Problem 7.6: Calculate the LarsonSkold Index LS for this

From Chapter 7 Problem Set. Level 2: Applied Thinking.

Question

Calculate the Larson–Skold Index (LS) for this water: Cl⁻ = 250 mg/L, SO₄²⁻ = 120 mg/L, M-alkalinity = 300 ppm as CaCO₃, P-alkalinity = 0. Interpret the result.

Answer

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Cl⁻ = 250 / 35.5 = 7.04 meq/L. SO₄²⁻ = 120 / 48 = 2.50 meq/L. P-alk = 0, so all alkalinity is bicarbonate: HCO₃⁻ = 300 / 50 = 6.00 meq/L. LS = (7.04 + 2.50) / 6.00 = 1.59. Interpretation: LS > 1.2 indicates high corrosion risk - aggressive anions significantly outweigh alkalinity protection. Passive film destabilization, pitting, and under-deposit corrosion are likely.