Calculate The Ph Of A 0.50 M Solution Of Nano2.

Use this interactive calculator to find the pH of aqueous sodium nitrite by modeling nitrite ion hydrolysis. The tool supports a quick approximation and a more rigorous quadratic solution using the acid dissociation constant of nitrous acid, HNO2.

How to calculate the pH of a 0.50 m solution of NaNO2

Sodium nitrite, NaNO2, is the salt of a strong base and a weak acid. The sodium ion, Na⁺, is essentially neutral in water, but the nitrite ion, NO2^–, acts as a weak base because it is the conjugate base of nitrous acid, HNO2. That means a solution of sodium nitrite is expected to be basic, with a pH greater than 7. To calculate the pH, you do not treat NaNO2 as if it were a strong base like NaOH. Instead, you use a weak base hydrolysis equilibrium.

The key reaction in water is:

NO2^– + H2O ⇌ HNO2 + OH^–

This equilibrium produces hydroxide ions, and those hydroxide ions determine the pOH and therefore the pH. Because the starting concentration is 0.50 m, which is often interpreted as 0.50 molal, many textbook and homework problems still approximate that value as 0.50 M when no density data are given. This calculator follows that standard classroom convention unless you provide a different context. The result is therefore an excellent estimate for most instructional chemistry problems.

Step 1: Identify the correct equilibrium constant

Most reference tables list the acid dissociation constant, Ka, for nitrous acid rather than the base dissociation constant, Kb, for nitrite. The relationship between them is:

Kb = Kw / Ka

At 25 C, a common tabulated value is:

• Ka(HNO2) = 4.5 × 10^-4
• Kw = 1.0 × 10^-14

So:

Kb = (1.0 × 10^-14) / (4.5 × 10^-4) = 2.22 × 10^-11

Step 2: Set up the ICE table

For a 0.50 m or approximately 0.50 M solution of sodium nitrite:

• Initial [NO2^–] = 0.50
• Initial [HNO2] = 0
• Initial [OH^–] = 0, ignoring water autoionization at first

Let x be the amount of NO2^– that reacts:

• [NO2^–] = 0.50 – x
• [HNO2] = x
• [OH^–] = x

Then the equilibrium expression is:

Kb = x² / (0.50 – x)

Step 3: Solve for hydroxide concentration

Because Kb is very small, x will be much smaller than 0.50, so the common weak base approximation is:

Kb ≈ x² / 0.50

Rearranging:

x = √(Kb × 0.50)

Substitute the numbers:

x = √((2.22 × 10^-11) × 0.50) = √(1.11 × 10^-11) ≈ 3.33 × 10^-6

This x value equals the hydroxide concentration:

[OH^–] ≈ 3.33 × 10^-6 M

Step 4: Convert [OH-] to pOH and pH

Use the definition:

pOH = -log[OH^–]

So:

pOH = -log(3.33 × 10^-6) ≈ 5.48

Then:

pH = 14.00 – 5.48 = 8.52

Therefore, the pH of a 0.50 m solution of NaNO2 is approximately 8.52 at 25 C, assuming ideal behavior and using the common Ka value of 4.5 × 10^-4 for HNO2.

Why sodium nitrite is basic

Many students first learn that salts are neutral because they come from an acid and a base. That rule is incomplete. The acid and base strengths matter. Sodium nitrite comes from sodium hydroxide, a strong base, and nitrous acid, a weak acid. The sodium ion is a spectator, but the nitrite ion can accept a proton from water. That proton transfer creates hydroxide ions, making the solution basic. This is a classic example of conjugate base hydrolysis.

The more weakly acidic the parent acid is, the more basic its conjugate base will be. Since nitrous acid is not a strong acid, the nitrite ion has some measurable basicity. Still, the basicity is weak, so the pH only rises modestly above neutral. A pH around 8.5 is basic, but not strongly caustic.

Approximation versus quadratic solution

For weak acid and weak base equilibria, the square root shortcut is widely used because it is fast and usually accurate. In this case, it works very well because x is tiny compared with 0.50. However, a senior level calculation often checks the exact quadratic form:

x² + Kb x – KbC = 0

Where C is the initial nitrite concentration. Solving this exactly gives a value for x that is nearly identical to the approximation. The reason is the same 5 percent rule that appears throughout equilibrium chemistry. If x/C is very small, replacing C – x with C is justified. Here, the percent ionization is far below 1 percent, so the approximation is excellent.

