Calculate the pH of 0.12 M KNO2
This interactive chemistry calculator determines the pH of a potassium nitrite solution by treating KNO2 as the salt of a strong base and a weak acid. Enter the concentration, Ka of nitrous acid, and your preferred solving method to get pH, pOH, Kb, hydroxide concentration, and a chart-based visualization.
KNO2 pH Calculator
Click Calculate pH to solve for the pH of 0.12 M KNO2.
Solution Profile
This chart compares key values used in the hydrolysis calculation for nitrite ion in water.
- KNO2 dissociates completely into K+ and NO2–.
- NO2– acts as a weak base because it is the conjugate base of HNO2.
- The solution is basic, so the final pH is greater than 7 at 25°C.
How to Calculate the pH of 0.12 M KNO2
If you need to calculate the pH of 0.12 M KNO2, the key idea is that potassium nitrite is not itself an acid. Instead, it is a salt made from the strong base KOH and the weak acid HNO2, also called nitrous acid. Because KNO2 contains the nitrite ion, NO2–, the solution undergoes basic hydrolysis in water. That means the solution will have a pH above 7. For the standard textbook value where the acid dissociation constant of HNO2 is approximately 4.0 × 10-4 at 25°C, the pH of 0.12 M KNO2 comes out to about 8.24.
Why KNO2 Makes a Basic Solution
When KNO2 dissolves in water, it dissociates almost completely:
KNO2 → K+ + NO2–
The potassium ion, K+, comes from the strong base KOH and does not significantly affect pH. The nitrite ion, however, is the conjugate base of nitrous acid. That means it can react with water and produce hydroxide ions:
NO2– + H2O ⇌ HNO2 + OH–
Because hydroxide ions are produced, the solution becomes basic. So, if you are asked to calculate the pH of 0.12 M KNO2, you should recognize immediately that this is a weak base hydrolysis problem, not a strong acid or strong base problem.
Step 1: Identify the Relevant Equilibrium Constant
Most chemistry problems provide the acid dissociation constant, Ka, for HNO2 rather than the base dissociation constant, Kb, for NO2–. To solve the problem, convert Ka to Kb using the relationship:
Kb = Kw / Ka
At 25°C:
- Kw = 1.0 × 10-14
- Ka(HNO2) = 4.0 × 10-4
So:
Kb = (1.0 × 10-14) / (4.0 × 10-4) = 2.5 × 10-11
Step 2: Set Up the ICE Table
Let the initial concentration of nitrite ion be 0.12 M. Since KNO2 dissociates fully, the starting concentration of NO2– is 0.12 M. Then define x as the amount that reacts with water:
- Initial: [NO2–] = 0.12, [HNO2] = 0, [OH–] = 0
- Change: [NO2–] = -x, [HNO2] = +x, [OH–] = +x
- Equilibrium: [NO2–] = 0.12 – x, [HNO2] = x, [OH–] = x
Substitute into the Kb expression:
Kb = x2 / (0.12 – x)
Using 2.5 × 10-11 for Kb:
2.5 × 10-11 = x2 / (0.12 – x)
Step 3: Solve for Hydroxide Ion Concentration
Because Kb is very small, x will also be very small compared with 0.12. That lets you use the weak base approximation:
0.12 – x ≈ 0.12
Then:
x2 = (2.5 × 10-11)(0.12) = 3.0 × 10-12
x = √(3.0 × 10-12) = 1.73 × 10-6
Since x = [OH–], the hydroxide concentration is:
[OH–] = 1.73 × 10-6 M
Step 4: Convert to pOH and pH
Now take the negative logarithm of the hydroxide concentration:
pOH = -log(1.73 × 10-6) ≈ 5.76
Finally:
pH = 14.00 – 5.76 = 8.24
That is the standard result for the problem statement “calculate the pH of 0.12 M KNO2.”
Exact Calculation vs Approximation
For very weak bases like nitrite, the approximation is usually excellent. Still, a precise calculator can solve the quadratic equation directly:
x2 + Kb x – Kb C = 0
With C = 0.12 M and Kb = 2.5 × 10-11, the exact solution produces nearly the same value for x, which means the pH remains approximately 8.24. This is why many general chemistry texts accept the square-root shortcut for this problem.
