration of NO2- be 0.010 M, and let x represent the amount that reacts with water. Then the ICE framework is:

• Initial: [NO2-] = 0.010, [HNO2] = 0, [OH-] = 0
• Change: [NO2-] = -x, [HNO2] = +x, [OH-] = +x
• Equilibrium: [NO2-] = 0.010 – x, [HNO2] = x, [OH-] = x

The equilibrium expression becomes:

Kb = x² / (0.010 – x)

Because Kb is very small, many instructors allow the approximation 0.010 – x ≈ 0.010. Then:

x = √(Kb × C) = √(2.22 × 10^-11 × 0.010) = 4.71 × 10^-7 M

This x value is the hydroxide concentration. From that:

• [OH-] = 4.71 × 10^-7 M
• pOH = -log(4.71 × 10^-7) = 6.33
• pH = 14.00 – 6.33 = 7.67

So the pH of a 0.010 M NaNO2 solution is approximately 7.67 at 25 degrees C when using Ka = 4.5 × 10^-4 for HNO2. The exact quadratic treatment gives essentially the same answer because x is tiny compared with 0.010 M. That is why this is a classic weak-base hydrolysis problem and not a strong-base calculation.

Key result: For a 0.010 M sodium nitrite solution, the expected pH is about 7.67 under standard classroom assumptions.

Why NaNO2 forms a basic solution

Students often wonder why a salt can change pH at all. The answer depends on the strengths of the parent acid and base from which the salt was formed. Sodium nitrite comes from sodium hydroxide, a strong base, and nitrous acid, a weak acid. The cation from a strong base, Na+, does not significantly react with water. The anion from the weak acid, NO2-, does react with water. That reaction creates OH-, which pushes the pH above neutral.

By contrast, a salt such as NaCl is neutral because neither Na+ nor Cl- hydrolyzes appreciably. A salt such as NH4Cl is acidic because NH4+ is the conjugate acid of the weak base NH3 and donates protons to water. Learning to classify salts this way is one of the most useful acid-base shortcuts in general chemistry.

Quick classification rule

• Strong acid + strong base salt: usually neutral
• Weak acid + strong base salt: usually basic
• Strong acid + weak base salt: usually acidic
• Weak acid + weak base salt: depends on relative Ka and Kb values

Step-by-step expert method

1. Write the dissociation of the salt: NaNO2 → Na+ + NO2-.
2. Identify NO2- as the conjugate base of HNO2.
3. Write the hydrolysis equation: NO2- + H2O ⇌ HNO2 + OH-.
4. Convert Ka of HNO2 to Kb of NO2- using Kb = Kw / Ka.
5. Set up an ICE table with initial concentration 0.010 M.
6. Solve for x, which equals [OH-].
7. Calculate pOH = -log[OH-].
8. Convert to pH using pH = 14.00 – pOH.

This process is robust and works for many salts of weak acids, including sodium acetate, sodium fluoride, and sodium cyanide. The main difference is the Ka value of the parent weak acid. A weaker parent acid has a smaller Ka and therefore a larger conjugate-base tendency to hydrolyze only if the relationship through Kw supports it. In practical terms, once you know the parent acid, you can almost always build the right equilibrium model.

Comparison table: pH of NaNO2 at different concentrations

The pH of sodium nitrite changes with concentration, but because it behaves as a weak base, the change is gradual rather than dramatic. Using Ka(HNO2) = 4.5 × 10^-4 and standard 25 degrees C conditions, the approximate values below illustrate the trend.

NaNO2 concentration (M)	Kb of NO2-	Approximate [OH-] (M)	Approximate pOH	Approximate pH
0.0010	2.22 × 10^-11	1.49 × 10^-7	6.83	7.17
0.010	2.22 × 10^-11	4.71 × 10^-7	6.33	7.67
0.050	2.22 × 10^-11	1.05 × 10^-6	5.98	8.02
0.100	2.22 × 10^-11	1.49 × 10^-6	5.83	8.17

Notice that even a tenfold increase in concentration does not send the pH to extremely high values. That is because nitrite is only a weak base. The amount of hydroxide formed is limited by equilibrium, not by complete dissociation as in a strong base like NaOH.

