Calculate the pH of a 0.050 M Solution of NaNO2
Use this interactive calculator to determine the pH of sodium nitrite solutions from weak acid-conjugate base chemistry. Adjust concentration, nitrous acid Ka, and temperature assumptions, then view the resulting pH, pOH, hydroxide concentration, and a responsive chart.
NaNO2 pH Calculator
Results
Enter your values and click Calculate pH to see the full solution.
pH Trend Chart
This chart compares pH against NaNO2 concentration using your selected Ka and temperature values.
How to Calculate the pH of a 0.050 M Solution of NaNO2
To calculate the pH of a 0.050 M solution of NaNO2, you need to recognize what sodium nitrite does in water. NaNO2 is a soluble ionic compound, so it dissociates essentially completely into Na+ and NO2–. The sodium ion is a spectator ion for acid-base purposes, but the nitrite ion is chemically important because it is the conjugate base of nitrous acid, HNO2. Since HNO2 is a weak acid, its conjugate base NO2– is basic enough to react with water and generate hydroxide ions. That means the solution will have a pH greater than 7.
This problem is a classic example of weak base hydrolysis from a salt of a weak acid and a strong base. Many students see a salt formula and initially wonder whether they should use a strong electrolyte model or an acid dissociation model. The right approach is to split the salt into ions first, then examine whether either ion reacts with water. For NaNO2, the answer is yes: nitrite reacts with water, and that hydrolysis determines the pH.
Step 1: Write the dissociation and hydrolysis reactions
The first equation is simple dissolution:
NaNO2 → Na+ + NO2–
Next, write the hydrolysis equilibrium for the nitrite ion:
NO2– + H2O ⇌ HNO2 + OH–
This equilibrium shows directly why the pH becomes basic. The production of OH– raises the pH above neutral.
Step 2: Convert Ka of HNO2 into Kb for NO2-
Most textbooks provide the acid dissociation constant for nitrous acid rather than the base dissociation constant for nitrite. A commonly used value at 25 C is:
Ka(HNO2) = 4.5 × 10-4
To get the base constant for nitrite, use:
Kb = Kw / Ka
At 25 C, Kw = 1.0 × 10-14, so:
Kb = (1.0 × 10-14) / (4.5 × 10-4) = 2.22 × 10-11
That is a small Kb, which tells you nitrite is a weak base. Even so, in a 0.050 M solution there is enough nitrite present to create a measurable hydroxide concentration.
Step 3: Set up the ICE table
Let the initial concentration of nitrite be 0.050 M. If x mol/L reacts with water, then:
- Initial: [NO2–] = 0.050, [HNO2] = 0, [OH–] = 0
- Change: [NO2–] = -x, [HNO2] = +x, [OH–] = +x
- Equilibrium: [NO2–] = 0.050 – x, [HNO2] = x, [OH–] = x
Substitute into the Kb expression:
Kb = x2 / (0.050 – x)
Now insert the value of Kb:
2.22 × 10-11 = x2 / (0.050 – x)
Step 4: Solve for hydroxide concentration
Because Kb is small relative to the initial concentration, the standard weak base approximation works very well:
0.050 – x ≈ 0.050
Then:
x2 = (2.22 × 10-11)(0.050) = 1.11 × 10-12
Taking the square root:
x = [OH–] = 1.05 × 10-6 M
Now compute pOH:
pOH = -log(1.05 × 10-6) ≈ 5.98
Finally:
pH = 14.00 – 5.98 = 8.02
Why the answer is only slightly basic
Students sometimes expect a much higher pH because the solution contains a base-producing ion. But nitrite is only a weak base. Its Kb is about 2.22 × 10-11, which is very small. That means only a tiny fraction of the 0.050 M nitrite actually reacts with water. The hydroxide concentration reaches the 10-6 M level, not the 10-3 or 10-2 M level typical of stronger bases. So the pH lands just above 8, not near 11 or 12.
Exact quadratic solution versus approximation
If you solve the full equilibrium expression without approximation, the equation becomes:
x2 + Kb x – KbC = 0
Using the quadratic formula gives nearly the same value for x because x is extremely small relative to 0.050 M. This is why the weak base shortcut is justified. In fact, the percent ionization is only:
(1.05 × 10-6 / 0.050) × 100 ≈ 0.0021%
Since the change is far less than 5%, the approximation is excellent.
