Calculate the pH of NaNO2 Solution Given Ka of HNO2
This premium calculator finds the pH of a sodium nitrite solution by converting the acid dissociation constant of nitrous acid into the corresponding base constant for nitrite. Enter the NaNO2 concentration and the Ka of HNO2, then generate the pH, pOH, hydroxide concentration, and a concentration sensitivity chart.
Calculator Inputs
Calculated Results
pH vs NaNO2 concentration
This chart shows how the predicted pH changes as the sodium nitrite concentration varies around your selected input while using the same Ka value.
How to calculate the pH of NaNO2 solution given Ka of HNO2
Sodium nitrite, NaNO2, is a salt that dissociates essentially completely in water into Na+ and NO2–. The sodium ion is a spectator ion for acid base purposes, but the nitrite ion is not. Nitrite is the conjugate base of nitrous acid, HNO2, which is a weak acid. Because nitrite can accept a proton from water, an aqueous NaNO2 solution is basic. That is the central reason the pH is above 7 for typical sodium nitrite solutions.
When a problem gives the Ka of HNO2, it is giving you the acidity information for the acid side of the conjugate acid base pair. To calculate the pH of the salt solution, you first convert that acid dissociation constant into the corresponding base dissociation constant for NO2–. At 25 degrees Celsius, the relationship is:
Kb = Kw / Ka
Here, Kw is the ion product of water, usually taken as 1.0 × 10-14 at 25 degrees Celsius. Once you know Kb, you treat the nitrite ion as a weak base in water:
NO2– + H2O ⇌ HNO2 + OH–
The amount of hydroxide formed determines the pOH, and then the pH follows from pH = 14.00 – pOH under standard textbook conditions.
Step by step method
- Write the base hydrolysis reaction for NO2–.
- Convert Ka of HNO2 into Kb of NO2– using Kb = Kw / Ka.
- Set up an ICE table with initial nitrite concentration equal to the NaNO2 molarity.
- Solve for x = [OH–] either by the weak base approximation or by the quadratic equation.
- Compute pOH = -log[OH–].
- Compute pH = 14 – pOH.
Worked example
Suppose the NaNO2 concentration is 0.100 M and the given Ka of HNO2 is 4.5 × 10-4. First calculate Kb:
Kb = (1.0 × 10-14) / (4.5 × 10-4) = 2.22 × 10-11
Now set up the equilibrium for the nitrite ion:
Initial: [NO2–] = 0.100, [HNO2] = 0, [OH–] = 0
Change: -x, +x, +x
Equilibrium: 0.100 – x, x, x
The equilibrium expression is:
Kb = x2 / (0.100 – x)
Because Kb is very small, many classes use the approximation 0.100 – x ≈ 0.100. Then:
x ≈ √(Kb × C) = √(2.22 × 10-11 × 0.100) = 1.49 × 10-6 M
So:
pOH = -log(1.49 × 10-6) ≈ 5.83
pH = 14.00 – 5.83 = 8.17
This is exactly the kind of result you should expect: sodium nitrite is basic, but only mildly so, because nitrite is a weak base.
Why NaNO2 forms a basic solution
The direction of hydrolysis depends on the acid and base strengths of the ions formed when a salt dissolves. NaNO2 comes from a strong base, NaOH, and a weak acid, HNO2. The cation from a strong base does not significantly affect pH, but the anion from a weak acid does. Nitrite reacts with water to form a small amount of hydroxide, shifting the pH above neutral.
- Na+ is effectively neutral in water.
- NO2– is a weak base because it is the conjugate base of HNO2.
- The larger the Ka of HNO2, the smaller the Kb of NO2–.
- The smaller the Ka of HNO2, the stronger nitrite is as a base.
