Calculate The Ph Of 0.15 M Na2So3

Use this premium calculator to estimate the pH of sodium sulfite solution from equilibrium chemistry. By default, it solves the classic case of 0.15 M Na2SO3 at 25 degrees Celsius using the sulfite base hydrolysis relation, where Kb = Kw / Ka2.

How to calculate the pH of 0.15 M Na2SO3

Sodium sulfite, Na2SO3, is not an acid. It is a salt that produces the sulfite ion, SO3^2-, in water. Because sulfite is the conjugate base of bisulfite, HSO3^–, it reacts with water to form hydroxide ions. That means a sodium sulfite solution is basic, not neutral. For the standard chemistry problem, the key reaction is:

SO3^2- + H2O ⇌ HSO3^– + OH^–

Once that hydrolysis reaction is written, the pH calculation follows the same logic used for weak bases. In most textbook treatments, you use the relation Kb = Kw / Ka2, where Ka2 is the second acid dissociation constant of sulfurous acid chemistry, often represented through the acid pair HSO3^– / SO3^2-.

Short answer for the classic problem

For 0.15 M Na2SO3 at 25 degrees Celsius, using Ka2 = 6.4 × 10^-8 and Kw = 1.0 × 10^-14:

1. Find the base constant of sulfite: Kb = Kw / Ka2 = 1.56 × 10^-7.
2. Use the weak base setup for initial concentration C = 0.15 M.
3. Approximate hydroxide concentration from x ≈ √(KbC).
4. This gives [OH^–] ≈ 1.53 × 10^-4 M.
5. Then pOH ≈ 3.82.
6. Finally, pH ≈ 10.18.

The exact quadratic solution gives essentially the same answer to normal reporting precision. So if your assignment is simply “calculate the pH of 0.15 M Na2SO3,” the expected result is usually about 10.18 to 10.19, depending on the Ka2 value chosen by your instructor or textbook.

Why Na2SO3 makes water basic

Na2SO3 dissociates completely in water:

Na2SO3 → 2 Na⁺ + SO3^2-

The sodium ion is a spectator ion because it comes from the strong base NaOH and does not significantly affect pH. The important species is sulfite, SO3^2-. Sulfite can accept a proton from water, which generates OH^–. More OH^– means the solution becomes basic.

This is a standard pattern in acid base chemistry:

• Salt of a strong acid and strong base, usually near neutral.
• Salt of a weak acid and strong base, usually basic.
• Salt of a strong acid and weak base, usually acidic.

Sodium sulfite falls into the second category. It contains the conjugate base of a weak acid system, so a pH above 7 is expected.

Step by step equilibrium method

1. Write the hydrolysis reaction

SO3^2- + H2O ⇌ HSO3^– + OH^–

2. Determine Kb from Ka2

If your chemistry source gives Ka2 for HSO3^–, then:

Kb = Kw / Ka2

Using Kw = 1.0 × 10^-14 and Ka2 = 6.4 × 10^-8:

Kb = 1.0 × 10^-14 / 6.4 × 10^-8 = 1.5625 × 10^-7

3. Set up an ICE table

Species	Initial (M)	Change (M)	Equilibrium (M)
SO3^2-	0.15	-x	0.15 – x
HSO3^–	0	+x	x
OH^–	0	+x	x

4. Write the equilibrium expression

Kb = [HSO3^–][OH^–] / [SO3^2-]

Substitute from the ICE table:

1.56 × 10^-7 = x² / (0.15 – x)

5. Use the weak base approximation

Because Kb is small and the concentration is relatively large, x is much smaller than 0.15. So:

1.56 × 10^-7 ≈ x² / 0.15

x ≈ √(1.56 × 10^-7 × 0.15) ≈ 1.53 × 10^-4 M

This x value is the hydroxide concentration.

6. Convert to pOH and pH

pOH = -log(1.53 × 10^-4) ≈ 3.82

pH = 14.00 – 3.82 = 10.18

7. Check the approximation

The percentage ionization is:

(1.53 × 10^-4 / 0.15) × 100 ≈ 0.10%

That is far below 5%, so the approximation is valid.

