Molecular Sulfur Dioxide in a White Wine: Add Sulfite or Lower the pH?
A small winery holds 4 500 dm³ of dry white wine at pH 3.75, with 32 mg dm⁻³ free SO₂ measured by Ripper titration and 95 mg dm⁻³ total SO₂.
The winemaker wants at least 0.8 mg dm⁻³ of molecular SO₂ - the uncharged form that actually inhibits yeasts and bacteria. Below that level the wine risks refermentation in bottle.
Two options are proposed:
- Add potassium metabisulfite, K₂S₂O₅ (Mr 222.3), which gives 2 mol SO₂ per mol of salt.
- Add tartaric acid to bring the pH to 3.20, adding no sulfite at all.
The legal limit for total SO₂ in this style of wine is 200 mg dm⁻³, and the winemaker does not want to exceed 140 mg dm⁻³ because tasters previously described the wine as "struck match".
Dissociation constants for sulfurous acid: Ka1 = 1.5 × 10⁻² (pKa1 1.82) and Ka2 = 6.3 × 10⁻⁸ (pKa2 7.20).
1. Calculate the percentage of free SO₂ present as molecular SO₂ at pH 3.75, and decide whether the wine currently meets the 0.8 mg dm⁻³ target.
Read the worked answer
Set up the fraction. Free SO₂ partitions as H₂SO₃ / HSO₃⁻ / SO₃²⁻. The fraction that is molecular is
f = [H⁺]² / ([H⁺]² + Ka1[H⁺] + Ka1Ka2)
At pH 3.75: [H⁺] = 10⁻³·⁷⁵ = 1.78 × 10⁻⁴ mol dm⁻³
| Term | Value |
|---|---|
| [H⁺]² | 3.16 × 10⁻⁸ |
| Ka1[H⁺] = (1.5 × 10⁻²)(1.78 × 10⁻⁴) | 2.67 × 10⁻⁶ |
| Ka1Ka2 = (1.5 × 10⁻²)(6.3 × 10⁻⁸) | 9.45 × 10⁻¹⁰ |
| Denominator | 2.70 × 10⁻⁶ |
f = 3.16 × 10⁻⁸ / 2.70 × 10⁻⁶ = 1.17 × 10⁻², i.e. 1.17 %
Check with the shortcut: because pH ≫ pKa1, almost everything is HSO₃⁻ and the Ka2 term is negligible, so f ≈ [H⁺]/Ka1 = 1.78 × 10⁻⁴ / 1.5 × 10⁻² = 1.19 × 10⁻². Agreement is good.
Molecular SO₂ present = 0.0117 × 32 = 0.375 mg dm⁻³
That is less than half the 0.8 mg dm⁻³ target - the wine is not protected, even though 32 mg dm⁻³ of "free SO₂" sounds generous. The titration measures HSO₃⁻ and SO₂ together; at this pH nearly all of it is the charged, membrane-impermeable bisulfite ion, which does not inhibit microbes.
2. Calculate what each option would require: the mass of K₂S₂O₅ needed to reach the target at pH 3.75, and the molecular SO₂ achieved if the pH is instead lowered to 3.20.
Read the worked answer
Option 1 - add metabisulfite at pH 3.75.
Free SO₂ required = 0.8 / 0.0117 = 68.4 mg dm⁻³ Additional SO₂ needed = 68.4 − 32 = 36.4 mg dm⁻³ Total SO₂ over 4 500 dm³ = 36.4 × 4 500 = 163 800 mg = 163.8 g SO₂
K₂S₂O₅ supplies 2 × 64.1 = 128.2 g SO₂ per 222.3 g salt: mass of salt = 163.8 × (222.3 / 128.2) = 284 g
Resulting total SO₂ = 95 + 36.4 = 131 mg dm⁻³ - legal, and just inside the winemaker's 140 mg dm⁻³ ceiling, but with little margin.
Option 2 - acidify to pH 3.20 with no sulfite.
[H⁺] = 10⁻³·²⁰ = 6.31 × 10⁻⁴ mol dm⁻³
| Term | Value |
|---|---|
| [H⁺]² | 3.98 × 10⁻⁷ |
| Ka1[H⁺] | 9.47 × 10⁻⁶ |
| Ka1Ka2 | 9.45 × 10⁻¹⁰ |
| Denominator | 9.87 × 10⁻⁶ |
f = 3.98 × 10⁻⁷ / 9.87 × 10⁻⁶ = 4.03 × 10⁻² (4.03 %)
Molecular SO₂ = 0.0403 × 32 = 1.29 mg dm⁻³
A drop of 0.55 pH units raises the active fraction by a factor of about 3.4 and meets the target with no sulfite added at all - total SO₂ stays at 95 mg dm⁻³.
3. Recommend an option to the winemaker, and identify the assumptions and risks that the calculation alone does not capture.
Read the worked answer
Recommendation: acidify rather than add sulfite, but not blindly to 3.20.
The calculation makes option 2 clearly preferable: it meets the 0.8 mg dm⁻³ target, costs far less than 284 g of metabisulfite, leaves total SO₂ at 95 mg dm⁻³ (well under both the legal 200 and the winemaker's 140 mg dm⁻³), and avoids adding any more of the compound responsible for the "struck match" note. A prudent target is pH 3.30–3.40 plus a small sulfite top-up, so that neither control is pushed to its limit.
What the arithmetic does not capture:
- Taste and texture. Tartaric acid addition raises titratable acidity and sharpens the palate; the wine may need cold stabilisation, and potassium hydrogen tartrate may precipitate later, releasing H⁺ and shifting pH unpredictably.
- pH is a moving target. Yeast and malolactic activity, and bitartrate precipitation, all buffer or shift pH; a wine adjusted to exactly 3.20 has no margin if pH drifts up. Molecular SO₂ should be checked weekly, not once at bottling.
- Bound SO₂. Ripper titration measures free SO₂, but acetaldehyde produced during fermentation binds SO₂ as a hydroxysulfonate adduct that is microbiologically inactive. "Free" is therefore an overestimate of what is available.
- Metabisulfite is the salt of a weak acid and raises pH slightly on dissolution, so option 1 would need marginally more salt than calculated, and the target would drift.
- Measurement uncertainty. Ripper titration is typically ±5–10 %, so 32 mg dm⁻³ could be 29–35, changing the answer appreciably when the safety margin is thin.
- Microbial load. 0.8 mg dm⁻³ molecular SO₂ suppresses most spoilage yeasts but not all; a wine with residual sugar and a high viable count needs more than the threshold, and filtration may be the better control.