Exact versus approximate calculation
Most chemistry courses first teach the square root shortcut, where [OH–] ≈ √(KbC). This works well when x is tiny compared with the initial concentration. For sodium nitrite, that approximation is usually excellent because Kb is quite small. However, an exact calculator should be able to solve the quadratic equation without approximation. That is what the calculator above can do when the exact method is selected.
| Given quantity | Formula used | Meaning in this problem |
|---|---|---|
| Ka of HNO2 | Kb = Kw / Ka | Converts weak acid data into weak base strength for NO2– |
| NaNO2 molarity | C = initial [NO2–] | Sets the starting concentration for the hydrolysis equilibrium |
| [OH–] | x from Kb = x2/(C – x) | Directly determines pOH and then pH |
| pOH | -log[OH–] | Intermediate acidity scale for basic solutions |
| pH | 14 – pOH | Final answer at 25 degrees Celsius |
Typical values and concentration trends
A useful insight is that sodium nitrite becomes more basic as the concentration increases, but the increase is not linear. Since the weak base approximation gives [OH–] proportional to the square root of concentration, the pH rises gradually rather than dramatically. This is why a tenfold increase in concentration only shifts pH by a modest amount.
| NaNO2 concentration (M) | Assumed Ka of HNO2 | Approximate Kb of NO2– | Predicted pH at 25 C |
|---|---|---|---|
| 0.001 | 4.5 × 10-4 | 2.22 × 10-11 | 7.67 |
| 0.010 | 4.5 × 10-4 | 2.22 × 10-11 | 7.92 |
| 0.100 | 4.5 × 10-4 | 2.22 × 10-11 | 8.17 |
| 1.000 | 4.5 × 10-4 | 2.22 × 10-11 | 8.42 |
The values in this table show a realistic trend often seen in weak base salt solutions. Even at 1.0 M, the pH is only moderately basic because the nitrite ion does not hydrolyze strongly. This is a valuable comparison against salts of stronger weak bases, which can produce a more pronounced pH shift.
Common student mistakes
- Using Ka directly for the salt. NaNO2 contains the conjugate base NO2–, so you need Kb, not Ka.
- Forgetting Kw. The bridge between acid and base constants is Kb = Kw / Ka.
- Assuming the solution is acidic. Sodium nitrite is basic because it comes from a strong base and a weak acid.
- Calculating pH from x immediately. x here is [OH–], so find pOH first, then convert to pH.
- Ignoring temperature assumptions. The common relation pH + pOH = 14.00 is for 25 C textbook conditions.
When the approximation is safe
The square root approximation is typically acceptable when x is less than 5 percent of the initial concentration. For sodium nitrite, x is usually orders of magnitude smaller than the starting concentration, so the approximation passes the 5 percent rule easily. Still, using the exact quadratic method removes doubt and is preferred for a calculator, exam checking, or higher precision work.
Quick check with the 5 percent rule
Using the earlier example, x ≈ 1.49 × 10-6 M and the initial concentration is 0.100 M. The percent change is:
(1.49 × 10-6 / 0.100) × 100 = 0.00149%
This is far below 5 percent, so the approximation is excellent.
Practical interpretation of the answer
Knowing the pH of sodium nitrite solutions matters in analytical chemistry, environmental chemistry, and industrial process control. Nitrite chemistry appears in corrosion inhibition systems, curing chemistry, water testing contexts, and discussions of nitrogen species in aqueous environments. In most classroom problems, though, the goal is to reinforce conjugate acid base relationships and weak electrolyte equilibria.
If your assignment gives a different Ka than the standard reference value, always use the value from your problem statement. Small changes in Ka produce measurable changes in Kb and therefore in pH. That is why this calculator accepts the Ka directly instead of assuming a single fixed constant.
Authoritative references for further study
- NIST Chemistry WebBook (.gov)
- Purdue University General Chemistry acid-base review (.edu)
- University of Wisconsin acid-base learning resource (.edu)
Final summary
To calculate the pH of NaNO2 solution given Ka of HNO2, treat nitrite as a weak base. Convert Ka to Kb using Kb = Kw / Ka, solve for [OH–] from the base hydrolysis equilibrium, then convert hydroxide concentration into pOH and pH. This sequence is reliable, chemically correct, and directly tied to the conjugate acid base concept that appears throughout general chemistry.