Calculate the pH of 0.15 M Na2SO3
Use this premium calculator to estimate the pH of sodium sulfite solutions by applying sulfite hydrolysis chemistry. The tool supports exact quadratic and approximation methods, lets you adjust Ka2 values, and visualizes the result with an interactive Chart.js graph.
How to calculate the pH of 0.15 M Na2SO3
To calculate the pH of 0.15 M Na2SO3, you treat sodium sulfite as a salt that dissociates completely in water to produce 2 Na+ ions and the sulfite ion, SO32-. The sodium ion is essentially a spectator ion for acid-base chemistry, but the sulfite ion is not. Sulfite is the conjugate base of bisulfite, HSO3–, and because it is a base, it reacts with water to produce hydroxide ions. That makes the solution basic, so the pH must be greater than 7 at 25 degrees C.
The key hydrolysis reaction is:
Once you recognize that hydrolysis reaction, the next step is to compute the base dissociation constant for sulfite. Because SO32- is the conjugate base of HSO3–, its Kb is related to the acid dissociation constant Ka2 of sulfurous acid chemistry by:
Using a common textbook value of Ka2 = 6.4 × 10-8 and Kw = 1.0 × 10-14, we get:
Now let the initial concentration of sulfite be C = 0.15 M, and let x be the amount of OH– formed. Then at equilibrium:
- [SO32-] = 0.15 – x
- [HSO3–] = x
- [OH–] = x
The equilibrium expression becomes:
Because Kb is much smaller than the concentration, most introductory chemistry courses use the approximation 0.15 – x ≈ 0.15. That gives:
Since x = [OH–], calculate pOH:
Then calculate pH:
So the standard answer for the pH of 0.15 M Na2SO3 is approximately 10.18 when you use Ka2 = 6.4 × 10-8 at 25 degrees C. If your textbook uses a slightly different Ka2 value, your final pH may differ by a few hundredths of a pH unit, which is completely normal in chemistry education and reference material.
Why sodium sulfite gives a basic pH
Sodium sulfite is the salt of a strong base and a weak acid. Specifically, NaOH is a strong base, while sulfurous acid chemistry is weak and stepwise. When a salt comes from a strong base and a weak acid, the anion often hydrolyzes water and raises the pH. This is exactly what sulfite does. The solution is not strongly basic like NaOH, but it is noticeably basic because the sulfite ion accepts a proton from water, creating OH–.
It helps to compare sodium sulfite with salts students already know:
- NaCl gives a nearly neutral solution because both ions come from strong electrolytes and neither hydrolyzes significantly.
- NH4Cl gives an acidic solution because NH4+ is the conjugate acid of a weak base.
- Na2CO3 and Na2SO3 give basic solutions because CO32- and SO32- are conjugate bases of weak acids.
That conceptual shortcut is useful on exams. Before doing any arithmetic, you can often predict whether the pH must be above 7, below 7, or close to 7. For 0.15 M Na2SO3, the pH should definitely be above 7.
Step-by-step worked solution
- Write the dissociation of the salt: Na2SO3 → 2 Na+ + SO32-.
- Identify the hydrolyzing ion: SO32-.
- Write the base hydrolysis reaction: SO32- + H2O ⇌ HSO3– + OH–.
- Convert Ka2 to Kb using Kb = Kw / Ka2.
- Set up an ICE table with initial sulfite concentration 0.15 M.
- Solve for x using either the approximation or the quadratic formula.
- Find pOH from [OH–] and then convert to pH.
Approximation check
Whenever you use x = √(KbC), you should verify that x is small relative to the initial concentration. Here, x ≈ 1.53 × 10-4 M. Dividing by 0.15 M gives about 0.10 percent, which is safely below the usual 5 percent rule. That means the approximation is excellent for this problem.
Exact quadratic solution
If you prefer the exact route, solve:
With Kb = 1.5625 × 10-7 and C = 0.15, the positive root still gives x very close to 1.53 × 10-4 M, so the pH remains about 10.18. This is why most instructors accept the square-root approximation for the pH of 0.15 M Na2SO3.
Comparison data table: pH at different Na2SO3 concentrations
The following values are calculated at 25 degrees C using Kw = 1.0 × 10-14 and Ka2 = 6.4 × 10-8. These numbers help you see how concentration affects the pH of sodium sulfite solutions.
| Na2SO3 concentration (M) | Kb for SO32- | Approximate [OH–] (M) | Approximate pH |
|---|---|---|---|
| 0.010 | 1.56 × 10-7 | 3.95 × 10-5 | 9.60 |
| 0.050 | 1.56 × 10-7 | 8.84 × 10-5 | 9.95 |
| 0.100 | 1.56 × 10-7 | 1.25 × 10-4 | 10.10 |
| 0.150 | 1.56 × 10-7 | 1.53 × 10-4 | 10.18 |
| 0.200 | 1.56 × 10-7 | 1.77 × 10-4 | 10.25 |
Constants table: values commonly used in general chemistry
One reason students sometimes get slightly different numerical answers is that the acid dissociation constants for sulfurous acid systems can vary between textbooks and databases. The table below shows representative values used in education and problem solving.
| Quantity | Representative value | Use in calculation | Impact on final pH |
|---|---|---|---|
| Kw at 25 degrees C | 1.0 × 10-14 | Converts Ka to Kb and relates pH to pOH | Sets the standard 25 degrees C pH scale |
| Ka2 for HSO3– | 6.4 × 10-8 | Most common classroom value for sulfite hydrolysis problems | Gives pH near 10.18 for 0.15 M Na2SO3 |
| Alternative Ka2 values in some references | About 1.0 × 10-7 | Used in some older or different reference sets | Can shift the pH downward by a few hundredths |
| Initial [SO32-] | 0.15 M | Starting concentration in the problem | Controls how much OH– can be generated |
Common mistakes when calculating the pH of 0.15 M Na2SO3
- Using Ka instead of Kb directly. The reacting species is SO32-, which behaves as a base, so you need Kb or a conversion from Ka2.
- Using Ka1 instead of Ka2. The relevant conjugate acid for SO32- is HSO3–, not H2SO3. Therefore, Ka2 is the correct acid constant to use.
- Forgetting that Na+ does not affect the pH in a meaningful way for this level of calculation.
- Assuming the solution is neutral because it is a salt. Many salts are not neutral. The parent acid and base strengths matter.
- Skipping the approximation check. It is good practice to verify that x is small before simplifying 0.15 – x to 0.15.
Practical interpretation of the answer
A pH around 10.18 means the solution is moderately basic. It is nowhere near as basic as a concentrated sodium hydroxide solution, but it is clearly alkaline enough to matter in laboratory handling, reaction conditions, and analytical chemistry. Sulfite chemistry is also important in industrial and environmental settings because sulfite and bisulfite species can participate in redox chemistry, oxygen scavenging, and sulfur cycling in water systems.
If your assignment only asks for a textbook answer, report pH = 10.18 with the assumptions stated. If your instructor wants a more rigorous answer, include the Ka2 value you used and note that small variations in data sources can slightly shift the result.
Authoritative references for acid-base and water chemistry
- U.S. Environmental Protection Agency: Water quality chemistry context
- Purdue chemistry educational reference on dissociation constants
- Princeton University: pH definition and acid-base fundamentals
Final answer summary
For the standard general chemistry problem, the pH of 0.15 M Na2SO3 is approximately 10.18 at 25 degrees C when using Ka2 = 6.4 × 10-8 for HSO3–. The solution is basic because SO32- hydrolyzes water to form OH–. The exact and approximate methods agree closely, so either is acceptable unless your instructor specifically requires the quadratic solution.