Calculate the pH of a 0.18 M Solution of KNO2
Use this interactive calculator to find the pH, pOH, hydroxide concentration, hydronium concentration, and weak-base hydrolysis behavior of potassium nitrite. The calculator defaults to a 0.18 M KNO2 solution at 25 degrees Celsius using a standard Ka value for nitrous acid.
Interactive KNO2 pH Calculator
NO2- + H2O ⇌ HNO2 + OH-
Kb = Kw / Ka
For a weak base approximation: [OH-] ≈ √(Kb × C)
Visual Output
This chart compares the calculated pH, pOH, pKa, and pKb values so you can quickly see why a potassium nitrite solution is basic rather than neutral or acidic.
Expert Guide: How to Calculate the pH of a 0.18 M Solution of KNO2
To calculate the pH of a 0.18 M solution of potassium nitrite, you need to recognize that KNO2 is a salt formed from a strong base and a weak acid. Potassium ion, K+, comes from potassium hydroxide, a strong base that does not significantly affect pH in aqueous solution. The nitrite ion, NO2-, is the conjugate base of nitrous acid, HNO2, which is a weak acid. That means nitrite can react with water to produce hydroxide ions, OH-, making the solution basic.
This is the central reason the pH is above 7. Students often make the mistake of assuming all salts are neutral in water. That is only true for salts made from a strong acid and a strong base, such as NaCl. For KNO2, the anion hydrolyzes in water. Once you see that clearly, the problem becomes a straightforward weak-base equilibrium calculation.
Step 1: Write the Dissociation and Hydrolysis Equations
Potassium nitrite dissociates completely in water:
- KNO2 → K+ + NO2-
The potassium ion is a spectator ion for acid-base purposes. The nitrite ion undergoes hydrolysis:
- NO2- + H2O ⇌ HNO2 + OH-
Because hydroxide ions are produced, the solution is basic.
Step 2: Convert Ka of HNO2 into Kb of NO2-
Most tables list the acid dissociation constant for nitrous acid rather than the base dissociation constant for nitrite. At 25 degrees Celsius, a commonly used value is:
- Ka(HNO2) = 4.5 × 10-4
- Kw = 1.0 × 10-14
The relationship between conjugate acid-base pairs is:
- Kb = Kw / Ka
So for nitrite:
- Kb = (1.0 × 10-14) / (4.5 × 10-4)
- Kb = 2.22 × 10-11
This very small Kb shows that nitrite is a weak base, but since the concentration is fairly high at 0.18 M, enough OH- is generated to make the solution noticeably basic.
Step 3: Set Up the ICE Table
Let the initial concentration of nitrite be 0.18 M. Then the hydrolysis equilibrium is:
- NO2- + H2O ⇌ HNO2 + OH-
An ICE table gives:
- Initial: [NO2-] = 0.18, [HNO2] = 0, [OH-] = 0
- Change: [NO2-] = -x, [HNO2] = +x, [OH-] = +x
- Equilibrium: [NO2-] = 0.18 – x, [HNO2] = x, [OH-] = x
The equilibrium expression is:
- Kb = x2 / (0.18 – x)
Step 4: Use the Weak Base Approximation
Because Kb is very small, x is much smaller than 0.18, so we approximate 0.18 – x as 0.18:
- Kb ≈ x2 / 0.18
- x2 ≈ Kb × 0.18
- x ≈ √(2.22 × 10-11 × 0.18)
- x ≈ √(4.00 × 10-12)
- x ≈ 2.00 × 10-6 M
This x value is the hydroxide concentration:
- [OH-] = 2.00 × 10-6 M
Step 5: Convert OH- to pOH and Then to pH
Now calculate pOH:
- pOH = -log(2.00 × 10-6)
- pOH ≈ 5.70
At 25 degrees Celsius:
- pH + pOH = 14.00
So:
- pH = 14.00 – 5.70 = 8.30
Final answer: the pH of a 0.18 M KNO2 solution is about 8.30.
Why the Approximation Works
In weak acid and weak base calculations, the square-root approximation is only valid when the amount that reacts is small relative to the initial concentration. Here, x is about 2.00 × 10-6, while the initial concentration is 0.18 M. The percent ionization is:
- (2.00 × 10-6 / 0.18) × 100 ≈ 0.0011%
Since this is much less than 5%, the approximation is excellent.
