Calculate the pH of a 0.30 M Solution of NaClO2
Use this interactive calculator to solve the pH of sodium chlorite, NaClO2, by treating the chlorite ion as the conjugate base of chlorous acid. The tool supports exact and approximation methods, shows the hydrolysis equilibrium, and visualizes the result with a responsive chart.
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How to Calculate the pH of a 0.30 M Solution of NaClO2
Calculating the pH of a 0.30 M solution of NaClO2 is a classic weak-base hydrolysis problem. Sodium chlorite dissociates essentially completely in water to give sodium ions and chlorite ions:
NaClO2(aq) → Na+(aq) + ClO2-(aq)
The sodium ion, Na+, is a spectator ion for acid-base chemistry because it comes from the strong base sodium hydroxide. The chemically important species is ClO2-, the conjugate base of chlorous acid, HClO2. Because chlorous acid is a weak acid, its conjugate base can react with water:
ClO2-(aq) + H2O(l) ⇌ HClO2(aq) + OH-(aq)
That reaction produces hydroxide ions, so the solution is basic. The goal is to find the hydroxide concentration generated by hydrolysis and then convert it to pOH and pH.
Step 1: Identify the correct equilibrium constant
Most textbook and laboratory problems provide or assume a value for the acid dissociation constant of chlorous acid. A commonly used value at 25°C is:
Ka(HClO2) ≈ 1.1 × 10^-2
Because chlorite is the conjugate base of chlorous acid, we find the base dissociation constant with the conjugate relationship:
Kb = Kw / Ka
At 25°C, Kw = 1.0 × 10^-14. Therefore:
Kb = (1.0 × 10^-14) / (1.1 × 10^-2) = 9.09 × 10^-13
This very small Kb tells us chlorite is only a very weak base, even though it still makes the solution slightly basic.
Step 2: Set up the ICE table
Start with a 0.30 M solution of sodium chlorite. Since it dissociates fully, the initial chlorite concentration is 0.30 M.
ClO2- + H2O ⇌ HClO2 + OH-
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| ClO2- | 0.30 | -x | 0.30 – x |
| HClO2 | 0 | +x | x |
| OH- | 0 | +x | x |
The equilibrium expression is:
Kb = [HClO2][OH-] / [ClO2-] = x^2 / (0.30 – x)
Step 3: Solve for x
Since Kb is extremely small, the value of x will be much smaller than 0.30. That allows the standard weak-base approximation:
0.30 – x ≈ 0.30
Then:
x^2 = Kb(0.30)
x = √(9.09 × 10^-13 × 0.30)
x = √(2.727 × 10^-13) ≈ 5.22 × 10^-7 M
This x is the hydroxide concentration:
[OH-] ≈ 5.22 × 10^-7 M
Step 4: Convert hydroxide concentration to pOH and pH
Use the standard logarithmic relationship:
pOH = -log[OH-]
pOH = -log(5.22 × 10^-7) ≈ 6.28
At 25°C:
pH = 14.00 – 6.28 = 7.72
So the pH of a 0.30 M NaClO2 solution is about 7.72, assuming Ka(HClO2) = 1.1 × 10^-2 and a temperature of 25°C.
Why the answer is only slightly basic
Students often expect a salt with a negatively charged ion to be strongly basic, but that depends entirely on the strength of the conjugate acid. Chlorous acid is much stronger than many common weak acids such as acetic acid or hypochlorous acid. Because HClO2 is comparatively stronger, its conjugate base ClO2- is comparatively weaker. That is why a 0.30 M sodium chlorite solution does not produce a pH of 10 or 11. It sits only modestly above neutral.
Exact method vs approximation
The exact method solves:
x^2 + Kb x – KbC = 0
With C = 0.30 and Kb = 9.09 × 10^-13, the physically meaningful root is:
x = (-Kb + √(Kb^2 + 4KbC)) / 2
Numerically, this gives essentially the same result as the approximation because the percent ionization is tiny:
% ionization = (x / 0.30) × 100 ≈ 0.000174%
Since that value is far below 5%, the approximation is fully justified for routine chemistry work.
