Calculate the pH of a Solution of 0.36 M NaNO2
Use this interactive sodium nitrite pH calculator to estimate the basicity of a nitrite salt solution from concentration and acid dissociation data for nitrous acid.
NaNO2 pH Calculator
Default value is 0.36 M.
Nitrite is the conjugate base of nitrous acid.
Common 25 C reference: Ka(HNO2) ≈ 4.0 × 10-4.
Default assumes standard aqueous conditions near 25 C.
The exact quadratic solution is preferred for accuracy, although both methods agree closely here.
Results
Ready to calculate
Enter or keep the default values, then click Calculate pH to see the sodium nitrite equilibrium result, hydroxide concentration, and a visual chart.
How to calculate the pH of a solution of 0.36 M NaNO2
To calculate the pH of a solution of 0.36 M sodium nitrite, you need to recognize what kind of substance NaNO2 is in water. Sodium nitrite is a salt formed from a strong base, sodium hydroxide, and a weak acid, nitrous acid (HNO2). Because the cation Na+ does not appreciably affect pH, the chemistry is controlled by the nitrite ion, NO2–. Nitrite acts as a weak base and reacts with water to produce hydroxide ions, making the solution basic.
Key idea: A 0.36 M NaNO2 solution is not neutral. It is basic because NO2– is the conjugate base of the weak acid HNO2.
Step 1: Write the hydrolysis reaction
When sodium nitrite dissolves, it dissociates essentially completely:
The nitrite ion then undergoes base hydrolysis:
This equilibrium is what determines the pH. Since hydroxide is produced, the final pH will be above 7 at standard conditions.
Step 2: Relate Kb for NO2- to Ka for HNO2
Most textbooks tabulate the acid dissociation constant of nitrous acid rather than the base dissociation constant of nitrite. The relation is:
Using common 25 C values:
- Ka(HNO2) ≈ 4.0 × 10-4
- Kw = 1.0 × 10-14
So the base dissociation constant for nitrite is:
Step 3: Set up the equilibrium table
Let the initial nitrite concentration be 0.36 M. If x mol/L of NO2– reacts, then:
- [NO2–] initial = 0.36
- [HNO2] initial = 0
- [OH–] initial ≈ 0
At equilibrium:
- [NO2–] = 0.36 – x
- [HNO2] = x
- [OH–] = x
The equilibrium expression is:
Step 4: Solve for hydroxide concentration
Because Kb is very small, x is tiny relative to 0.36, so many instructors use the weak-base approximation:
Substitute the values:
Thus:
- [OH–] ≈ 3.0 × 10-6 M
- pOH = -log(3.0 × 10-6) ≈ 5.52
- pH = 14.00 – 5.52 ≈ 8.48
Final answer: The pH of a 0.36 M NaNO2 solution is approximately 8.48 at 25 C when Ka(HNO2) = 4.0 × 10-4.
Why sodium nitrite solutions are basic
Students often wonder why a salt like sodium nitrite changes the pH at all. The answer lies in the parent acid and base. If a salt comes from a strong acid and a strong base, the ions are usually spectators and the solution is near neutral. If a salt comes from a weak acid and a strong base, the anion is basic. That is exactly the case for NaNO2.
Nitrous acid is not fully ionized in water, which means its conjugate base, nitrite, has measurable affinity for protons from water. When nitrite accepts a proton, it leaves behind hydroxide. The production of OH– is what pushes the pH upward.
This pattern is broadly useful in chemistry:
- NaCl from HCl and NaOH gives a nearly neutral solution.
- NH4Cl from NH3 and HCl gives an acidic solution.
- NaF, CH3COONa, and NaNO2 from weak acids and strong bases give basic solutions.
Exact result versus approximation
For this specific problem, the approximation works extremely well because the hydrolysis is weak and the fraction of nitrite that reacts is very small. Still, in a premium calculator it is better to use the exact quadratic when possible. Starting from:
Rearranging gives:
The physically meaningful solution is:
Using C = 0.36 M and Kb = 2.5 × 10-11, the exact x value is essentially 3.0 × 10-6 M, confirming the approximation. The percent hydrolysis is only about 0.00083%, which is far below the common 5% guideline used to justify equilibrium approximations.
| Parameter | Approximation method | Exact quadratic method |
|---|---|---|
| Initial [NO2–] | 0.36 M | 0.36 M |
| Kb used | 2.5 × 10-11 | 2.5 × 10-11 |
| [OH–] at equilibrium | 3.0 × 10-6 M | 2.99999 × 10-6 M |
| pOH | 5.52 | 5.52 |
| pH | 8.48 | 8.48 |
| Percent hydrolysis | 0.00083% | 0.00083% |
How the concentration affects pH
Although this page focuses on 0.36 M NaNO2, it helps to understand how concentration changes the answer. For a weak base salt with fixed Kb, the hydroxide concentration rises roughly with the square root of concentration. That means pH increases as the solution becomes more concentrated, but not in a simple linear way. Doubling the concentration does not double the pH increase.
