Calculate the pH of a 0.245 M Solution of NaNO2
Use this interactive calculator to determine the pH of sodium nitrite solutions from weak base hydrolysis. The tool uses the relationship between nitrite ion, nitrous acid, and water to compute pOH, pH, hydroxide concentration, and related values.
Default data assume sodium nitrite completely dissociates, producing NO2−, the conjugate base of weak acid HNO2.
Calculated Results
How to calculate the pH of a 0.245 M solution of NaNO2
To calculate the pH of a 0.245 M solution of sodium nitrite, you need to recognize what kind of salt NaNO2 is. Sodium nitrite is formed from the strong base sodium hydroxide and the weak acid nitrous acid, HNO2. Because the cation Na+ does not hydrolyze in water to any meaningful extent, the chemistry that matters comes from the nitrite ion, NO2−. That ion acts as a weak base in water and produces hydroxide ions according to the equilibrium:
Since hydroxide is generated, the solution becomes basic, so the pH will be above 7. The key to solving the problem is finding the base dissociation constant for nitrite, Kb. Most textbook and laboratory references report the acid dissociation constant for nitrous acid instead, typically around Ka = 4.5 × 10−4 at room temperature. Once you know Ka, you can use the standard relationship:
At approximately 25 C, Kw is 1.0 × 10−14, so:
Next, let the initial concentration of nitrite ion be 0.245 M. Because sodium nitrite dissociates essentially completely in water, the starting concentration of NO2− is the same as the formal concentration of the salt. Set up an ICE table for the hydrolysis equilibrium. If x is the amount of hydroxide produced:
- Initial: [NO2−] = 0.245, [HNO2] = 0, [OH−] = 0
- Change: [NO2−] = −x, [HNO2] = +x, [OH−] = +x
- Equilibrium: [NO2−] = 0.245 − x, [HNO2] = x, [OH−] = x
Substituting these terms into the equilibrium expression gives:
Because Kb is very small and the concentration is fairly large, x is much smaller than 0.245. That means the common approximation 0.245 − x ≈ 0.245 is valid. Then:
This x value is the hydroxide concentration. Now calculate pOH:
Finally, use pH + pOH = 14.00:
Therefore, the pH of a 0.245 M solution of NaNO2 is approximately 8.37 under standard 25 C conditions when Ka for HNO2 is taken as 4.5 × 10−4. This is the result most instructors expect in general chemistry unless a different Ka value is specified.
Why sodium nitrite makes water basic
Understanding the origin of the basicity matters more than memorizing one numerical answer. Sodium nitrite is not itself a strong base in the same way sodium hydroxide is. Instead, it is the salt of a weak acid. The nitrite ion is willing to accept a proton from water because it is the conjugate base of HNO2. When it pulls a proton from water, hydroxide ions are left behind, and that is why the pH rises above neutral.
This idea is part of a larger pattern in acid-base chemistry:
- Salt from strong acid + strong base: usually neutral
- Salt from strong acid + weak base: usually acidic
- Salt from weak acid + strong base: usually basic
- Salt from weak acid + weak base: depends on relative strengths
NaNO2 fits the third category. Sodium comes from a strong base and behaves as a spectator ion, while nitrite contributes the hydrolysis chemistry.
Step by step method you can use on homework or exams
- Write the dissociation of the salt: NaNO2 → Na+ + NO2−.
- Identify the reactive ion: NO2− is the conjugate base of HNO2.
- Write the hydrolysis equation: NO2− + H2O ⇌ HNO2 + OH−.
- Convert Ka of HNO2 into Kb of NO2− using Kb = Kw / Ka.
- Set up the ICE table and define x as [OH−] formed.
- Use either the approximation method or the quadratic formula.
- Find pOH from [OH−].
- Convert pOH to pH.
- Check that the approximation is valid by comparing x with the initial concentration.
For this specific problem, the approximation works extremely well because x is about 2.33 × 10−6 while the starting concentration is 0.245. The ratio is far below 5%, so the simplifying assumption is justified.
Approximation versus exact quadratic
Students are often told to use the square root shortcut for weak acids and weak bases. That shortcut is excellent when the equilibrium constant is small and the concentration is not tiny. However, if you want maximum precision, the exact solution comes from rearranging:
into the quadratic form:
Solving this gives nearly the same hydroxide concentration for 0.245 M NaNO2 because the equilibrium lies far to the left. In practical classroom terms, both methods produce a pH near 8.37.