Common mistake: treating NaClO2 as neutral
Not every sodium salt is neutral. A sodium salt is neutral only when its anion comes from a strong acid, such as chloride from hydrochloric acid or nitrate from nitric acid. In this case, chlorite comes from chlorous acid, which is a weak acid. Therefore the anion hydrolyzes and the solution becomes basic. That single conceptual checkpoint prevents many pH errors on homework, exams, and laboratory reports.
Comparison table: chlorine oxyacid strength and conjugate-base behavior
The acidity of chlorine oxyacids increases as the oxygen count rises. That trend helps explain why chlorite is only a weak base. The values below are commonly cited approximate pKa statistics used in general and analytical chemistry discussions.
| Acid | Formula | Approximate pKa at 25°C | Conjugate Base | Relative Basicity of Conjugate Base |
|---|---|---|---|---|
| Hypochlorous acid | HClO | 7.5 | ClO- | Much stronger base than ClO2- |
| Chlorous acid | HClO2 | 1.96 | ClO2- | Weak base |
| Chloric acid | HClO3 | About -1 | ClO3- | Negligibly basic in water |
| Perchloric acid | HClO4 | About -10 | ClO4- | Essentially nonbasic spectator ion |
This table makes the trend intuitive. As the acid gets stronger, the conjugate base gets weaker. Chlorous acid is much stronger than hypochlorous acid, so chlorite is much less basic than hypochlorite.
How concentration affects the pH of sodium chlorite
For weak bases, the pH rises gradually with concentration, not dramatically. Because hydroxide concentration depends roughly on the square root of concentration, increasing the salt concentration tenfold raises pH by only about 0.5 unit. That is a useful estimation rule when checking whether an answer is realistic.
| NaClO2 Concentration (M) | Kb Used | Calculated [OH-] (M) | pOH | pH at 25°C |
|---|---|---|---|---|
| 0.01 | 9.09 × 10^-13 | 9.53 × 10^-8 | 7.02 | 6.98 to 7.00 after including water autoionization effects |
| 0.10 | 9.09 × 10^-13 | 3.02 × 10^-7 | 6.52 | 7.48 |
| 0.30 | 9.09 × 10^-13 | 5.22 × 10^-7 | 6.28 | 7.72 |
| 1.00 | 9.09 × 10^-13 | 9.53 × 10^-7 | 6.02 | 7.98 |
Detailed reasoning in plain language
If you want a fast conceptual shortcut, think through the species in three layers. First, sodium chlorite is soluble and dissociates fully. Second, sodium does not alter pH because it is the cation of a strong base. Third, chlorite can take a proton from water because it is the conjugate base of a weak acid. Once you know those three facts, you already know the pH must be greater than 7.
The only remaining question is how much greater. Since chlorous acid has a fairly large acid dissociation constant compared with many weak acids, chlorite must be a correspondingly weak base. So the pH will not be strongly basic. The actual number, around 7.72 for a 0.30 M solution, fits that logic perfectly: basic, but only mildly so.
Checklist for solving similar salt-hydrolysis problems
- Write the dissociation of the salt in water.
- Identify whether the cation or anion comes from a weak acid or weak base.
- Choose Ka or Kb appropriately.
- Convert using KaKb = Kw if needed.
- Build an ICE table for the hydrolysis reaction.
- Solve for x exactly or by approximation.
- Convert to pOH or pH and check whether the answer is chemically reasonable.
Final answer for this problem
Using Ka(HClO2) = 1.1 × 10^-2 at 25°C, the chlorite ion has Kb = 9.09 × 10^-13. For a 0.30 M NaClO2 solution, the equilibrium hydroxide concentration is approximately 5.22 × 10^-7 M, which gives:
pOH ≈ 6.28
pH ≈ 7.72
That is the value most instructors expect unless a different Ka is specified.