Parameter	Value used	Meaning
Initial NaNO2 concentration	0.50	Approximate aqueous concentration of nitrite ion
Ka of HNO2	4.5 × 10^-4	Acid strength of nitrous acid at 25 C
Kw	1.0 × 10^-14	Ion product of water at 25 C
Kb of NO2^–	2.22 × 10^-11	Base strength of nitrite ion
[OH^–]	3.33 × 10^-6 M	Hydroxide produced by hydrolysis
pOH	5.48	Negative log of hydroxide concentration
pH	8.52	Final estimated pH of the solution

Important note about molality versus molarity

The symbol m means molality, while M means molarity. These are not exactly the same. Molality is moles of solute per kilogram of solvent. Molarity is moles of solute per liter of solution. For precise solution chemistry, especially at higher ionic strengths, density and activity effects can matter. But in many educational problems, a 0.50 m aqueous solution is treated as roughly 0.50 M if no density information is supplied. That is why the classic textbook answer for this problem is around pH 8.52.

If you were doing a very high precision physical chemistry analysis, you would account for:

• Solution density to convert molality into molarity
• Temperature dependence of Kw and Ka
• Activity coefficients, especially as ionic strength rises
• Possible deviations from ideality in concentrated electrolyte solutions

For most general chemistry and analytical chemistry settings, these corrections are not required to get the intended answer.

Comparison with other salts: what the pH tells you

One useful way to understand sodium nitrite is to compare it with salts made from strong or weak acids and bases. This gives a quick intuition for whether a salt solution will be acidic, neutral, or basic.

Salt	Parent acid	Parent base	Expected pH behavior	Typical classroom interpretation
NaCl	HCl, strong	NaOH, strong	Near neutral	Neither ion hydrolyzes significantly
NH4Cl	HCl, strong	NH3, weak	Acidic	NH4^+ donates a proton to water
CH3COONa	CH3COOH, weak	NaOH, strong	Basic	Acetate accepts a proton from water
NaNO2	HNO2, weak	NaOH, strong	Basic	Nitrite hydrolysis forms OH^–

Common mistakes to avoid

1. Treating NaNO2 as a strong base. Sodium nitrite is not like sodium hydroxide. It only becomes basic through partial hydrolysis of the nitrite ion.
2. Using Ka directly instead of Kb. The hydrolysis reaction is a base equilibrium, so you must convert Ka to Kb using Kw/Ka.
3. Forgetting the log conversion. The equilibrium gives [OH^–], not pH directly. You must calculate pOH first, then pH.
4. Confusing m and M. If a problem states molality but supplies no density, note the approximation clearly.
5. Dropping significant figures too early. Keep extra digits through the intermediate calculations for a cleaner final answer.

When the answer can shift slightly

You may see pH values from about 8.50 to 8.54 in different resources. That small spread usually comes from different reference values for Ka of nitrous acid, different rounding conventions, or whether the author uses an exact quadratic solution rather than the square root approximation. All of these answers are chemically consistent with the same equilibrium framework.

Typical sensitivity to Ka choices

If the tabulated Ka is a bit larger, then Kb becomes smaller, [OH^–] drops slightly, and the pH falls a little. If the tabulated Ka is smaller, the opposite happens. The pH of a weak base salt is therefore somewhat sensitive to the chosen equilibrium constant, but not enough to change the qualitative conclusion that sodium nitrite solution is basic.

Authoritative references for deeper study

If you want to verify equilibrium constants, water quality context, or acid-base theory from high quality sources, these references are useful:

• NIST Chemistry WebBook for reliable thermodynamic and chemical reference data.
• U.S. Environmental Protection Agency drinking water regulations for nitrite related public health context.
• Purdue Chemistry topic reviews for academic explanations of equilibrium and acid-base calculations.

Final answer for the standard problem

Using the standard classroom assumptions at 25 C, a 0.50 m solution of NaNO2 has a pH of about 8.52. The solution is basic because nitrite is the conjugate base of the weak acid nitrous acid, so it hydrolyzes in water to generate a small but measurable amount of hydroxide ion.

If you want to explore how the answer changes with a different Ka or a different starting concentration, use the calculator above. It will recompute the pH instantly and plot how pH changes across nearby sodium nitrite concentrations.