Calculate the pH of 0.90 M KNO2
Use this premium acid-base calculator to find the pH of potassium nitrite solution from equilibrium chemistry. The tool uses the hydrolysis of the nitrite ion, exact equilibrium math, and a live Chart.js visualization.
KNO2 pH Calculator
NO2^- + H2O ⇌ HNO2 + OH^-
Kb = Kw / Ka
x = [OH^-]
Exact: x = (-Kb + sqrt(Kb^2 + 4KbC)) / 2
pOH = -log10(x)
pH = 14 - pOH
Quick Answer
For a 0.90 M KNO2 solution using Ka(HNO2) = 4.5 × 10^-4 and Kw = 1.0 × 10^-14, the pH is about 8.65.
- KNO2 is a salt of a strong base and weak acid.
- The nitrite ion behaves as a weak base in water.
- The solution is basic, so pH is greater than 7.
What the chart shows
The graph compares the starting nitrite concentration with the equilibrium concentrations of OH^- and HNO2. This helps visualize why the pH increase is measurable even though only a small fraction hydrolyzes.
How to calculate the pH of 0.90 M KNO2
To calculate the pH of 0.90 M KNO2, you must recognize what kind of compound potassium nitrite is and how it behaves in water. KNO2 is the salt of potassium hydroxide, a strong base, and nitrous acid, HNO2, a weak acid. Because potassium ions do not significantly react with water, the chemistry is controlled by the nitrite ion, NO2^-. Nitrite is the conjugate base of nitrous acid, so it undergoes base hydrolysis and produces hydroxide ions. That hydroxide generation is what makes the solution basic.
The key equilibrium is:
NO2^- + H2O ⇌ HNO2 + OH^-
This means the problem is not solved by plugging 0.90 directly into a pH formula for strong bases. Instead, you first convert the acid dissociation constant of nitrous acid into a base dissociation constant for nitrite:
Kb = Kw / Ka
If you use a typical 25 C textbook value of Ka(HNO2) = 4.5 × 10^-4 and Kw = 1.0 × 10^-14, then:
Kb = 1.0 × 10^-14 / 4.5 × 10^-4 = 2.22 × 10^-11
With an initial nitrite concentration of 0.90 M, let x be the hydroxide concentration produced at equilibrium. The ICE setup becomes:
- Initial: [NO2^-] = 0.90, [HNO2] = 0, [OH^-] = 0
- Change: [NO2^-] = -x, [HNO2] = +x, [OH^-] = +x
- Equilibrium: [NO2^-] = 0.90 – x, [HNO2] = x, [OH^-] = x
Substitute into the equilibrium expression:
Kb = x^2 / (0.90 – x)
Since Kb is very small relative to 0.90, many instructors allow the approximation 0.90 – x ≈ 0.90. Then:
x ≈ sqrt(KbC) = sqrt((2.22 × 10^-11)(0.90)) = 4.47 × 10^-6 M
This x is the hydroxide concentration. So:
pOH = -log(4.47 × 10^-6) = 5.35
pH = 14.00 – 5.35 = 8.65
Why KNO2 gives a basic solution
Students often ask why a neutral-looking salt like KNO2 does not simply have pH 7. The answer comes from the strengths of the parent acid and base. Potassium comes from KOH, which is a strong base and dissociates completely. The cation K^+ is therefore effectively neutral in water. Nitrite, however, comes from nitrous acid, which is weak and does not fully ionize. Because HNO2 is weak, its conjugate base NO2^- has enough affinity for protons to pull a proton from water and generate OH^-.
That is the same pattern used for many salt hydrolysis problems:
- Identify the parent acid and base.
- Determine whether the ions are spectators or reactive.
- Write the hydrolysis equation for the reactive ion.
- Use Ka and Kw to calculate Kb if needed.
- Solve for [OH^-] or [H3O^+], then convert to pH.
If the salt were from a weak base and strong acid, the solution would be acidic instead. If the salt came from a strong acid and strong base, the solution would usually be near neutral. This classification step saves time and prevents using the wrong equation.
Step-by-step method for homework and exams
- Write the dissociation of KNO2: KNO2 → K^+ + NO2^-.
- Ignore K^+ because it is the conjugate acid of a strong base and does not hydrolyze significantly.
- Write the nitrite hydrolysis equation: NO2^- + H2O ⇌ HNO2 + OH^-.
- Calculate Kb from Ka using Kb = Kw / Ka.
- Set up an ICE table with initial concentration 0.90 M for NO2^-.
- Solve for x, which equals [OH^-] at equilibrium.
- Find pOH from pOH = -log[OH^-].
- Calculate pH with pH = 14 – pOH.