Exact solution versus approximation

Students often wonder whether they need the quadratic equation. In this problem, the approximation is excellent. The exact equation is:

x² + Kb x – Kb C = 0

For C = 0.15 and Kb = 1.5625 × 10^-7, the positive root gives almost the same hydroxide concentration as the square root shortcut. The difference in pH is tiny, which is why most instructors accept either method if your setup is chemically correct.

Method	[OH-] (M)	pOH	pH	Difference from exact
Weak base approximation	1.5309 × 10^-4	3.8150	10.1850	Very small
Exact quadratic	1.5301 × 10^-4	3.8152	10.1848	Reference

The values above show why the approximation is trusted here. The difference is only in the fourth decimal place of pH, much smaller than the uncertainty introduced by different tabulated Ka values.

How concentration changes the pH of sodium sulfite

As sodium sulfite concentration increases, the hydroxide concentration generated by hydrolysis increases too. However, pH does not rise linearly with concentration because pH is logarithmic. This is why a tenfold increase in concentration does not produce a tenfold increase in pH.

Na2SO3 concentration (M)	Approx. [OH-] (M)	Approx. pOH	Approx. pH
0.010	3.95 × 10^-5	4.40	9.60
0.050	8.84 × 10^-5	4.05	9.95
0.100	1.25 × 10^-4	3.90	10.10
0.150	1.53 × 10^-4	3.82	10.18
0.500	2.80 × 10^-4	3.55	10.45

These values are based on the same Ka2 and Kw assumptions. They illustrate a practical point: sodium sulfite is a moderately basic salt solution, but not a strong base like concentrated NaOH. Even at 0.15 M, the pH is around 10.18, not 13 or 14.

Common mistakes students make

• Treating Na2SO3 as a strong base. It is basic, but it is not the same as adding 0.15 M OH^– directly.
• Using Ka1 instead of Ka2. The relevant conjugate pair for SO3^2- is HSO3^– / SO3^2-, so Ka2 is the needed constant.
• Forgetting that sodium is a spectator ion. Na⁺ does not control the pH here.
• Confusing HSO3^– with SO3^2-. The species present initially from Na2SO3 is sulfite, not bisulfite.
• Skipping the pOH step. The hydrolysis gives OH^–, so pOH is often calculated first, then converted to pH.
• Ignoring the approximation check. The 5% rule is a good habit, even when the approximation obviously works.

Advanced note on sulfur(IV) equilibria

In more advanced analytical or environmental chemistry, sulfur(IV) systems can be more complex because dissolved sulfur dioxide, hydrated sulfurous acid species, bisulfite, and sulfite are linked through coupled equilibria. At a basic pH near 10, sulfite becomes much more important than bisulfite. For introductory problems, however, the dominant calculation is still the hydrolysis of SO3^2- using Kb = Kw / Ka2.

This simplified approach is entirely appropriate for general chemistry, AP Chemistry style work, and many first year university problems. If a course specifically asks for a full speciation model, then multiple equilibrium constants, charge balance, and mass balance may be required. For the stated question, that level of complexity is normally unnecessary.

Authoritative references for acid base constants and water chemistry

For students who want higher confidence in the equilibrium framework, these authoritative educational and government resources are useful:

• LibreTexts Chemistry for educational explanations of weak acid and weak base equilibrium methods.
• U.S. Environmental Protection Agency for water chemistry and pH background relevant to aqueous systems.
• NIST Chemistry WebBook for authoritative chemical data and reference information.

Although not every source lists exactly the same Ka values, the procedure remains the same. Small differences in tabulated constants may shift the final pH by a few hundredths.

Final takeaway

If you are asked to calculate the pH of 0.15 M Na2SO3, the reliable chemistry answer is that the solution is basic. Using a standard Ka2 value of 6.4 × 10^-8, the pH comes out to approximately 10.18. The reasoning is straightforward:

1. Na2SO3 dissociates to give SO3^2-.
2. Sulfite hydrolyzes water to form OH^–.
3. Find Kb from Kw / Ka2.
4. Solve for [OH^–].
5. Convert pOH to pH.

This calculator lets you test both the exact quadratic and the approximation so you can see why the expected answer stays very close to 10.18 under standard assumptions.