Calculate the pH of a 0.010 M NaNO2 Solution
Use this interactive calculator to find the pH, pOH, hydroxide concentration, and percent hydrolysis for aqueous sodium nitrite. The default setup evaluates a 0.010 M NaNO2 solution at 25 degrees Celsius using the acid dissociation constant of nitrous acid.
Results
Enter values and click Calculate pH to see the answer for your sodium nitrite solution.
How to calculate the pH of a 0.010 M NaNO2 solution
Sodium nitrite, NaNO2, is the salt of a strong base and a weak acid. The sodium ion does not appreciably affect the pH of water, but the nitrite ion, NO2–, does. Because nitrite is the conjugate base of nitrous acid, HNO2, it reacts with water and produces hydroxide ions. That means a sodium nitrite solution is basic, not neutral.
If you want to calculate the pH of a 0.010 M NaNO2 solution correctly, the core chemistry is hydrolysis of the conjugate base:
NO2- + H2O ⇌ HNO2 + OH-
The equilibrium constant for this base reaction is the base dissociation constant, Kb. Since nitrite comes from nitrous acid, you obtain Kb from the relationship:
Kb = Kw / Ka
At 25 degrees Celsius, Kw is commonly taken as 1.0 × 10-14. A typical textbook value for the acid dissociation constant of nitrous acid is about 4.5 × 10-4. Using those values:
Kb = (1.0 × 10^-14) / (4.5 × 10^-4) = 2.22 × 10^-11
Now let the initial concentration of nitrite be 0.010 M. If x is the amount of OH– formed, the equilibrium expression is:
Kb = x^2 / (0.010 – x)
Because Kb is very small, x is much smaller than 0.010, so the weak base approximation is valid. This gives:
x ≈ √(Kb × C)
x ≈ √((2.22 × 10^-11)(0.010)) = 4.71 × 10^-7 M
This x value is the hydroxide concentration generated by nitrite hydrolysis. Then:
- Find pOH: pOH = -log[OH-] = -log(4.71 × 10^-7) ≈ 6.33
- Find pH: pH = 14.00 – 6.33 = 7.67
So the pH of a 0.010 M NaNO2 solution is approximately 7.67 at 25 degrees Celsius using Ka = 4.5 × 10-4 for HNO2.
Why NaNO2 gives a basic solution
Many students first expect a salt solution to be neutral, but that is only true for salts derived from strong acids and strong bases, such as NaCl. Sodium nitrite behaves differently because nitrous acid is weak. The nitrite ion still has enough basic character to pull a proton from water, forming HNO2 and OH–. Even though this basicity is weak, it is enough to push the pH above 7.
- Na+ is a spectator ion from the strong base NaOH.
- NO2– is the conjugate base of weak acid HNO2.
- Conjugate bases of weak acids hydrolyze and make solutions basic.
Exact quadratic method versus approximation
For this problem, the approximation method works very well because the equilibrium shift is tiny compared with the starting concentration. However, a premium calculator should be able to do both the approximation and the exact method.
Starting from:
Kb = x^2 / (C – x)
Rearrange to:
x^2 + Kb x – Kb C = 0
The physically meaningful root is:
x = (-Kb + √(Kb^2 + 4KbC)) / 2
With this exact expression, the result is almost identical for 0.010 M NaNO2 because Kb is so small. In practical chemistry classes, the approximation is usually accepted, but using the exact quadratic confirms the answer and is especially helpful at very low concentrations or when the weak base is stronger.
| Method | [OH-] produced | pOH | pH | Comment |
|---|---|---|---|---|
| Weak base approximation | 4.71 × 10-7 M | 6.33 | 7.67 | Fast and accurate for this concentration |
| Exact quadratic solution | 4.71 × 10-7 M | 6.33 | 7.67 | Preferred when you want the most rigorous result |
Step by step logic for exam problems
If this appears on a chemistry exam, the biggest challenge is usually choosing the right equilibrium model. A reliable decision sequence is:
- Identify whether the compound is an acid, base, or salt.
- Determine whether the ions come from strong or weak parents.