Exact Versus Approximate Calculation
If you solve the quadratic equation exactly, the result is essentially the same to practical significant figures. That is why chemistry instructors often expect the approximation method on homework and exams. It is fast, accurate, and reveals the underlying logic of the equilibrium system. The exact solution differs only in the farthest decimal places for a problem like this.
| Quantity | Value for 0.18 M KNO2 | Meaning |
|---|---|---|
| Ka of HNO2 | 4.5 × 10-4 | Weak acid strength of nitrous acid |
| Kb of NO2- | 2.22 × 10-11 | Weak base strength of nitrite |
| [OH-] | 2.00 × 10-6 M | Hydroxide formed by hydrolysis |
| pOH | 5.70 | Basicity expressed on pOH scale |
| pH | 8.30 | Final solution acidity-basicity |
How KNO2 Compares with Other Common Salts
A good way to understand this result is to compare potassium nitrite with salts that produce neutral, acidic, and basic solutions. The cation and anion both matter. If neither hydrolyzes, the pH stays near 7. If the anion is the conjugate base of a weak acid, the solution becomes basic. If the cation is the conjugate acid of a weak base, the solution becomes acidic.
| Salt | Parent Acid | Parent Base | Expected pH Trend in Water |
|---|---|---|---|
| NaCl | HCl, strong acid | NaOH, strong base | Approximately neutral, near pH 7 |
| NH4Cl | HCl, strong acid | NH3, weak base | Acidic |
| CH3COONa | CH3COOH, weak acid | NaOH, strong base | Basic |
| KNO2 | HNO2, weak acid | KOH, strong base | Basic, about pH 8.30 at 0.18 M |
Common Mistakes in This Problem
- Treating KNO2 as a strong base. KNO2 is not a base like KOH. It is a salt whose anion behaves as a weak base.
- Using Ka directly instead of converting to Kb. Since the reacting species is NO2-, you need the base constant.
- Forgetting that K+ is a spectator ion. Potassium does not hydrolyze appreciably in water.
- Calculating pH directly from [OH-]. First compute pOH, then convert to pH.
- Assuming all salts are neutral. Salt hydrolysis is one of the most tested conceptual topics in aqueous equilibria.
Why Nitrite Is a Weak Base
Nitrite is the conjugate base of nitrous acid. Because HNO2 is not a strong acid, its conjugate base retains some tendency to accept a proton from water. This proton-accepting process generates OH-. The stronger the parent acid, the weaker its conjugate base. Since nitrous acid is only moderately weak, nitrite is a weak base, though still much weaker than classic strong bases.
Practical Context for Nitrite Chemistry
Nitrite chemistry matters in environmental science, analytical chemistry, food chemistry, and water treatment. Nitrite and nitrate species are routinely monitored in natural waters and engineered systems because they participate in oxidation-reduction chemistry, biological nitrogen cycling, and public health related testing. In many educational settings, nitrite salts are used to teach the link between equilibrium constants and pH behavior of conjugate acid-base pairs.
For deeper reference material, consult authoritative educational and government sources such as the LibreTexts chemistry resource for equilibrium concepts, the U.S. Environmental Protection Agency for water chemistry context, and the NIST Chemistry WebBook for chemical data. For academic support, many university general chemistry pages, including those from MIT Chemistry, also explain conjugate acid-base relationships and equilibrium methods.
Short Exam-Style Solution
If you need a concise version for classwork, use this format:
- KNO2 is a salt of a strong base and weak acid, so the solution is basic.
- NO2- + H2O ⇌ HNO2 + OH-
- Kb = Kw / Ka = (1.0 × 10-14) / (4.5 × 10-4) = 2.22 × 10-11
- [OH-] ≈ √(KbC) = √[(2.22 × 10-11)(0.18)] = 2.00 × 10-6 M
- pOH = 5.70
- pH = 14.00 – 5.70 = 8.30
That complete sequence is enough to earn full credit in many chemistry courses, provided your instructor permits the approximation method.
Final Takeaway
When asked to calculate the pH of a 0.18 M solution of KNO2, the key insight is that the nitrite ion is the conjugate base of nitrous acid. After converting Ka to Kb and applying the weak-base equilibrium approximation, you find a hydroxide concentration of about 2.00 × 10-6 M, a pOH of 5.70, and a final pH of 8.30. That makes the solution mildly basic, exactly as expected for a salt derived from a strong base and a weak acid.