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Ca and Mg concentration, alkalinity (300 mg/l hardness)

Short answer

Let Ca = Mg = x mg/L. Total hardness = x(50/20.04 + 50/12.15) = 6.610x = 300, so x = 45.38 mg/L. Alkalinity = 150 × (50/61.02) = 122.91 mg/L as CaCO3. Carbonate hardness = alkalinity = 122.91 mg/L; non-carbonate hardness = 300 – 122.91 = 177.09 mg/L.

The hardness of a water sample was found to be 300 mg/l as CaCO3_3. The hardness was found due to Ca and Mg ions only. The concentrations of these ions are equal in water. The water analysis showed the concentration of HCO3−_3^- was 150 mg/l. Calculate (i) the concentrations of Ca and Mg, (ii) alkalinity of water and (iii) carbonate hardness and non-carbonate hardness of water.

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.

(i) Concentration of Ca and Mg

Let Ca = Mg = xx mg/l.

TH=x(5020.04+5012.15)=x(2.495+4.115)=6.610x=300x=45.38 mg/l\begin{aligned} \text{TH} &= x\left(\frac{50}{20.04}+\frac{50}{12.15}\right) = x(2.495+4.115) = 6.610x = 300\\ x &= 45.38\ \text{mg/l} \end{aligned}

Ca2+^{2+} = Mg2+^{2+} = 45.38 mg/l.

(ii) Alkalinity

Only HCO3−_3^- is present (no CO32−_3^{2-} or OH−^-), so

Alkalinity=150×5061.02=122.91 mg/l as CaCO3\text{Alkalinity} = 150\times\frac{50}{61.02} = 122.91\ \text{mg/l as CaCO}_3

(iii) Carbonate and non-carbonate hardness

The alkalinity (122.91) is less than the total hardness (300), so carbonate hardness equals alkalinity.

CH=122.91 mg/lNCH=TH−CH=300−122.91=177.09 mg/l\begin{aligned} \text{CH} &= 122.91\ \text{mg/l}\\ \text{NCH} &= \text{TH}-\text{CH} = 300-122.91 = 177.09\ \text{mg/l} \end{aligned}

Answer: Ca = Mg = 45.38 mg/l; alkalinity = 122.91 mg/l as CaCO3_3; CH = 122.91 mg/l; NCH = 177.09 mg/l.

✓ Answer checked by Claude Opus 5.5 (high effort)

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