Exact versus approximate calculation
In many general chemistry settings, the square root approximation is acceptable because the hydrolysis is small. However, premium problem solving means understanding both the fast approximation and the exact quadratic approach. The exact equilibrium equation is:
x^2 + Kbx – KbC = 0
Solving for the positive root gives:
x = (-Kb + sqrt(Kb^2 + 4KbC)) / 2
For this specific problem, the exact and approximate answers are practically the same because x is tiny compared with 0.90 M. That agreement also passes the common 5 percent rule check. Since x / 0.90 is far below 5 percent, the approximation is fully justified. In other words, the chemistry is weak enough that the starting nitrite concentration barely changes, but the generated hydroxide is still large enough to push the pH above neutral.
| Parameter | Value used | Meaning in the calculation |
|---|---|---|
| KNO2 concentration | 0.90 M | Initial concentration of nitrite ion from complete salt dissociation |
| Ka of HNO2 | 4.5 × 10^-4 | Acid strength of nitrous acid, used to derive Kb of nitrite |
| Kw at 25 C | 1.0 × 10^-14 | Water ion product used in the relation KaKb = Kw |
| Kb of NO2^- | 2.22 × 10^-11 | Base strength that determines OH^- production |
| [OH^-] at equilibrium | 4.47 × 10^-6 M | Hydroxide concentration generated by hydrolysis |
| Final pH | 8.65 | Basic solution result for 0.90 M KNO2 |
Comparison table: how concentration affects pH for KNO2
One excellent way to understand this problem is to compare several KNO2 concentrations using the same Ka value for HNO2. Because [OH^-] is approximately proportional to the square root of concentration for a weak base, the pH rises more slowly than concentration itself. A tenfold increase in concentration does not create a tenfold increase in pH. Instead, the pH shifts modestly.
| KNO2 concentration (M) | Approximate [OH^-] (M) | Approximate pOH | Approximate pH at 25 C |
|---|---|---|---|
| 0.010 | 4.71 × 10^-7 | 6.33 | 7.67 |
| 0.050 | 1.05 × 10^-6 | 5.98 | 8.02 |
| 0.10 | 1.49 × 10^-6 | 5.83 | 8.17 |
| 0.50 | 3.33 × 10^-6 | 5.48 | 8.52 |
| 0.90 | 4.47 × 10^-6 | 5.35 | 8.65 |
| 1.00 | 4.71 × 10^-6 | 5.33 | 8.67 |
Common mistakes when solving for the pH of KNO2
- Treating KNO2 as a strong base. KNO2 is not OH^- and does not fully produce hydroxide. It is a weakly basic salt.
- Using Ka directly in the ICE table. The reacting species is NO2^-, so you need Kb, not Ka.
- Forgetting that potassium is a spectator ion. K^+ does not control the pH here.
- Confusing pOH with pH. Once you find [OH^-], you must calculate pOH first, then convert to pH.
- Ignoring temperature assumptions. The common relation pH + pOH = 14 is tied to Kw = 1.0 × 10^-14 at 25 C.
Why the result depends on Ka data
Different textbooks and reference sources may report slightly different Ka values for nitrous acid, often due to rounding, data set differences, or temperature conditions. You may see values near 4.0 × 10^-4, 4.5 × 10^-4, or similar. That means your final pH can vary slightly, often by only a few hundredths. For high-quality work, always cite the Ka value you used. If your instructor gives a specific value, use that one instead of a generic table value.
For example, if you used Ka = 4.0 × 10^-4 instead of 4.5 × 10^-4, Kb would be a little larger, and the pH would be a little higher. The qualitative conclusion does not change: the solution remains basic. But in graded chemistry problems, matching the expected reference value can matter.
Practical significance of pH around 8.65
A pH of 8.65 is moderately basic, not strongly caustic. This level is far less basic than concentrated sodium hydroxide solutions, but it is clearly above neutral and absolutely large enough to affect acid-base indicators, equilibrium behavior, and many water chemistry measurements. In laboratory settings, weakly basic salt solutions like KNO2 are useful examples because they show how conjugate bases shift the pH even when they are not strong bases themselves.
In environmental and analytical chemistry, pH values in the upper 8 range can influence metal solubility, biological systems, and indicator color transitions. The underlying lesson is bigger than this one exercise: pH is often determined by equilibrium chemistry, not just by whether a compound “looks acidic” or “looks basic” at first glance.
Authoritative references for acid-base and pH concepts
If you want to cross-check pH fundamentals and water chemistry concepts, these authoritative resources are useful:
- USGS: pH and Water
- U.S. EPA: Alkalinity and Acid Neutralizing Capacity
- Purdue University: Acid-Base Equilibrium Help
Final takeaway
To calculate the pH of 0.90 M KNO2, first identify KNO2 as a salt containing the conjugate base of a weak acid. Then convert the known Ka of HNO2 into Kb for NO2^-, solve the hydrolysis equilibrium, find [OH^-], and convert to pH. Using Ka(HNO2) = 4.5 × 10^-4 at 25 C, the result is approximately pH = 8.65. That value makes sense chemically because nitrite is a weak base, so the solution should be basic but not extremely basic.
If you are studying for chemistry exams, this problem is a perfect example of a broader pattern: salts of strong bases and weak acids produce basic solutions. Once you can classify the salt, write the correct hydrolysis reaction, and connect Ka to Kb, problems like this become systematic rather than intimidating.