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Hardness of well water with strontium 25.26 mg/l

Short answer

Using equivalent factors, total hardness = sum of Ca, Mg and Sr hardness = 397.57 mg/L as CaCO3. Alkalinity from HCO3- and CO3-- is 262.05 mg/L. Carbonate hardness = 262.05, non-carbonate = 135.52 mg/L.

The analysis of water from a well showed the following result in mg/l: Na+^+ = 205, Mg++^{++} = 45, Ca++^{++} = 75, K+^+ = 21.5, Cl−^- = 40, HCO3−_3^- = 92, SO4−−_4^{--} = 22.8, NO3−_3^- = 12, CO3−−_3^{--} = 112. The concentration of strontium (Sr) is equivalent to a hardness of 25.26 mg/l. Calculate the carbonate hardness, non-carbonate hardness and total hardness.

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 is given as 25.26 mg/l hardness. Na+^+, K+^+, Cl−^-, SO42−_4^{2-} and NO3−_3^- do not cause hardness or alkalinity.

Total hardness

Ionmg/lHardness (mg/l as CaCO3_3)
Ca2+^{2+}75187.13
Mg2+^{2+}45185.19
Sr25.26
Total397.57

Alkalinity

HCO3−: 92×5061.02=75.39CO32−: 112×5030=186.67Total alkalinity=262.05 mg/l as CaCO3\begin{aligned} \text{HCO}_3^-:&\ 92\times\frac{50}{61.02} = 75.39\\ \text{CO}_3^{2-}:&\ 112\times\frac{50}{30} = 186.67\\ \text{Total alkalinity} &= 262.05\ \text{mg/l as CaCO}_3 \end{aligned}

Carbonate and non-carbonate hardness

Alkalinity (262.05) is less than total hardness (397.57), so CH = alkalinity and

NCH=397.57−262.05=135.52 mg/l\text{NCH} = 397.57-262.05 = 135.52\ \text{mg/l}

Answer: Carbonate hardness = 262.05 mg/l; non-carbonate hardness = 135.52 mg/l; total hardness = 397.57 mg/l (as CaCO3_3).

✓ Answer checked by Claude Opus 5.5 (high effort)

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