Comparison table: NaNO2 versus common salt types

The table below helps place sodium nitrite in context with other salts students frequently study. These are representative classroom trends under dilute aqueous conditions, not universal values for every concentration and temperature.

Salt	Parent acid	Parent base	Expected aqueous behavior	Typical pH tendency
NaNO2	HNO2, weak acid	NaOH, strong base	Basic due to NO2- hydrolysis	Above 7
NaCl	HCl, strong acid	NaOH, strong base	Essentially neutral	Near 7
NH4Cl	HCl, strong acid	NH3, weak base	Acidic due to NH4+ hydrolysis	Below 7
CH3COONa	CH3COOH, weak acid	NaOH, strong base	Basic due to acetate hydrolysis	Above 7

When should you use the exact quadratic instead of the approximation?

For many weak acid and weak base problems, the approximation method is acceptable if the percent ionization or hydrolysis is small, often less than 5 percent. In the NaNO2 example, x is on the order of 10^-7 while the initial concentration is 10^-2. The ratio is far below 1 percent, so the shortcut is excellent.

However, the exact quadratic method is still worth understanding because it is mathematically complete and avoids approximation errors in more dilute solutions. As concentration decreases, the contribution from water autoionization can become more important, and the simplifications used in standard homework examples may need refinement. That is one reason the calculator above provides an exact option for the salt hydrolysis step.

Situations where exact treatment is preferred

• Very dilute solutions
• Problems where the 5 percent rule is not satisfied
• High-precision lab calculations
• Cases where multiple equilibria interact strongly

Common mistakes students make

• Treating NaNO2 as neutral. It is not neutral because NO2- is a weak base.
• Using Ka directly in the hydrolysis expression. You need Kb for NO2-.
• Assuming the solution is acidic because nitrous acid is an acid. The solution contains its conjugate base, not the acid itself.
• Forgetting to convert pOH to pH. The hydrolysis gives OH-, so pOH comes first.
• Applying strong-base logic. NO2- does not fully generate OH- like NaOH would.

Why the literature value of Ka matters

The exact numeric pH you calculate depends on the Ka value chosen for nitrous acid. Textbooks and databases may list slightly different values, often because of rounding, temperature differences, or reference conventions. A Ka around 4.0 × 10^-4 to 4.5 × 10^-4 is common for introductory work. That means your final pH may differ slightly in the second decimal place depending on the source, but it will still clearly indicate a mildly basic solution.

If your instructor provides a specific Ka or pKa, always use that value rather than a general reference number. In graded chemistry work, following the given data is more important than using a different accepted constant from another source.

Authoritative references for acid-base data and water chemistry

For students who want to verify equilibrium constants, review acid-base fundamentals, or explore water chemistry in more depth, these sources are reliable starting points:

• LibreTexts Chemistry for broad instructional chemistry content.
• U.S. Environmental Protection Agency (.gov) for water quality chemistry context.
• NIST Chemistry WebBook (.gov) for chemical reference data.
• Princeton University Chemistry (.edu) for university-level chemistry resources.

Final takeaway

To calculate the pH of a 0.010 M NaNO2 solution, recognize that sodium nitrite is the salt of a weak acid and a strong base. The nitrite ion hydrolyzes water, producing hydroxide and making the solution basic. With Ka(HNO2) = 4.5 × 10^-4, the conjugate base constant is Kb = 2.22 × 10^-11. Solving the equilibrium gives [OH-] ≈ 4.71 × 10^-7 M, pOH ≈ 6.33, and pH ≈ 7.67. That value is the standard answer expected in most general chemistry courses.

Educational note: real measured pH can vary slightly with temperature, ionic strength, and the equilibrium constant source used.