Calculate the pH of 0.12 M KNO2
Use this interactive chemistry calculator to determine the pH, pOH, hydroxide concentration, and hydrolysis behavior of potassium nitrite in water at 25 degrees Celsius.
KNO2 pH Calculator
For a salt such as KNO2, K+ is a spectator ion from the strong base KOH, while NO2- acts as a weak base because it is the conjugate base of the weak acid HNO2.
Results
Default inputs are set for 0.12 M KNO2 with Ka(HNO2) = 4.5 × 10-4 at 25 degrees Celsius.
How to calculate the pH of 0.12 M KNO2
To calculate the pH of 0.12 M KNO2, you need to recognize the chemistry of the salt first. Potassium nitrite, KNO2, dissociates completely in water into K+ and NO2-. The potassium ion does not significantly affect pH because it comes from the strong base potassium hydroxide. The nitrite ion, however, is the conjugate base of nitrous acid, HNO2, which is a weak acid. That means NO2- reacts with water to produce a small amount of OH-, making the solution basic.
The key hydrolysis reaction is:
NO2- + H2O ⇌ HNO2 + OH-
Because hydroxide ions are produced, the pH of a KNO2 solution is greater than 7. For a 0.12 M solution, the basicity is moderate rather than extreme. This type of problem is a classic example of calculating the pH of a salt derived from a strong base and a weak acid.
Step 1: Identify the correct equilibrium constant
You are usually given or expected to know the acid dissociation constant for nitrous acid. A commonly used value at 25 degrees Celsius is:
- Ka for HNO2 = 4.5 × 10-4
- Kw = 1.0 × 10-14
Since nitrite is acting as a base, you need the base dissociation constant, Kb. Convert Ka to Kb using the relation:
Kb = Kw / Ka
Substitute the values:
Kb = (1.0 × 10-14) / (4.5 × 10-4) = 2.22 × 10-11
Step 2: Set up the ICE table
For the reaction NO2- + H2O ⇌ HNO2 + OH-, start with 0.12 M NO2- and assume initially that HNO2 and OH- from hydrolysis are approximately zero compared with the salt concentration. Then define x as the amount of hydroxide formed.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| NO2- | 0.12 | -x | 0.12 – x |
| HNO2 | 0 | +x | x |
| OH- | 0 | +x | x |
The equilibrium expression becomes:
Kb = [HNO2][OH-] / [NO2-] = x2 / (0.12 – x)
Step 3: Solve for x
Because Kb is very small, x will be much smaller than 0.12. So the common approximation is:
0.12 – x ≈ 0.12
This gives:
x2 / 0.12 = 2.22 × 10-11
x2 = 2.664 × 10-12
x = 1.63 × 10-6 M
That x value is the hydroxide concentration:
[OH-] = 1.63 × 10-6 M
Step 4: Convert to pOH and pH
Now calculate pOH:
pOH = -log(1.63 × 10-6) = 5.79
Then use the relationship:
pH = 14.00 – pOH
So:
pH = 14.00 – 5.79 = 8.21
Why KNO2 gives a basic solution
Students often ask why a salt can change pH at all. The reason is that salts are made from acids and bases, and the strengths of those parent species matter. KNO2 comes from KOH, a strong base, and HNO2, a weak acid. The cation K+ does not hydrolyze noticeably in water, but the anion NO2- does. It steals a proton from water, generating OH-. That is why the solution becomes basic.
If the salt had instead come from a strong acid and a strong base, such as KCl, the solution would be nearly neutral. If it came from a weak base and a strong acid, the solution would be acidic. This classification approach helps you quickly predict pH behavior before you do any math.
Quick classification of salt solutions
- Strong acid + strong base: approximately neutral
- Weak acid + strong base: basic
- Strong acid + weak base: acidic
- Weak acid + weak base: depends on relative Ka and Kb
Approximation versus exact quadratic solution
For weak base hydrolysis problems, the approximation method is often more than adequate. In this case, the amount of nitrite that reacts is tiny compared with the starting concentration. The percent ionization is only about 0.0014 percent, which strongly justifies the approximation.
