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Alkalinity and Ca, Mg: TH 280, CH 75

Short answer

Total alkalinity is 75 mg/l as CaCO3 because carbonate hardness is less than total hardness. With Ca and Mg equal, solve 280 = x(2.5 + 4.1), giving Ca = Mg = 42.36 mg/l.

Determine total alkalinity and concentration of calcium and magnesium in the water sample if both calcium and magnesium ions were found equal. Total hardness is 280 mg/l and carbonate hardness is 75 mg/l.

Total hardness TH =280= 280 mg/l, carbonate hardness CH =75= 75 mg/l, Ca2+^{2+} = Mg2+^{2+} =x= x mg/l.

Total alkalinity

TH is greater than CH, so the alkalinity is fully used in carbonate hardness: total alkalinity = CH =75= 75 mg/l as CaCO3_3. (Non-carbonate hardness =280−75=205= 280-75 = 205 mg/l.)

Concentration of Ca and Mg

TH=x(5020.04+5012.15)=6.610x=280x=42.36 mg/l\begin{aligned} \text{TH} &= x\left(\frac{50}{20.04}+\frac{50}{12.15}\right) = 6.610x = 280\\ x &= 42.36\ \text{mg/l} \end{aligned}

Answer: Total alkalinity = 75 mg/l as CaCO3_3; Ca2+^{2+} = Mg2+^{2+} = 42.36 mg/l.

✓ Answer checked by Claude Opus 5.5 (high effort)

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