Calculate The Ph Of A 0.50 M Solution Of Na2So3

Use this premium chemistry calculator to estimate the pH of sodium sulfite solutions from hydrolysis equilibrium. The default setup solves the classic problem for a 0.50 M Na2SO3 solution and shows pH, pOH, hydroxide concentration, and the hydrolysis constant used.

Core chemistry used

1. Na2SO3 dissociates essentially completely into 2 Na+ and SO3^2-.
2. Sulfite is the conjugate base of HSO3-.
3. For sulfite hydrolysis, Kb = Kw / Ka2.
4. Equilibrium setup: SO3^2- + H2O ⇌ HSO3- + OH-.
5. If the initial sulfite concentration is C and x = [OH-], then Kb = x² / (C – x).
6. From x, compute pOH = -log10[OH-] and pH = 14 – pOH.

For the default values, the expected answer is mildly basic: pH ≈ 10.45.

This is because sulfite is a weak base in water, generated from the second dissociation step of sulfurous acid.

Expert Guide: How to Calculate the pH of a 0.50 M Solution of Na2SO3

To calculate the pH of a 0.50 M solution of Na2SO3, you need to recognize what sodium sulfite does in water. Sodium ions, Na+, are spectator ions for this problem, while sulfite, SO3^2-, is the chemically active species. Because SO3^2- is the conjugate base of hydrogen sulfite, HSO3-, it reacts with water to produce hydroxide ions. That hydrolysis makes the solution basic. The result is not strongly basic like sodium hydroxide, but clearly above neutral pH.

This question appears often in general chemistry, AP Chemistry, and introductory analytical chemistry because it tests several important ideas at once: salt hydrolysis, conjugate acid-base relationships, equilibrium constants, and approximation versus exact solutions. A student who simply sees “salt in water” and assumes neutrality will miss the fact that Na2SO3 comes from a strong base component and a weak acid component. The weak acid piece matters here.

Step 1: Identify the relevant acid-base behavior

Na2SO3 dissociates in water:

Na2SO3 → 2 Na+ + SO3^2-

The sodium ion does not meaningfully affect pH in this context. The sulfite ion does. Sulfite is the conjugate base of HSO3-, so it hydrolyzes in water according to:

SO3^2- + H2O ⇌ HSO3- + OH-

Because hydroxide ions are produced, the solution is basic. Therefore, the key quantity is the base ionization constant, Kb, for sulfite.

Step 2: Convert Ka2 into Kb

For a conjugate acid-base pair, the relationship is:

Kb = Kw / Ka2

At 25 C, a common value used for the second acid dissociation constant of sulfurous acid is:

Ka2 = 6.4 × 10^-8

And the ionic product of water is:

Kw = 1.0 × 10^-14

So:

Kb = (1.0 × 10^-14) / (6.4 × 10^-8) = 1.56 × 10^-7

Quantity	Symbol	Typical value at 25 C	Why it matters
Second dissociation constant of sulfurous acid	Ka2	6.4 × 10^-8	Connects HSO3- and SO3^2- in equilibrium calculations
Ion-product constant for water	Kw	1.0 × 10^-14	Used to derive Kb from Ka
Base hydrolysis constant for sulfite	Kb	1.56 × 10^-7	Determines how much OH- forms
Initial sodium sulfite concentration	C	0.50 M	Sets the equilibrium starting point

Step 3: Set up the ICE table

Let the initial concentration of sulfite be 0.50 M and let x be the amount that reacts.

• Initial: [SO3^2-] = 0.50, [HSO3-] = 0, [OH-] = 0
• Change: [SO3^2-] = -x, [HSO3-] = +x, [OH-] = +x
• Equilibrium: [SO3^2-] = 0.50 – x, [HSO3-] = x, [OH-] = x

Substitute into the base equilibrium expression:

Kb = [HSO3-][OH-] / [SO3^2-] = x² / (0.50 – x)

Using Kb = 1.56 × 10^-7:

1.56 × 10^-7 = x² / (0.50 – x)

Step 4: Solve for hydroxide concentration

Because Kb is small relative to the initial concentration, many courses use the weak base approximation and assume that 0.50 – x ≈ 0.50. Then:

x² = (1.56 × 10^-7)(0.50) = 7.8 × 10^-8

x = √(7.8 × 10^-8) ≈ 2.79 × 10^-4 M

So:

[OH-] = 2.79 × 10^-4 M

If you prefer the exact quadratic solution, you solve:

x² + (1.56 × 10^-7)x – (7.8 × 10^-8) = 0

The positive root is essentially the same as the approximate answer for this concentration because x is tiny compared with 0.50 M. The percent ionization is only about 0.056%, so the approximation is excellent.

Step 5: Convert hydroxide concentration to pOH and pH

Now compute pOH:

pOH = -log(2.79 × 10^-4) ≈ 3.55

Then calculate pH:

pH = 14.00 – 3.55 = 10.45

Final answer for the default problem: the pH of a 0.50 M Na2SO3 solution is approximately 10.45 at 25 C.