Still, if you solve the equilibrium exactly using the quadratic formula, you get essentially the same result. That is why this calculator offers both methods. The exact method is useful when concentrations are lower, equilibrium constants are larger, or when your course specifically requires rigorous algebra.
| Method | Equation Used | Calculated [OH-] | Calculated pH |
|---|---|---|---|
| Approximation | x = √(KbC) | 1.63 × 10-6 M | 8.21 |
| Quadratic | x2 + Kb x – KbC = 0 | 1.63 × 10-6 M | 8.21 |
Effect of concentration on pH
As the concentration of KNO2 increases, the hydroxide concentration generated by hydrolysis also increases, but not in a simple one to one way. Because hydrolysis depends on an equilibrium relation with a square root behavior under the weak base approximation, a large increase in concentration produces a smaller increase in pH than many students expect.
The table below shows how pH changes for several concentrations of KNO2 using Ka(HNO2) = 4.5 × 10-4 and Kw = 1.0 × 10-14 at 25 degrees Celsius.
| KNO2 Concentration (M) | Kb for NO2- | Approximate [OH-] (M) | pOH | pH |
|---|---|---|---|---|
| 0.010 | 2.22 × 10-11 | 4.71 × 10-7 | 6.33 | 7.67 |
| 0.050 | 2.22 × 10-11 | 1.05 × 10-6 | 5.98 | 8.02 |
| 0.120 | 2.22 × 10-11 | 1.63 × 10-6 | 5.79 | 8.21 |
| 0.500 | 2.22 × 10-11 | 3.33 × 10-6 | 5.48 | 8.52 |
Common mistakes when solving KNO2 pH problems
- Treating KNO2 as a strong base. KNO2 is not KOH. It is a salt whose anion hydrolyzes weakly, so the pH is only mildly basic.
- Using Ka directly instead of converting to Kb. The reacting species is NO2-, so you need the base constant for nitrite.
- Forgetting that K+ is a spectator ion. Potassium generally does not affect pH in aqueous solution.
- Confusing M with m. In chemistry, capital M usually means molarity. If your instructor writes 0.12 m instead, check whether they mean molal concentration. Many classroom examples still intend molarity in quick pH exercises, but it is important to verify the notation.
- Skipping the final pH conversion. Once you have [OH-], calculate pOH first, then pH.
What if the problem literally says 0.12 m instead of 0.12 M?
This is an important nuance. In strict chemical notation, lowercase m means molality, while uppercase M means molarity. Molality is moles of solute per kilogram of solvent, whereas molarity is moles of solute per liter of solution. For dilute aqueous solutions, these can be numerically close, especially near room temperature, but they are not identical definitions.
If the problem is from a general chemistry homework set, many instructors and websites really intend 0.12 M when asking for pH of a salt solution. If the concentration is truly 0.12 m, then a more rigorous treatment would require solution density information to convert to exact molarity. In the absence of that information, many practical calculations assume the values are close enough for an introductory estimate. This calculator is built around the standard aqueous pH approach using molarity, but it allows you to enter any concentration value you want.
How this topic connects to broader acid base chemistry
The KNO2 example teaches several core equilibrium ideas. First, conjugate acid base pairs are central to predicting pH. Second, the strength of an acid determines the weakness of its conjugate base. Since HNO2 is a weak acid, NO2- is a weak base. Third, equilibrium calculations often benefit from approximations, but those approximations should always be checked for validity. Finally, this kind of problem shows how pH is governed by chemical speciation, not simply by whether a compound contains a metal or nonmetal.
Understanding these patterns helps with related compounds such as sodium acetate, sodium fluoride, ammonium chloride, and sodium cyanide. In each case, identifying the conjugate acid or conjugate base tells you whether the solution will be acidic, basic, or nearly neutral.
Authoritative references for acid base constants and aqueous chemistry
If you want to validate equilibrium constants or review acid base principles from trusted sources, these references are useful:
- National Institute of Standards and Technology (NIST)
- Chemistry LibreTexts
- United States Environmental Protection Agency (EPA)
Final takeaway
To calculate the pH of 0.12 M KNO2, treat nitrite as a weak base, convert the acid constant of HNO2 to Kb, solve for hydroxide concentration, and then convert to pOH and pH. Using Ka = 4.5 × 10-4 at 25 degrees Celsius gives Kb = 2.22 × 10-11, [OH-] = 1.63 × 10-6 M, pOH = 5.79, and pH = 8.21. The result is a mildly basic solution, exactly what you should expect for the salt of a strong base and a weak acid.