Calculate the pH of a 0.200 M NaClO2 Solution
Use this premium sodium chlorite pH calculator to estimate hydroxide concentration, pOH, pH, and equilibrium composition from the base hydrolysis of the chlorite ion, ClO2-. The default values are set for a 0.200 M NaClO2 solution at 25 C.
NaClO2 pH Calculator
- Base hydrolysis: ClO2- + H2O ⇌ HClO2 + OH-
- For sodium chlorite, the cation Na+ is a spectator ion.
- Default acid constant used here: Ka(HClO2) = 1.1 × 10^-2
Calculated Results
Ready to calculate. Click the button to compute the pH of a 0.200 M NaClO2 solution.
How to calculate the pH of a 0.200 M NaClO2 solution
To calculate the pH of a 0.200 M sodium chlorite solution, you need to recognize that NaClO2 is a soluble ionic salt composed of Na+ and ClO2-. The sodium ion does not affect the acid base chemistry in water, but the chlorite ion does. Chlorite is the conjugate base of chlorous acid, HClO2. That means it can react with water to produce hydroxide ions, making the solution mildly basic.
The essential equilibrium is ClO2- + H2O ⇌ HClO2 + OH-. Once you know that, the rest of the problem becomes a standard weak base hydrolysis calculation. Many students mistakenly assume all salts are neutral. That is only true when both ions come from a strong acid and a strong base. Sodium chlorite does not fit that pattern because chlorous acid is a weak acid, so its conjugate base has measurable basicity.
Short answer: using Ka(HClO2) = 1.1 × 10^-2 and Kw = 1.0 × 10^-14 at 25 C, a 0.200 M NaClO2 solution has an estimated pH of about 7.63 when solved by the standard weak base equilibrium method.
Step 1: Identify the relevant acid base pair
When NaClO2 dissolves, it dissociates essentially completely:
NaClO2 → Na+ + ClO2-
The species that matters for pH is chlorite, ClO2-. Its conjugate acid is chlorous acid, HClO2. The acid dissociation relationship for chlorous acid is:
HClO2 ⇌ H+ + ClO2-
If the acid constant is known, then the base constant for chlorite is found from:
Kb = Kw / Ka
With common textbook values at 25 C:
- Ka(HClO2) ≈ 1.1 × 10^-2
- Kw = 1.0 × 10^-14
So:
Kb = (1.0 × 10^-14) / (1.1 × 10^-2) = 9.09 × 10^-13
Step 2: Set up the weak base equilibrium expression
Now write the hydrolysis reaction:
ClO2- + H2O ⇌ HClO2 + OH-
Start with an initial concentration of 0.200 M chlorite ion. Before hydrolysis, assume:
- [ClO2-] = 0.200 M
- [HClO2] = 0
- [OH-] = 0 from hydrolysis setup
If x is the amount that reacts, then at equilibrium:
- [ClO2-] = 0.200 – x
- [HClO2] = x
- [OH-] = x
Substitute into the equilibrium expression:
Kb = [HClO2][OH-] / [ClO2-] = x^2 / (0.200 – x)
Step 3: Solve for hydroxide concentration
Because Kb is very small, the hydrolysis is weak, so x is tiny compared with 0.200. That allows the common approximation:
0.200 – x ≈ 0.200
Then:
x^2 / 0.200 = 9.09 × 10^-13
x^2 = 1.818 × 10^-13
x = 4.26 × 10^-7 M
So the hydroxide ion concentration is approximately:
[OH-] = 4.26 × 10^-7 M
Step 4: Convert hydroxide concentration into pOH and pH
Now calculate pOH:
pOH = -log(4.26 × 10^-7) = 6.37
Then use:
pH = 14.00 – 6.37 = 7.63
That gives the commonly accepted weak base equilibrium result:
pH ≈ 7.63
Why the pH is only slightly above 7
This often surprises learners. Even though chlorite is a base, it is the conjugate base of a relatively stronger weak acid than many students expect. Since chlorous acid has a Ka around 10^-2, its conjugate base is correspondingly quite weak. That means hydrolysis generates only a small amount of OH-. As a result, the solution is basic, but only mildly basic.
In practical terms, that means a 0.200 M NaClO2 solution is not strongly alkaline like a sodium hydroxide solution of the same concentration. The chemistry is governed by equilibrium, not complete reaction with water.
