Larson–Skold Index

Also known as larson ratio · chloride sulfate ratio

LS=Cl35.45+SO448.03Alk50.04\mathrm{LS} = \dfrac{\frac{\mathrm{Cl}}{35.45} + \frac{\mathrm{SO_4}}{48.03}}{\frac{\mathrm{Alk}}{50.04}}

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Learning zone

Langelier and Ryznar only ever look at calcium carbonate. Larson and Skold, working on Great Lakes water for the Illinois State Water Survey in 1958, went after the other half of the story: chloride and sulphate are penetrating anions that break down the passive film on mild steel, while bicarbonate alkalinity helps repair it. Their index is a straight equivalents ratio — divide each ion by its equivalent weight (35.45 for Cl⁻, 48.03 for SO₄²⁻, 50.04 for alkalinity as CaCO₃) and compare aggressors to protectors. A water with 50 mg/L chloride, 80 mg/L sulphate and 150 mg/L alkalinity gives (1.410 + 1.666)/2.998 = 1.03.

Read it in thirds: below 0.8 chloride and sulphate are unlikely to interfere with film formation, 0.8 to 1.2 means corrosion rates run higher than open-recirculating norms, and above 1.2 you should expect localized pitting rather than general wastage. This is the index that explains why a softened, high-chloride makeup can hold a perfect +0.2 LSI and still perforate a carbon-steel line in eighteen months, and why blowing down a tower on conductivity alone concentrates exactly the ions Larson–Skold cares about. Note that it is silent on stainless and copper — for those, chloride's absolute concentration and temperature matter more than the ratio.

Larson–Skold Index
LS=Cl35.45+SO448.03Alk50.04\mathrm{LS} = \dfrac{\frac{\mathrm{Cl}}{35.45} + \frac{\mathrm{SO_4}}{48.03}}{\frac{\mathrm{Alk}}{50.04}}
Where
  • LS\mathrm{LS}= Larson–Skold Index
  • Cl\mathrm{Cl}= Chloride as Cl⁻
  • SO4\mathrm{SO_4}= Sulphate as SO₄²⁻
  • Alk\mathrm{Alk}= Total alkalinity as CaCO₃
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