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Hardness from ion analysis with strontium

Short answer

Multiply each hardness ion concentration by (50/equivalent weight); sum for total hardness. Alkalinity from HCO3- only. Total hardness = 101.02 mg/L, carbonate hardness = 59.00 mg/L, non-carbonate hardness = 42.02 mg/L as CaCO3.

The analysis of a water sample shows the following results in mg/l: Na+^+ = 15; K+^+ = 25; Ca++^{++} = 10; Mg++^{++} = 18; Cl−^- = 35; HCO3−_3^- = 72; SO4−−_4^{--} = 8; NO3−_3^- = 15. The concentration of strontium (Sr) is equivalent to a hardness of 2.0 mg/l and carbonate alkalinity in this water is zero. Calculate the total hardness, carbonate hardness and non-carbonate hardness in mg/l as CaCO3_3.

Hardness (or alkalinity) due to an ion =concentration (mg/l)×50equivalent weight= \text{concentration (mg/l)}\times\dfrac{50}{\text{equivalent weight}} as CaCO3_3. Equivalent weights: Ca = 20.04, Mg = 12.15, Sr = 43.81, HCO3_3 = 61.02, CO3_3 = 30. Strontium contributes 2.0 mg/l as given. Na+^+ and K+^+ do not cause hardness.

Total hardness

Ionmg/lFactorHardness (mg/l as CaCO3_3)
Ca2+^{2+}1050/20.04 = 2.49524.95
Mg2+^{2+}1850/12.15 = 4.11574.07
Srgiven2.00
Total101.02

Alkalinity and carbonate hardness

Carbonate alkalinity is zero, so the alkalinity is due to HCO3−_3^-:

Alkalinity=72×5061.02=59.00 mg/l as CaCO3\text{Alkalinity} = 72\times\frac{50}{61.02} = 59.00\ \text{mg/l as CaCO}_3

Alkalinity is less than the total hardness, so CH = alkalinity. NCH =101.02−59.00=42.02= 101.02-59.00 = 42.02 mg/l.

Answer: Total hardness = 101.02 mg/l; carbonate hardness = 59.00 mg/l; non-carbonate hardness = 42.02 mg/l (all as CaCO3_3).

✓ Answer checked by Claude Opus 5.5 (high effort)

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