Comparison table: chlorine oxyacids and conjugate base strength
The table below helps place chlorous acid and chlorite in context. As the acid gets stronger, the conjugate base gets weaker. This trend explains why ClO2- only raises the pH slightly.
| Acid | Formula | Approximate pKa | Conjugate Base | Relative Base Strength |
|---|---|---|---|---|
| Hypochlorous acid | HClO | 7.5 | ClO- | Moderately weak base |
| Chlorous acid | HClO2 | 1.96 | ClO2- | Very weak base |
| Chloric acid | HClO3 | about -1 | ClO3- | Negligible basicity |
| Perchloric acid | HClO4 | about -10 | ClO4- | Essentially nonbasic |
Exact method vs approximation
The standard weak base approximation works well here because x is much smaller than the initial concentration. A more exact treatment solves the quadratic expression:
x^2 + Kb x – KbC = 0
For this problem, the exact and approximate answers are nearly the same because the amount hydrolyzed is tiny. In other words, the approximation is chemically sound. The calculator above lets you compare methods instantly.
One subtle issue is that the [OH-] generated by hydrolysis is only a few times larger than 10^-7 M, which is the order of pure water autoionization. Introductory chemistry problems usually still solve this type of question with the hydrolysis equilibrium model shown above. If your instructor emphasizes very high precision near neutral pH, mention that pure water effects can make a small numerical difference.
Common mistakes when solving NaClO2 pH problems
- Treating NaClO2 as neutral. Sodium is neutral, but chlorite is a weak base.
- Using Ka directly instead of converting to Kb. For the base reaction, you need Kb = Kw / Ka.
- Using the wrong parent acid. The conjugate acid of ClO2- is HClO2, not HClO or HClO3.
- Forgetting the pOH to pH step. If you calculate [OH-], you must find pOH first, then pH.
- Assuming strong base behavior. NaClO2 does not produce OH- quantitatively like NaOH.
How pH changes with concentration
Because sodium chlorite is a weak base salt, increasing concentration raises the pH, but not dramatically. The relationship is logarithmic and moderated by the small Kb value. The data below use the same constants as the calculator and the standard hydrolysis model.
| NaClO2 concentration (M) | Kb of ClO2- | Approximate [OH-] (M) | Approximate pOH | Approximate pH |
|---|---|---|---|---|
| 0.010 | 9.09 × 10^-13 | 9.53 × 10^-8 | 7.02 | 6.98 to 7.00 by simple model limits |
| 0.050 | 9.09 × 10^-13 | 2.13 × 10^-7 | 6.67 | 7.33 |
| 0.100 | 9.09 × 10^-13 | 3.02 × 10^-7 | 6.52 | 7.48 |
| 0.200 | 9.09 × 10^-13 | 4.26 × 10^-7 | 6.37 | 7.63 |
| 0.500 | 9.09 × 10^-13 | 6.74 × 10^-7 | 6.17 | 7.83 |
What this means in a lab or industrial context
Sodium chlorite is important in bleaching, oxidation chemistry, and water treatment related applications. In these environments, pH can strongly influence speciation, oxidation behavior, and safety protocols. A mildly basic pH for sodium chlorite solutions can affect stability and downstream reactions. For that reason, professionals often combine equilibrium calculations with measured pH and process specific operating data.
If you are working outside a classroom, always verify whether your system contains additional acids, buffers, ionic strength effects, or decomposition pathways. Those can shift the observed pH from the idealized textbook value.
Authoritative references and further reading
- U.S. Environmental Protection Agency on chlorite in drinking water
- Chemistry LibreTexts educational chemistry resource
- Michigan State University acid base equilibrium tutorial
Final answer for a 0.200 M NaClO2 solution
Using the usual equilibrium approach for the chlorite ion in water and the commonly cited value Ka(HClO2) = 1.1 × 10^-2, the chlorite ion has Kb = 9.09 × 10^-13. Solving the weak base equilibrium gives [OH-] ≈ 4.26 × 10^-7 M, so pOH ≈ 6.37 and pH ≈ 7.63.
So, the best standard textbook result is: the pH of a 0.200 M NaClO2 solution is approximately 7.63.
Note: exact values can vary slightly depending on the Ka source used for chlorous acid, temperature, and whether water autoionization is explicitly included in the model.