Comparison Table: Key Constants for the KNO2 pH Calculation
| Parameter | Typical Value | Why It Matters |
|---|---|---|
| Concentration of KNO2 | 0.12 M | This is the initial concentration of NO2– in solution. |
| Ka of HNO2 | 4.0 × 10-4 | Used to derive Kb for nitrite ion. |
| Kw at 25°C | 1.0 × 10-14 | Needed for converting Ka to Kb. |
| Kb of NO2– | 2.5 × 10-11 | Determines how much OH– forms. |
| [OH–] | 1.73 × 10-6 M | Intermediate concentration used for pOH. |
| Calculated pH | 8.24 | Final answer for the standard problem setup. |
How Concentration Changes the pH of KNO2
One of the best ways to understand weak base salts is to compare different concentrations. Because pH depends on the square root of concentration in the approximation, doubling or halving concentration does not produce dramatic changes. The solution becomes more basic as concentration increases, but the pH shifts gradually rather than explosively.
| KNO2 Concentration (M) | Approximate [OH–] (M) | Approximate pOH | Approximate pH |
|---|---|---|---|
| 0.010 | 5.00 × 10-7 | 6.30 | 7.70 |
| 0.050 | 1.12 × 10-6 | 5.95 | 8.05 |
| 0.120 | 1.73 × 10-6 | 5.76 | 8.24 |
| 0.500 | 3.54 × 10-6 | 5.45 | 8.55 |
| 1.000 | 5.00 × 10-6 | 5.30 | 8.70 |
Common Mistakes Students Make
- Treating KNO2 as neutral. Many salts are neutral, but not all. KNO2 is basic because nitrite is the conjugate base of a weak acid.
- Using Ka directly instead of Kb. The reacting species is NO2–, so you need the base constant or must convert Ka to Kb.
- Forgetting that K+ is a spectator ion. Potassium does not hydrolyze in water enough to affect pH under normal conditions.
- Mixing up pOH and pH. Once you find [OH–], you first calculate pOH, then convert to pH.
- Rounding too early. Keep several digits through the logarithm steps, then round the final pH appropriately.
When the Approximation Is Valid
The approximation x << C is valid when the percent ionization is small, usually less than 5%. In this problem, x is about 1.73 × 10-6 M while the initial concentration is 0.12 M. The percent change is tiny, far below 5%, so the approximation is completely justified. That is why classroom solutions and exam keys frequently use the square-root shortcut for this exact question.
Why This Problem Matters in General Chemistry
Weak acid and weak base salt problems test your understanding of conjugate pairs, hydrolysis, and equilibrium constants. KNO2 is a classic example because it forces you to recognize that the salt comes from a strong base and weak acid. Once you identify that pattern, the problem becomes much easier. The same logic applies to other salts such as sodium acetate, potassium cyanide, or ammonium chloride, except the acidic or basic direction depends on which ion hydrolyzes.
Authoritative Chemistry References
For deeper study of acid-base equilibria, hydrolysis, and equilibrium constants, consult these reliable educational and government resources:
- Chemistry LibreTexts educational resource
- National Institute of Standards and Technology (NIST)
- United States Environmental Protection Agency (EPA)
Practical Interpretation of the Final Answer
A pH of about 8.24 means the solution is mildly basic, not strongly alkaline. In a laboratory setting, that matters because indicator color changes, reaction conditions, and analyte stability may all depend on pH. If you are preparing a nitrite-containing solution for analytical chemistry, environmental testing, or equilibrium demonstrations, even a modest pH shift can influence side reactions and measurement accuracy. That is why it is useful to know not just that KNO2 is basic, but how basic it is at a specific molarity.
Summary of the Full Process
- Write the dissociation of KNO2 into K+ and NO2–.
- Identify NO2– as the conjugate base of weak acid HNO2.
- Use Kb = Kw / Ka to find the base constant.
- Set up the hydrolysis equilibrium NO2– + H2O ⇌ HNO2 + OH–.
- Solve for [OH–] using either the approximation or the exact quadratic method.
- Calculate pOH, then convert to pH.
Following those steps for a 0.12 M solution gives a pH of approximately 8.24, which is the accepted answer under standard 25°C conditions using Ka(HNO2) = 4.0 × 10-4.