Calculate the pH of 0.150 M NaClO2

Use this premium hydrolysis calculator to estimate the pH, pOH, Kb, hydroxide concentration, and equilibrium behavior of sodium chlorite solutions. The default setup is configured for 0.150 M NaClO2 at 25 degrees Celsius.

Weak base salt hydrolysis Quadratic equilibrium support Interactive Chart.js visualization

NaClO2 pH Calculator

Sodium chlorite dissociates completely into Na⁺ and ClO₂^–. The chlorite ion acts as a weak base because it is the conjugate base of chlorous acid, HClO₂.

NaClO2 concentration (M)

Default example: 0.150 M

Ka of HClO2

Common textbook value at 25 degrees Celsius: 1.1 × 10^-2

Default water ion-product at 25 degrees Celsius: 1.0 × 10^-14

Calculation method

Exact mode is recommended when you want the full equilibrium result.

Enter values and click Calculate pH to see the result.

How to calculate the pH of 0.150 M NaClO2

To calculate the pH of 0.150 M NaClO2, you treat sodium chlorite as a soluble ionic salt that fully dissociates in water into sodium ions and chlorite ions. The sodium ion, Na⁺, is a spectator ion for acid-base purposes because it comes from the strong base sodium hydroxide. The chlorite ion, ClO₂^–, is the important species because it is the conjugate base of chlorous acid, HClO₂. That means the solution is expected to be basic, although only weakly basic, because chlorite hydrolyzes water to generate hydroxide ions.

The first chemical idea is dissociation:

NaClO₂(aq) → Na⁺(aq) + ClO₂^–(aq)

The second idea is the base hydrolysis equilibrium of the chlorite ion:

ClO₂^–(aq) + H₂O(l) ⇌ HClO₂(aq) + OH^–(aq)

Because chlorite is the conjugate base of chlorous acid, you can find its base dissociation constant from the acid dissociation constant of HClO₂. At 25 degrees Celsius, many general chemistry references use a Ka value near 1.1 × 10^-2 for chlorous acid. Then:

Kb = Kw / Ka = (1.0 × 10^-14) / (1.1 × 10^-2) = 9.09 × 10^-13

Now set up the ICE table for a 0.150 M NaClO2 solution. Initially, the chlorite concentration is 0.150 M, while HClO₂ and OH^– formed from hydrolysis are approximately zero. Let x be the amount of OH^– produced at equilibrium:

Initial: [ClO₂^–] = 0.150, [HClO₂] = 0, [OH^–] = 0
Change: [ClO₂^–] = -x, [HClO₂] = +x, [OH^–] = +x
Equilibrium: [ClO₂^–] = 0.150 – x, [HClO₂] = x, [OH^–] = x

The equilibrium expression is:

Kb = x² / (0.150 – x)

Since Kb is very small, the approximation method is usually acceptable:

x ≈ √(Kb × C) = √((9.09 × 10^-13)(0.150)) ≈ 3.69 × 10^-7 M

This gives the hydroxide concentration:

[OH^–] ≈ 3.69 × 10^-7 M

Then calculate pOH and pH:

pOH = -log(3.69 × 10^-7) ≈ 6.43
pH = 14.00 – 6.43 ≈ 7.57

So the expected pH of 0.150 M NaClO2 is approximately 7.57 at 25 degrees Celsius when Ka for HClO₂ is taken as 1.1 × 10^-2. This result is only slightly above neutral because chlorite is the conjugate base of a relatively strong weak acid. In other words, HClO₂ is much stronger than acetic acid, so its conjugate base is much weaker than acetate. That is why the pH is basic, but not dramatically basic.

Why NaClO2 is basic in water

Students often ask why a salt can produce a basic solution. The answer lies in the parent acid and parent base. Sodium chlorite comes from sodium hydroxide, a strong base, and chlorous acid, a weak acid. Whenever a salt contains the cation of a strong base and the anion of a weak acid, the anion can react with water and make OH^–. That shifts the solution above pH 7.

This classification rule is useful:

Strong acid + strong base salt: usually neutral
Strong acid + weak base salt: acidic
Weak acid + strong base salt: basic
Weak acid + weak base salt: depends on relative Ka and Kb

NaClO2 belongs in the third category. However, because chlorous acid is not an extremely weak acid, its conjugate base is very weak. The consequence is a pH only modestly above 7.

Exact quadratic method versus approximation

For many weak acid and weak base problems, chemistry students use the shortcut x ≈ √(KC). This is usually valid when x is much smaller than the starting concentration. In this problem, that condition is definitely satisfied because x is on the order of 10^-7, while the initial concentration is 0.150 M. The percent change is tiny, so the approximation is extremely good.