Comparison table: pH at different NaNO2 concentrations
The concentration of sodium nitrite affects the pH, but because the base is weak, the pH rises gradually rather than dramatically. The following values are calculated at 25 C using Ka(HNO2) = 4.5 × 10-4.
| NaNO2 Concentration (M) | Kb for NO2- | Approx. [OH-] (M) | Approx. pOH | Approx. pH |
|---|---|---|---|---|
| 0.005 | 2.22 × 10-11 | 3.33 × 10-7 | 6.48 | 7.52 |
| 0.010 | 2.22 × 10-11 | 4.71 × 10-7 | 6.33 | 7.67 |
| 0.025 | 2.22 × 10-11 | 7.45 × 10-7 | 6.13 | 7.87 |
| 0.050 | 2.22 × 10-11 | 1.05 × 10-6 | 5.98 | 8.02 |
| 0.100 | 2.22 × 10-11 | 1.49 × 10-6 | 5.83 | 8.17 |
Temperature matters because Kw changes
Strictly speaking, pH calculations depend on temperature because the ion product of water changes. In many classroom problems, you assume 25 C, where pKw = 14.00. However, if temperature shifts, the relationship between pH and pOH changes slightly. The calculator above includes common temperature assumptions so you can see the effect directly.
| Temperature | Kw | pKw | Implication for NaNO2 pH Work |
|---|---|---|---|
| 20 C | 6.81 × 10-15 | 14.167 | Neutral pH is slightly above 7, so computed pH values shift upward a bit for the same hydroxide level. |
| 25 C | 1.00 × 10-14 | 14.000 | Standard textbook reference point used in most general chemistry problems. |
| 30 C | 1.47 × 10-14 | 13.833 | Neutral pH is slightly below 7, so final pH for the same pOH is a bit lower than the 25 C value. |
Common mistakes to avoid
- Treating NaNO2 as neutral. This would be wrong because nitrite is the conjugate base of a weak acid and therefore hydrolyzes water.
- Using Ka directly to find pH. You need Kb for NO2–, not Ka for HNO2, unless you reformulate the problem carefully through conjugate relationships.
- Assuming sodium ion affects pH. Na+ is from the strong base NaOH and does not significantly hydrolyze in water.
- Forgetting to convert pOH to pH. Once you find [OH–], the next step is pOH, then pH.
- Using the wrong value of Kw or pKw for temperature. At 25 C, pH + pOH = 14.00. At other temperatures, use the correct pKw.
How this problem fits into acid-base chemistry
This calculation sits at the intersection of salt hydrolysis, weak acid chemistry, and equilibrium approximations. It is useful because it teaches a reliable decision process:
- Identify whether the dissolved species are acidic, basic, or neutral.
- Determine whether the ion comes from a weak acid or weak base.
- Convert Ka to Kb or Kb to Ka when necessary.
- Use an ICE table and equilibrium expression.
- Evaluate whether the weak equilibrium approximation is valid.
Once you master this pattern, you can solve similar problems such as the pH of sodium acetate, ammonium chloride, potassium cyanide, and ammonium nitrite. The chemistry is the same in spirit even though the constants differ.
Authoritative resources for deeper study
If you want to verify properties of sodium nitrite or review acid-base concepts at a higher level, these sources are useful: the NIH PubChem sodium nitrite record summarizes compound identity and chemical data; Purdue University acid-base equilibrium materials provide a strong conceptual review; and the U.S. EPA pH reference page helps place pH in a broader scientific context.
Bottom line
To calculate the pH of a 0.050 M solution of NaNO2, treat nitrite as a weak base. Use the Ka of nitrous acid to find Kb for nitrite, set up the hydrolysis equilibrium, solve for hydroxide concentration, and convert to pH. With Ka(HNO2) = 4.5 × 10-4 at 25 C, the result is about pH = 8.02. That answer makes chemical sense because sodium nitrite is a salt of a strong base and a weak acid, so its aqueous solution is mildly basic rather than strongly alkaline.