Calculate The Ph Of 0.250 M Hno2

Use this premium calculator to find the pH of nitrous acid solutions with full weak-acid equilibrium steps. The default setup solves the classic chemistry problem for 0.250 M HNO2 using the accepted Ka value near room temperature.

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How to calculate the pH of 0.250 M HNO2

To calculate the pH of 0.250 M HNO2, you must recognize that nitrous acid is a weak acid, not a strong acid. That means it does not fully ionize in water. Instead, it establishes an equilibrium:

HNO2 + H2O ⇌ H3O+ + NO2-

Because HNO2 only partially dissociates, the hydrogen ion concentration is not simply equal to 0.250 M. You need to use the acid dissociation constant, Ka. A common room-temperature value used in many general chemistry problems is Ka = 4.5 × 10^-4. With that Ka and an initial concentration of 0.250 M, the pH comes out to approximately 1.98 when solved accurately.

This page is designed for students, tutors, lab instructors, and anyone reviewing acid-base equilibrium calculations. Below, you will find the theory, the equilibrium setup, the exact algebra, approximation checks, and practical interpretation of the result.

Step 1: Write the balanced acid dissociation equation

For nitrous acid in water:

HNO2 + H2O ⇌ H3O+ + NO2-

Water is the solvent, so it is not included in the Ka expression. The acid dissociation constant is therefore:

Ka = [H3O+][NO2-] / [HNO2]

Since one molecule of HNO2 produces one hydronium ion and one nitrite ion, the equilibrium concentrations of H3O+ and NO2- will be equal if the solution starts with only the acid and pure water.

Step 2: Set up an ICE table

An ICE table tracks Initial, Change, and Equilibrium concentrations:

Species	Initial (M)	Change (M)	Equilibrium (M)
HNO2	0.250	-x	0.250 – x
H3O+	0	+x	x
NO2-	0	+x	x

Substitute these equilibrium expressions into the Ka formula:

4.5 × 10^-4 = x² / (0.250 – x)

At this point, you have two ways to proceed: use the weak-acid approximation or solve the quadratic equation exactly.

Step 3: Solve using the weak-acid approximation

If x is small compared with 0.250, then 0.250 – x ≈ 0.250. This simplifies the equation to:

4.5 × 10^-4 = x² / 0.250

Rearrange:

x² = (4.5 × 10^-4)(0.250) = 1.125 × 10^-4

x = √(1.125 × 10^-4) ≈ 0.01061 M

This x value equals the hydronium ion concentration:

[H3O+] ≈ 0.01061 M

Now calculate pH:

pH = -log(0.01061) ≈ 1.97

This is already a good result, but we should verify whether the approximation is valid. The 5 percent rule says x should be less than 5 percent of the initial concentration.

(0.01061 / 0.250) × 100 ≈ 4.24%

Because 4.24 percent is below 5 percent, the approximation is acceptable.

Step 4: Solve using the exact quadratic equation

For a more accurate answer, solve:

4.5 × 10^-4 = x² / (0.250 – x)

Multiply both sides:

4.5 × 10^-4(0.250 – x) = x²

1.125 × 10^-4 – 4.5 × 10^-4x = x²

Rearrange into standard quadratic form:

x² + 4.5 × 10^-4x – 1.125 × 10^-4 = 0

Using the quadratic formula gives the physically meaningful positive root:

x ≈ 0.01039 M

Then:

pH = -log(0.01039) ≈ 1.983

That exact answer is slightly higher than the approximation result because the denominator was not held constant at 0.250 M.

Final answer for 0.250 M HNO2

Using Ka = 4.5 × 10^-4, the pH of 0.250 M nitrous acid is:

pH ≈ 1.98

If your textbook uses a slightly different Ka value, your final pH may differ by a few hundredths. That is normal. Chemistry references sometimes list HNO2 Ka values in the range of roughly 4.0 × 10^-4 to 4.6 × 10^-4 depending on source and temperature.