Why the solution is basic

This is a good place to reinforce the underlying logic. Sodium sulfite is a salt formed from sodium hydroxide, a strong base, and sulfurous acid, a weak diprotic acid. Salts of strong bases and weak acids generally produce basic solutions because the anion hydrolyzes water. Sulfite is basic enough to generate a measurable hydroxide concentration, but not so basic that the pH climbs to 13 or 14 at this concentration. That is why the answer lands around 10.45.

Common student mistakes

1. Treating Na2SO3 as neutral. This overlooks sulfite hydrolysis.
2. Using Ka1 instead of Ka2. The conjugate pair relevant to SO3^2- is HSO3-/SO3^2-, so Ka2 is the proper constant.
3. Forgetting to convert Kb to pH through pOH. Since the base reaction makes OH-, pOH comes first.
4. Using 0.50 M directly as [OH-]. The salt concentration is not the hydroxide concentration.
5. Ignoring temperature dependence. If temperature changes enough, Kw and equilibrium constants shift.

Approximation versus exact solution

For many weak acid and weak base problems, the approximation method saves time. Here it works very well because the equilibrium shift is small compared with the starting concentration. Still, advanced classes may expect you to justify the approximation. A quick check is to compare x to the initial concentration:

(2.79 × 10^-4 / 0.50) × 100 ≈ 0.056%

This is far below the common 5% guideline, so the approximation is valid.

Na2SO3 concentration (M)	Predicted [OH-] (M)	Predicted pOH	Predicted pH	Approximate percent hydrolyzed
0.010	3.95 × 10^-5	4.40	9.60	0.395%
0.050	8.83 × 10^-5	4.05	9.95	0.177%
0.10	1.25 × 10^-4	3.90	10.10	0.125%
0.50	2.79 × 10^-4	3.55	10.45	0.056%
1.00	3.95 × 10^-4	3.40	10.60	0.0395%

What these numbers tell you

The comparison table shows a pattern common to weak base hydrolysis. As concentration rises, pH increases, but not linearly. A tenfold concentration increase does not raise the pH by a full unit because the equilibrium relation depends on the square root of concentration in the approximation region. That is a useful mental shortcut for quick estimation. If concentration increases by a factor of 100, the hydroxide concentration rises by roughly a factor of 10, corresponding to about 1 pH unit.

Connection to polyprotic acid chemistry

The sulfite system comes from sulfurous acid, H2SO3, a diprotic acid. Diprotic acids lose protons in steps, and each step has its own Ka. The first dissociation is significantly stronger than the second. Because SO3^2- is formed after the second deprotonation, its conjugate acid is HSO3-, and the proper constant is Ka2, not Ka1. That distinction is fundamental in equilibrium chemistry and appears in many exam questions involving carbonate, sulfite, phosphate, and similar ions.

Real-world context for sulfite chemistry

Sulfite chemistry matters beyond textbooks. Sulfite and bisulfite species appear in industrial processes, environmental chemistry, water treatment contexts, and food preservation discussions. Their acid-base behavior affects stability, reactivity, and speciation. In aqueous systems, pH can shift the balance among sulfurous acid related species. While the simplified classroom model here focuses on sulfite hydrolysis alone, the same equilibrium thinking extends to more advanced studies in geochemistry, environmental engineering, and industrial chemistry.

Authoritative references for equilibrium constants and acid-base concepts

• National Institute of Standards and Technology (NIST)
• Chemistry LibreTexts
• U.S. Environmental Protection Agency (EPA)

For .gov and .edu sources specifically relevant to acid-base chemistry and aqueous equilibria, these are useful starting points:

• nist.gov for standards and chemical data resources.
• LibreTexts chemistry course content, widely used in college chemistry education.
• EPA pH overview for broader pH context in water systems.

Fast exam strategy

1. Write the dissociation of Na2SO3 and identify SO3^2- as the basic species.
2. Write the hydrolysis reaction with water.
3. Compute Kb = Kw / Ka2.
4. Use an ICE table with x = [OH-].
5. Apply the approximation if justified.
6. Calculate pOH, then convert to pH.

If you remember only one number from this page, remember this: a 0.50 M sodium sulfite solution has a pH of about 10.45 at 25 C when Ka2 = 6.4 × 10^-8. That answer follows directly from sulfite hydrolysis and is a classic example of a basic salt solution.

Final summary

To calculate the pH of a 0.50 M solution of Na2SO3, treat sulfite as a weak base. Use the equilibrium relation for the hydrolysis of SO3^2- and derive Kb from Ka2 and Kw. Solving the equilibrium gives an hydroxide concentration of roughly 2.79 × 10^-4 M, which corresponds to pOH ≈ 3.55 and pH ≈ 10.45. The answer is basic because sulfite consumes water and produces hydroxide. With this framework, you can solve the same type of problem for many other salts of weak acids and strong bases.