Still, the exact quadratic approach is conceptually stronger because it avoids assumptions. Starting from:

Kb = x² / (0.150 – x)

Rearrange to:

x² + Kb x – 0.150 Kb = 0

Then solve with the quadratic formula. Because Kb is so small, the positive root is almost identical to the square-root estimate. For classroom work, both methods lead to essentially the same pH for 0.150 M NaClO2.

Quantity	Value Used	Meaning in the Calculation
NaClO2 concentration	0.150 M	Initial concentration of ClO₂^– after complete dissociation
Ka of HClO2	1.1 × 10^-2	Acid strength of chlorous acid
Kw at 25 degrees Celsius	1.0 × 10^-14	Used to convert Ka to Kb
Kb of ClO₂^–	9.09 × 10^-13	Base strength of chlorite ion
[OH^–]	3.69 × 10^-7 M	Hydroxide formed by hydrolysis
pOH	6.43	Negative logarithm of hydroxide concentration
pH	7.57	Final solution acidity-basicity estimate

Interpretation of the result

A pH around 7.57 tells you that the solution is basic, but not strongly basic. This matters in lab interpretation. If you compare NaClO2 to sodium acetate, carbonate, or phosphate salts, sodium chlorite often appears much less basic than students initially expect. The reason is purely thermodynamic. The stronger the parent acid, the weaker its conjugate base. Since chlorous acid has a Ka around 10^-2, chlorite has a correspondingly tiny Kb around 10^-12.

That relationship can be summarized simply:

Larger Ka for the parent acid means smaller Kb for the conjugate base.
Smaller Kb means less OH^– generated from hydrolysis.
Less OH^– means a pH closer to neutral.

How concentration affects the pH

The concentration of the salt still matters even when the base is weak. If you lower NaClO2 concentration, the generated hydroxide decreases and the pH moves closer to 7. If you increase concentration, pH rises, but only gradually because the dependence is through a square root relationship under the common approximation. This is why doubling concentration does not double pH change.

For weak base salts such as NaClO2, the rough trend is:

Find Kb from Ka.
Estimate [OH^–] using √(KbC).
Convert to pOH and pH.
Recognize that concentration changes produce logarithmic pH shifts, not linear ones.

Reference or Regulatory Statistic	Reported Figure	Why It Matters
EPA maximum residual disinfectant level for chlorite in drinking water	1.0 mg/L	Shows chlorite chemistry has real-world water treatment relevance beyond classroom equilibrium problems.
EPA maximum residual disinfectant level for chlorine dioxide	0.8 mg/L	Important because chlorine dioxide use can generate chlorite as a byproduct.
Standard Kw at 25 degrees Celsius	1.0 × 10^-14	Sets the basis for converting between Ka and Kb in aqueous solution chemistry.
Representative Ka for chlorous acid used in general chemistry calculations	1.1 × 10^-2	Directly controls the predicted pH of sodium chlorite solutions.

Common mistakes when solving this problem

Even strong chemistry students can make avoidable errors on this exact question. Here are the most common ones:

Treating NaClO2 as neutral. It is not neutral because ClO₂^– is the conjugate base of a weak acid.
Using Ka directly in the ICE table. The dissolved species is chlorite, so you need Kb for the base hydrolysis reaction.
Forgetting to convert pOH to pH. If you compute [OH^–], your first logarithm gives pOH, not pH.
Assuming a strong base. Chlorite does not dissociate to release OH^– directly the way NaOH does.
Ignoring temperature dependence. The default calculation assumes 25 degrees Celsius, where Kw is 1.0 × 10^-14.

Authority sources for chlorite and water chemistry

If you want to verify chlorite-related regulatory values and broader water chemistry context, these sources are excellent starting points:

Final answer for 0.150 M NaClO2

Using a representative Ka of 1.1 × 10^-2 for chlorous acid and Kw = 1.0 × 10^-14 at 25 degrees Celsius, the chlorite ion has Kb = 9.09 × 10^-13. Solving the weak-base hydrolysis equilibrium for a 0.150 M sodium chlorite solution gives an hydroxide concentration of about 3.69 × 10^-7 M, a pOH of about 6.43, and a pH of about 7.57.

That means a 0.150 M NaClO2 solution is slightly basic. If your textbook uses a slightly different Ka value for HClO₂, your final pH may vary a little, but it should still land close to the upper-7 range rather than in the strongly basic region.

Calculate The Ph 0.150 M Naclo2