- Recognize that NaNO2 is a salt of strong base NaOH and weak acid HNO2.
- Write the hydrolysis equation for NO2–.
- Use Ka of HNO2 to calculate Kb of NO2–.
- Set up an ICE table or weak base expression.
- Find [OH–], then pOH, then pH.
That reasoning pattern works for many conjugate base salts, including sodium acetate, sodium fluoride, and sodium cyanide, although each one has a different Kb value because each parent acid has a different Ka.
Common mistakes when calculating the pH of sodium nitrite
- Treating the solution as neutral. NaNO2 is not neutral because nitrite hydrolyzes.
- Using Ka directly instead of converting to Kb. The reacting species is NO2–, so you need the base constant.
- Forgetting that pH is found from pOH. Since the equilibrium produces OH–, calculate pOH first.
- Ignoring temperature if a nonstandard Kw is given. Kw changes with temperature, so pH = 7 is not always neutral.
- Rounding too early. Keep several significant digits through the logarithm step.
How concentration changes the pH
The pH of sodium nitrite depends on concentration. More concentrated solutions produce slightly more hydroxide and therefore have a higher pH. Because this is a weak base hydrolysis system, the increase is not linear. Doubling concentration does not double pH. Instead, [OH–] scales approximately with the square root of concentration when the weak base approximation is valid.
| NaNO2 concentration (M) | Kb of NO2- | Approx. [OH-] (M) | Approx. pH at 25 degrees C |
|---|---|---|---|
| 0.0010 | 2.22 × 10-11 | 1.49 × 10-7 | 7.17 |
| 0.010 | 2.22 × 10-11 | 4.71 × 10-7 | 7.67 |
| 0.050 | 2.22 × 10-11 | 1.05 × 10-6 | 8.02 |
| 0.100 | 2.22 × 10-11 | 1.49 × 10-6 | 8.17 |
How Ka data affect the final answer
Different textbooks and data tables may list slightly different Ka values for nitrous acid. Because Kb is calculated from Ka, any change in Ka changes the predicted pH. The effect is not huge for classroom calculations, but it is real. When your instructor gives a specific Ka, always use that number rather than a memorized value from another source.
| Ka for HNO2 | Calculated Kb for NO2- | Estimated pH of 0.010 M NaNO2 | Interpretation |
|---|---|---|---|
| 4.0 × 10-4 | 2.50 × 10-11 | 7.70 | Weaker acid gives slightly stronger conjugate base |
| 4.5 × 10-4 | 2.22 × 10-11 | 7.67 | Common classroom reference value |
| 5.0 × 10-4 | 2.00 × 10-11 | 7.65 | Stronger acid gives slightly weaker conjugate base |
Interpretation of the result
A pH near 7.67 means the solution is mildly basic. It is not strongly alkaline like sodium hydroxide, but it is clearly above neutral under standard conditions. This modest basicity is exactly what you expect from the conjugate base of a weak acid with a small Kb value. The hydrolysis is real but limited, and the percent of nitrite converted into HNO2 is very low.
For a 0.010 M sodium nitrite solution, the fraction hydrolyzed is approximately:
% hydrolysis = (x / C) × 100 = (4.71 × 10^-7 / 0.010) × 100 ≈ 0.0047%
That tiny percentage is why the approximation method works so well here. Since much less than 5% of the original nitrite reacts, replacing 0.010 – x with 0.010 is justified.
Useful authoritative references
If you want to check equilibrium concepts, acid-base theory, and water ionization constants, these sources are excellent starting points:
- chem.libretexts.org for broad acid-base equilibrium explanations used in university chemistry instruction.
- www.epa.gov for water chemistry context and pH background in environmental systems.
- webbook.nist.gov for authoritative physical chemistry data and reference standards.
Final answer
Using Ka(HNO2) = 4.5 × 10^-4 and Kw = 1.0 × 10^-14 at 25 degrees Celsius, the pH of a 0.010 M NaNO2 solution is approximately 7.67. This occurs because the nitrite ion is a weak base that hydrolyzes in water to generate a small amount of hydroxide.