The following comparison uses Ka(HNO2) = 4.0 × 10-4 and standard aqueous conditions:
| NaNO2 concentration (M) | Estimated [OH–] (M) | pOH | Estimated pH |
|---|---|---|---|
| 0.010 | 5.00 × 10-7 | 6.30 | 7.70 |
| 0.050 | 1.12 × 10-6 | 5.95 | 8.05 |
| 0.100 | 1.58 × 10-6 | 5.80 | 8.20 |
| 0.360 | 3.00 × 10-6 | 5.52 | 8.48 |
| 0.500 | 3.54 × 10-6 | 5.45 | 8.55 |
| 1.000 | 5.00 × 10-6 | 5.30 | 8.70 |
Common mistakes when solving this problem
This kind of question looks straightforward, but several predictable errors appear in homework, quizzes, and exam settings. If you want a reliable answer, avoid the following traps:
- Treating NaNO2 as a strong base. Sodium nitrite is not NaOH. The solution is basic, but only mildly so.
- Using Ka directly instead of converting to Kb. Since nitrite is acting as a base, you must use Kb for NO2–, or convert from Ka correctly.
- Forgetting the parent acid-base logic. Na+ comes from a strong base and is negligible; NO2– controls the pH.
- Mixing up pH and pOH. The hydrolysis calculation gives [OH–] first. You calculate pOH before converting to pH.
- Rounding too early. Keep enough significant figures through intermediate steps, especially for logarithms.
When to use the 5% approximation rule
The 5% rule is a fast way to judge whether you can ignore x in expressions like C – x. For this NaNO2 problem, x is around 3.0 × 10-6 M while C is 0.36 M. The ratio is:
That is nowhere near 5%, so the approximation is exceptionally safe. In more weakly concentrated solutions or in systems with larger K values, the exact quadratic can become more important. A calculator like the one above lets you compare both methods instantly.
What this calculation means in practical terms
A pH of about 8.48 means the solution is mildly basic, not strongly caustic. In laboratory settings, that matters for indicator selection, buffer compatibility, and reaction planning. If you were preparing a nitrite-containing solution for analytical work, this pH could affect metal ion speciation, oxidation-reduction side chemistry, and the relative stability of acid-sensitive reagents.
In broader chemical education, this problem is a classic demonstration of salt hydrolysis and conjugate acid-base relationships. It reinforces several central ideas at once:
- Salts can change pH depending on the acid and base they come from.
- The conjugate base of a weak acid is basic.
- Equilibrium constants connect through Kw.
- Weak-acid and weak-base approximations can simplify calculations dramatically.
Reference chemistry workflow for exams and homework
If you need a repeatable system for similar pH questions, use this sequence:
- Identify whether the species is acidic, basic, or neutral.
- Write the relevant hydrolysis or dissociation equation.
- Determine whether Ka, Kb, pKa, or pKb is needed.
- Convert constants if necessary using Kw.
- Set up an ICE table.
- Solve using the quadratic or the justified approximation.
- Convert to pOH or pH as needed.
- Check whether the answer is chemically reasonable.
For 0.36 M NaNO2, the chemically reasonable answer must be above 7 but nowhere near the pH of a strong base. A result around 8.5 passes that reality check.
Authoritative sources for acid-base and aqueous equilibrium concepts
For deeper study, consult these reputable educational and government resources:
- University of Illinois chemistry resources on equilibrium and calculations
- University of Wisconsin acid-base equilibrium tutorial
- U.S. Environmental Protection Agency overview of pH in aqueous systems
Bottom line
If you are asked to calculate the pH of a solution of 0.36 M NaNO2, the core idea is that nitrite is a weak base in water. Use the Ka of nitrous acid to find Kb for NO2–, solve for hydroxide concentration, then convert to pH. With Ka(HNO2) = 4.0 × 10-4 at 25 C, the solution comes out to approximately pH 8.48. That answer is consistent with a mildly basic salt solution and is a textbook example of salt hydrolysis from the conjugate base of a weak acid.