Reference values used in nitrite pH calculations
| Quantity | Typical value at about 25 C | Why it matters |
|---|---|---|
| Ka for HNO2 | 4.5 × 10−4 | Determines how weak the acid is, which sets the basicity of NO2− |
| Kw | 1.0 × 10−14 | Used to convert Ka to Kb through Kb = Kw / Ka |
| Kb for NO2− | 2.22 × 10−11 | Controls the hydrolysis equilibrium in water |
| Formal concentration of NaNO2 | 0.245 M | Initial concentration of NO2− for the ICE table |
Comparison of pH at different sodium nitrite concentrations
One useful way to understand this system is to compare pH across a range of sodium nitrite concentrations while keeping Ka fixed at 4.5 × 10−4. Notice that the pH changes slowly because weak base hydrolysis depends on the square root of concentration rather than changing in a one-to-one way.
| NaNO2 concentration (M) | Estimated [OH−] (M) | Estimated pOH | Estimated pH |
|---|---|---|---|
| 0.010 | 4.71 × 10−7 | 6.33 | 7.67 |
| 0.050 | 1.05 × 10−6 | 5.98 | 8.02 |
| 0.100 | 1.49 × 10−6 | 5.83 | 8.17 |
| 0.245 | 2.33 × 10−6 | 5.63 | 8.37 |
| 0.500 | 3.33 × 10−6 | 5.48 | 8.52 |
| 1.000 | 4.71 × 10−6 | 5.33 | 8.67 |
Common mistakes when solving this problem
1. Treating NaNO2 as a neutral salt
This is the most common conceptual mistake. NaNO2 is not neutral in water because NO2− hydrolyzes. If you assume it is neutral simply because it contains sodium, you miss the entire equilibrium process.
2. Using Ka directly instead of converting to Kb
The hydrolysis reaction involves the base NO2−, not the acid HNO2. That means you need Kb, not Ka, unless you rewrite the problem in a different equivalent form. Students often plug Ka into a weak acid formula by mistake and get a pH below 7, which is chemically inconsistent.
3. Forgetting that the solution is basic
Before doing any arithmetic, predict whether the answer should be acidic, neutral, or basic. Since nitrite is a weak base, your final pH must be greater than 7. This quick mental check helps catch calculator and logarithm errors.
4. Using the wrong concentration
If the problem states a 0.245 M solution of NaNO2, then the initial nitrite concentration is 0.245 M because the salt dissociates fully. You should not halve it or otherwise alter it unless stoichiometry in a previous step requires that.
5. Rounding too early
Keep several significant digits during intermediate steps, especially for Kb and [OH−]. Rounding only at the end will give a more stable final pH value, usually around 8.36 to 8.37 depending on the exact constants used.
How temperature and data source can change the answer slightly
If your textbook or instructor uses a slightly different Ka for HNO2, your final pH can shift by a few hundredths of a pH unit. Temperature also affects Kw and equilibrium constants. At standard general chemistry conditions, the answer near 8.37 is entirely reasonable. In advanced analytical work, one might also account for activity effects at higher ionic strength, but that level of correction is usually beyond introductory chemistry.
For deeper reference data on water chemistry, acid-base constants, and equilibrium methods, consult authoritative educational and government sources such as U.S. Environmental Protection Agency water quality resources, chemistry instructional materials hosted on university-backed educational platforms, the NIST Chemistry WebBook, and OpenStax Chemistry 2e. If you need strictly .gov or .edu sources, the NIST site and many university chemistry departments provide reliable support for equilibrium concepts and constants.
Practical interpretation of the result
A pH of about 8.37 means the solution is only moderately basic, not strongly caustic. That is exactly what you expect from the conjugate base of a weak acid. The hydroxide concentration is on the order of 10−6 M, which is much greater than pure water but still tiny compared with the total salt concentration. In other words, only a very small fraction of nitrite ions actually hydrolyze at equilibrium.
This small fraction is another reason the approximation method works well. Even though the salt concentration is 0.245 M, the amount converted into HNO2 and OH− is only a few millionths of a mole per liter. The equilibrium stays heavily weighted toward NO2−.
Final answer summary
For a 0.245 M NaNO2 solution, assuming Ka(HNO2) = 4.5 × 10−4 and Kw = 1.0 × 10−14 at 25 C:
- Kb(NO2−) = 2.22 × 10−11
- [OH−] ≈ 2.33 × 10−6 M
- pOH ≈ 5.63
- pH ≈ 8.37
If you are asked on an exam to “calculate the pH of a 0.245 M solution of NaNO2,” the concise final response is usually: The solution is basic, and its pH is approximately 8.37.