Why HNO2 is not treated like a strong acid

A common mistake is to assume all acids fully dissociate. Strong acids such as HCl, HBr, and HNO3 ionize essentially completely in dilute aqueous solution. Nitrous acid, however, is weak. Its Ka value is much smaller than 1, which means the equilibrium strongly favors undissociated HNO2.

• Strong acid: [H3O+] is approximately equal to the initial acid concentration.
• Weak acid: [H3O+] must be found from an equilibrium calculation.
• HNO2: only a small fraction of the 0.250 M initial concentration ionizes.

In this problem, only about 0.01039 M out of 0.250 M dissociates at equilibrium, which is just over 4 percent.

Comparison table: exact versus approximation results

Method	Ka used	[H3O+] (M)	pH	Percent ionization
Weak-acid approximation	4.5 × 10^-4	0.01061	1.974	4.24%
Exact quadratic solution	4.5 × 10^-4	0.01039	1.983	4.16%

The difference between methods is very small here, which is why the approximation is often accepted in introductory chemistry courses. Still, if you are using software, a calculator, or presenting a polished lab report, the exact quadratic solution is best.

How concentration affects pH for nitrous acid

For weak acids, the pH does not change in a one-to-one way with concentration because ionization is governed by equilibrium. As concentration decreases, percent ionization tends to increase. As concentration increases, the acid is still more acidic in absolute terms, but the fraction that ionizes often decreases.

Initial HNO2 concentration (M)	Approximate pH using Ka = 4.5 × 10^-4	Approximate percent ionization	Interpretation
0.010	2.18	21.2%	Dilute weak acid with noticeably higher ionization fraction
0.050	1.83	9.5%	Moderately acidic solution
0.100	1.68	6.7%	Typical textbook weak-acid example
0.250	1.98 exact result from this page	4.16%	Concentrated enough that exact equilibrium is recommended
0.500	1.53	3.0%	Lower ionization fraction but higher hydronium concentration

Important chemistry concepts behind the answer

1. Monoprotic behavior: HNO2 donates one acidic proton per molecule.
2. Equilibrium control: Weak acids require Ka or pKa to predict pH.
3. ICE tables: These organize the algebra for equilibrium calculations.
4. Approximations: If x is small enough, calculations become simpler with little loss of accuracy.
5. pH relation: pH is always calculated from hydronium concentration using pH = -log[H3O+].

Common mistakes when solving this problem

• Assuming HNO2 is a strong acid and setting pH = -log(0.250), which would be incorrect.
• Using the wrong acid constant, such as the Ka for nitric acid or another unrelated acid.
• Forgetting that x appears in the denominator as 0.250 – x.
• Accepting the approximation without checking the 5 percent rule.
• Rounding too early, which can shift the final pH by a few hundredths.

How this result compares with stronger acids

If a 0.250 M solution were made from a strong monoprotic acid instead, the hydronium concentration would be close to 0.250 M and the pH would be about 0.60. For HNO2, the pH is near 1.98. That difference is substantial and clearly demonstrates why acid strength matters just as much as concentration.

Useful authoritative references

If you want to verify acid-base concepts, aqueous equilibrium behavior, or standard pH methods, these government and university resources are excellent starting points:

• LibreTexts Chemistry for detailed acid-base equilibrium tutorials.
• U.S. Environmental Protection Agency for practical background on pH in aqueous systems.
• Michigan State University chemistry acid-base reference for instructional equilibrium discussions.
• U.S. Geological Survey pH and water overview for a reliable pH primer.

Practical takeaway

When asked to calculate the pH of 0.250 M HNO2, the correct strategy is to treat HNO2 as a weak acid, write the equilibrium expression using Ka, solve for hydronium concentration, and then convert that value to pH. With Ka = 4.5 × 10^-4, the exact pH is about 1.98. This is a classic example showing why concentration alone does not determine acidity. The degree of ionization is the key piece of the chemistry.

Educational note: Ka values can vary slightly by source and temperature, so tiny differences in the reported pH are expected in textbooks, homework systems, and laboratory references.