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Non-carbonate hardness: TH 150, CH 60

Short answer

Non-carbonate hardness = 150 – 60 = 90 mg/L; alkalinity = 60 mg/L as CaCO3. Let equal concentration x of Ca, Mg, Sr; total hardness factor sum 7.752×x = 150, so x = 19.35 mg/L.

The total hardness of water is 150 mg/l and carbonate hardness is 60 mg/l. All the three bi-valent metallic ions causing hardness are same. Determine the non-carbonate hardness, alkalinity and concentration of bi-valent metallic ions.

Total hardness (TH) =150= 150 mg/l, carbonate hardness (CH) =60= 60 mg/l. The three bivalent metallic ions are taken as Ca2+^{2+}, Mg2+^{2+} and Sr2+^{2+} present in equal concentration xx mg/l (equivalent weights 20.04, 12.15 and 43.81).

Non-carbonate hardness

NCH=TH−CH=150−60=90 mg/l as CaCO3\text{NCH} = \text{TH}-\text{CH} = 150-60 = 90\ \text{mg/l as CaCO}_3

Alkalinity

Total hardness is greater than carbonate hardness, so all alkalinity is used in carbonate hardness: alkalinity = CH =60= 60 mg/l as CaCO3_3.

Concentration of the bivalent ions

TH=x(5020.04+5012.15+5043.81)=x(2.495+4.115+1.141)=7.752x=150x=19.35 mg/l\begin{aligned} \text{TH} &= x\left(\frac{50}{20.04}+\frac{50}{12.15}+\frac{50}{43.81}\right) = x(2.495+4.115+1.141) = 7.752x = 150\\ x &= 19.35\ \text{mg/l} \end{aligned}

Answer: NCH = 90 mg/l as CaCO3_3; alkalinity = 60 mg/l as CaCO3_3; each bivalent ion = 19.35 mg/l.

✓ Answer checked by Claude Opus 5.5 (high effort)

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