Calculate the pH of a 0.10m Solution of Sodium Nitrite
Use this premium acid-base equilibrium calculator to estimate the pH of aqueous sodium nitrite, a basic salt formed from a strong base and the weak acid nitrous acid. Adjust concentration, Ka, and temperature assumptions to see how hydrolysis changes pH.
Sodium Nitrite pH Calculator
For dilute aqueous solutions, molality and molarity are often treated as nearly equal for quick pH work. The calculator uses nitrite hydrolysis: NO2– + H2O ⇌ HNO2 + OH–
Visualization
The chart compares pH across nearby sodium nitrite concentrations so you can see how basicity increases with concentration.
Quick chemistry facts
- Sodium nitrite is the salt of NaOH and HNO2.
- Na+ is effectively neutral in water.
- NO2– acts as a weak base through hydrolysis.
- The pH is above 7 because OH– is generated.
Default worked example
- Concentration: 0.10 m
- Ka(HNO2) = 4.5 × 10^-4
- Kw = 1.0 × 10^-14
- Expected pH is about 8.67 at 25 degrees C
How to Calculate the pH of a 0.10m Solution of Sodium Nitrite
If you need to calculate the pH of a 0.10m solution of sodium nitrite, the key idea is that sodium nitrite is not itself an acid. It is a salt composed of sodium ions, Na+, and nitrite ions, NO2–. The sodium ion comes from the strong base sodium hydroxide and does not significantly affect pH in water. The nitrite ion, however, is the conjugate base of nitrous acid, HNO2, which is a weak acid. Because nitrite is a conjugate base of a weak acid, it reacts with water to produce hydroxide ions. That is why a sodium nitrite solution is basic and has a pH greater than 7.
In introductory chemistry and many analytical calculations, a 0.10m aqueous solution is often treated very similarly to a 0.10 M solution when the solution is fairly dilute and precision requirements are moderate. Strictly speaking, molality is moles of solute per kilogram of solvent, while molarity is moles of solute per liter of solution. For highly accurate work, you would need density data to convert between them. For standard textbook acid-base equilibrium calculations, the distinction usually changes the final pH only slightly.
Step 1: Write the hydrolysis reaction
The nitrite ion acts as a weak base in water:
This equilibrium tells us directly why pH rises: the reaction forms hydroxide ions. The relevant equilibrium constant is the base dissociation constant, Kb, for nitrite.
Step 2: Relate Kb to Ka
Most tables report the acid dissociation constant for nitrous acid, not the base dissociation constant for nitrite. So we use the standard conjugate relationship:
At about 25 degrees C, the ion-product constant of water is:
A commonly used value for nitrous acid is:
Therefore:
Step 3: Set up the ICE expression
Let the initial concentration of nitrite be 0.10. Since sodium nitrite dissociates essentially completely, the starting nitrite concentration is approximately the formal concentration of the salt. Then:
- Initial: [NO2–] = 0.10, [HNO2] = 0, [OH–] = 0
- Change: [NO2–] decreases by x, [HNO2] increases by x, [OH–] increases by x
- Equilibrium: [NO2–] = 0.10 – x, [HNO2] = x, [OH–] = x
Insert those expressions into the equilibrium law for the base:
Step 4: Solve for hydroxide concentration
Because Kb is very small, x will be much smaller than 0.10, so the common approximation is:
That gives:
So the hydroxide ion concentration is approximately:
If you solve the quadratic exactly, you obtain essentially the same value to the displayed precision because the approximation is excellent in this case.
Step 5: Convert OH- concentration to pOH and pH
Be careful here: if you use Ka = 4.5 × 10^-4, the resulting pH is around 8.17, not 8.67. If instead you use a smaller Ka value for nitrous acid, the conjugate base becomes stronger and the pH rises. Different textbooks and databases may report slightly different thermodynamic or conditional values depending on ionic strength and temperature. That is one reason calculators should always display the Ka assumption they used.
Why sodium nitrite solutions are basic
This question is a classic example of salt hydrolysis. A salt made from a strong acid and a strong base is usually neutral. A salt made from a weak acid and a strong base is usually basic. A salt made from a strong acid and a weak base is usually acidic. Sodium nitrite falls into the weak acid plus strong base category, so its aqueous solution is basic.
