Hardness with Ca, Mg, Sr equal, TH 150

Hardness is expressed as CaCO$_3$ equivalent. Each cation of concentration $C$ (mg/l) contributes $C \times \dfrac{50}{\text{eq. wt}}$ to hardness, which gives the standard factors: Ca 2.5, Mg 4.1, Sr 1.14.

$$TH = 2.5\,Ca + 4.1\,Mg + 1.14\,Sr$$

(i) Non-carbonate hardness

$$NCH = TH - CH = 150 - 77 = 73\ \text{mg/l as CaCO}_3$$

(ii) Concentration of principal cations

Let $Ca = Mg = Sr = x$.

$$\begin{aligned} 150 &= (2.5 + 4.1 + 1.14)\,x = 7.74\,x \\ x &= \frac{150}{7.74} = 19.38\ \text{mg/l} \end{aligned}$$

So Ca = Mg = Sr = 19.38 mg/l each.

(iii) Total alkalinity

Carbonate hardness is the part of hardness balanced by alkalinity. When alkalinity is less than total hardness, carbonate hardness equals alkalinity (the alkalinity is due to bicarbonate). Since CH = 77 < TH = 150:

$$\text{Total alkalinity} = CH = 77\ \text{mg/l as CaCO}_3$$

Answer: NCH = 73 mg/l; Ca = Mg = Sr = 19.38 mg/l; total alkalinity = 77 mg/l (as CaCO$_3$).