Calculate the pH of 6 x 10^-4 NaNO2
Use this premium nitrite hydrolysis calculator to determine the pH of a sodium nitrite solution. The tool models NaNO2 as the salt of a weak acid, HNO2, and a strong base, NaOH, then solves the equilibrium with water autoionization for a reliable pH value at 25 C.
How to calculate the pH of 6 x 10^-4 NaNO2
If you need to calculate the pH of 6 x 10^-4 NaNO2, the key idea is that sodium nitrite is not an acid itself. It is a salt made from a strong base, sodium hydroxide, and a weak acid, nitrous acid, HNO2. Because the cation Na+ is essentially neutral in water, the chemistry is controlled by the nitrite ion, NO2-. Nitrite acts as a weak base by reacting with water to form a small amount of HNO2 and OH-. That small amount of hydroxide pushes the solution slightly above neutral, so the pH is expected to be a little greater than 7.
For the specific concentration 6 x 10^-4 M NaNO2, many students first use the quick weak base approximation. That method is often fine at moderate concentrations, but here the solution is dilute enough that the hydroxide generated from hydrolysis is only a bit larger than the hydroxide that comes from water itself. Because of that, the most defensible answer uses an exact equilibrium treatment that includes water autoionization. When you do that, the pH comes out to about 7.18 at 25 C when you use Ka(HNO2) = 4.5 x 10^-4.
Why NaNO2 gives a basic solution
NaNO2 dissociates almost completely in water:
The sodium ion does not significantly affect pH, but the nitrite ion does. Nitrite is the conjugate base of nitrous acid:
That means nitrite can pull a proton from water:
Because hydroxide ions are produced, the solution becomes basic. The strength of that basicity depends on the base constant for nitrite, Kb, which is related to the acid constant of nitrous acid, Ka, by:
At 25 C, Kw is 1.0 x 10^-14. If we use Ka for HNO2 of 4.5 x 10^-4, then:
Quick approximation method
The most common classroom method treats nitrite as a weak base of concentration C and assumes the hydroxide formed from hydrolysis is x. Then:
If x is small compared with C, then C – x is approximately C, and:
For 6 x 10^-4 M NaNO2:
Now find pOH and pH:
This result is chemically sensible because it is slightly basic. However, the approximation has a weakness: the calculated hydroxide concentration is extremely close to 1.0 x 10^-7 M, which is the hydroxide level in pure water at 25 C. When the equilibrium contribution from the salt is on the same order as water itself, the approximation becomes less reliable.
Exact method with water autoionization
For a better answer, use mass balance, charge balance, and the acid equilibrium of HNO2. Let the formal concentration of sodium nitrite be C = 6 x 10^-4 M. Then:
- Mass balance: [NO2-] + [HNO2] = C
- Acid equilibrium: Ka = [H+][NO2-] / [HNO2]
- Charge balance: [Na+] + [H+] = [NO2-] + [OH-]
- Since NaNO2 dissociates fully, [Na+] = C
- Water autoionization: [H+][OH-] = Kw
Combining these equations leads to a solvable expression in [H+]. Numerically solving that system at 25 C with Ka = 4.5 x 10^-4 gives:
This exact value is more trustworthy for a dilute basic salt. It still confirms the core idea: sodium nitrite at 6 x 10^-4 M is basic, but only slightly basic.
Step by step summary for students
- Recognize that NaNO2 is a salt of a strong base and a weak acid.
- Conclude that the solution will be basic because NO2- hydrolyzes to make OH-.
- Use Ka(HNO2) to get Kb(NO2-) from Kb = Kw / Ka.
- For a rough estimate, calculate [OH-] from sqrt(KbC).
- For a more accurate answer at this dilute concentration, include water autoionization and solve the equilibrium exactly.
- State the pH with the assumptions used, because different literature values of Ka can shift the answer slightly.
Comparison table: approximation versus exact calculation
| Method | Main assumption | Calculated value | Resulting pH | Comment |
|---|---|---|---|---|
| Weak base approximation | Ignore water autoionization and assume x is small | [OH-] = 1.15 x 10^-7 M | 7.06 | Fast and often taught first, but less accurate here |
| Exact charge balance | Includes Ka, Kw, mass balance, and charge balance | [H+] = 6.55 x 10^-8 M | 7.18 | Better choice for very dilute solutions |
How concentration changes the pH of sodium nitrite
One reason this problem can feel tricky is that the pH of a weakly basic salt does not rise very dramatically at low concentration. The hydrolysis is weak, so even a tenfold concentration change shifts pH by a fairly modest amount. The table below shows approximate exact pH values for NaNO2 using Ka(HNO2) = 4.5 x 10^-4 at 25 C.
| NaNO2 concentration | Approximate exact pH | Chemical interpretation |
|---|---|---|
| 6 x 10^-2 M | 8.06 | Clearly basic because nitrite hydrolysis dominates water |
| 6 x 10^-3 M | 7.57 | Mildly basic |
| 6 x 10^-4 M | 7.18 | Slightly basic, close to neutral |
| 6 x 10^-5 M | 7.02 | Very close to neutral because water matters strongly |
Common mistakes when solving this problem
1. Treating NaNO2 like a neutral salt
Students sometimes assume every sodium salt is neutral. That is only true when the anion comes from a strong acid, such as Cl- from HCl or NO3- from HNO3. Nitrite, NO2-, comes from the weak acid HNO2, so it reacts with water and raises pH.
2. Using the wrong equilibrium constant
For nitrite in water, the reacting species is NO2-, so the relevant constant is Kb. If you are given Ka for HNO2, convert it using Kb = Kw / Ka.
3. Forgetting water autoionization at very low concentration
When your calculated hydroxide or hydronium is close to 1 x 10^-7 M, pure water contributes a nontrivial amount. That is why the exact answer for 6 x 10^-4 M NaNO2 is noticeably different from the quick estimate.
4. Rounding too aggressively
Because the final pH is only slightly above 7, rough rounding can change the reported answer by a few hundredths. It is good practice to carry extra digits until the final step.
Which Ka value should you use for HNO2?
Different textbooks and databases may list slightly different values for Ka of nitrous acid depending on temperature and data source. Common values lie near 4.0 x 10^-4 to 4.5 x 10^-4 at room temperature, corresponding to a pKa around 3.35 to 3.40. That variation changes the pH of 6 x 10^-4 M NaNO2 only slightly. In practical coursework, your instructor may specify which constant to use. If no value is specified, 4.5 x 10^-4 is a common and defensible choice.
Authoritative chemistry references
If you want to verify equilibrium concepts, pH definitions, or acid base fundamentals from authoritative educational and government sources, these references are useful:
- U.S. Environmental Protection Agency on acidity, pH, and alkalinity
- Purdue University general chemistry acid base review
- UC Davis educational material on hydrolysis of salt solutions
Bottom line
To calculate the pH of 6 x 10^-4 NaNO2, identify nitrite as the conjugate base of the weak acid HNO2. Using Ka(HNO2) near 4.5 x 10^-4 gives Kb(NO2-) near 2.22 x 10^-11. The quick approximation produces pH about 7.06, but because the concentration is dilute and the hydroxide formed is near the level present in pure water, the exact equilibrium treatment is better. That calculation gives a pH of about 7.18 at 25 C. So the solution is slightly basic, not strongly basic and not neutral.