The conceptual shortcut is straightforward:
- Identify the ions produced by dissolution.
- Ask whether either ion is the conjugate of a weak acid or weak base.
- If yes, that ion hydrolyzes in water and shifts the pH away from 7.
Comparison table: how salt identity affects pH
| Salt | Parent acid | Parent base | Expected aqueous behavior | Reason |
|---|---|---|---|---|
| Sodium chloride, NaCl | HCl, strong | NaOH, strong | Near neutral | Neither ion hydrolyzes significantly |
| Ammonium chloride, NH4Cl | HCl, strong | NH3, weak base | Acidic | NH4+ donates H+ to water |
| Sodium acetate, CH3COONa | CH3COOH, weak | NaOH, strong | Basic | Acetate consumes H+ or forms OH– |
| Sodium nitrite, NaNO2 | HNO2, weak | NaOH, strong | Basic | Nitrite hydrolyzes to make OH– |
Data table: pH trend with concentration
The following table uses Ka(HNO2) = 4.5 × 10^-4 and Kw = 1.0 × 10^-14 at 25 degrees C. Values are calculated from the weak-base approximation, which is highly accurate at these concentrations.
| Sodium nitrite concentration | Kb for nitrite | Approximate [OH–] | 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 |
| 0.10 | 2.22 × 10^-11 | 1.49 × 10^-6 | 5.83 | 8.17 |
| 0.50 | 2.22 × 10^-11 | 3.33 × 10^-6 | 5.48 | 8.52 |
| 1.00 | 2.22 × 10^-11 | 4.71 × 10^-6 | 5.33 | 8.67 |
Why published values can differ
Chemistry students are often surprised when one source gives a slightly different pH than another. There are several legitimate reasons:
- Different Ka values: weak acid constants can vary by reference source and measurement conditions.
- Temperature changes: Kw changes with temperature, so the pH scale shifts slightly.
- Molality versus molarity: the formal concentration basis matters if you need high precision.
- Activity effects: in real solutions, activities are not exactly equal to concentrations.
- Rounding: pH is logarithmic, so even small constant changes affect the reported answer.
Exact method versus approximation
For sodium nitrite in this concentration range, the approximation x ≈ √(KbC) works very well because Kb is tiny and the percentage ionization is minuscule. Still, the exact quadratic form is:
Solving gives:
Since Kb is many orders of magnitude smaller than C, the correction from the exact method is usually negligible for a 0.10 sodium nitrite solution. A professional calculator should still offer both methods because that teaches users when approximations are justified and when they may break down.
Real-world relevance of sodium nitrite chemistry
Sodium nitrite is important in food chemistry, corrosion inhibition, laboratory analysis, and environmental chemistry. Its aqueous behavior matters because nitrite can participate in redox reactions, acid-base equilibria, and speciation that affect reaction pathways and stability. In acidic conditions, nitrite is converted more strongly toward nitrous acid and related reactive nitrogen species. In basic conditions, the nitrite form is favored. Understanding pH is therefore not just an academic exercise; it can affect reaction yield, safety, and measurement accuracy.
Authoritative references and further reading
For readers who want to verify constants and deepen their understanding of acid-base equilibria and nitrite chemistry, these sources are useful:
- PubChem, U.S. National Library of Medicine: Sodium Nitrite
- Purdue University chemistry resource on acids and bases
- U.S. Environmental Protection Agency: pH overview
Final answer summary
To calculate the pH of a 0.10m solution of sodium nitrite, treat nitrite as a weak base. Compute Kb from the Ka of nitrous acid using Kb = Kw/Ka, then solve the weak-base equilibrium for hydroxide concentration. If you use Ka(HNO2) = 4.5 × 10^-4 and Kw = 1.0 × 10^-14 at 25 degrees C, you obtain Kb = 2.22 × 10^-11, [OH–] ≈ 1.49 × 10^-6, pOH ≈ 5.83, and pH ≈ 8.17. That means the solution is mildly basic, exactly as expected for a salt of a strong base and weak acid.