Hardness from ion analysis (Ca 65, Mg 35)

Hardness (or alkalinity) due to an ion $= \text{concentration (mg/l)}\times\dfrac{50}{\text{equivalent weight}}$ as CaCO$_3$. Equivalent weights: Ca = 20.04, Mg = 12.15, Sr = 43.81, HCO$_3$ = 61.02, CO$_3$ = 30. Na$^+$, K$^+$ and SO$_4^{2-}$ do not contribute to hardness or alkalinity.

Total hardness

$$\begin{aligned} \text{Ca hardness} &= 65\times\frac{50}{20.04} = 162.18\\ \text{Mg hardness} &= 35\times\frac{50}{12.15} = 144.03\\ \text{TH} &= 306.21\ \text{mg/l as CaCO}_3 \end{aligned}$$

Carbonate hardness

Alkalinity due to HCO$_3^-$ $= 248\times\dfrac{50}{61.02} = 203.21$ mg/l as CaCO$_3$. It is less than TH, so

$$\text{CH} = 203.21\ \text{mg/l},\qquad \text{NCH} = 306.21-203.21 = 103.00\ \text{mg/l}$$

Answer: Total hardness = 306.21 mg/l; carbonate hardness = 203.21 mg/l; non-carbonate hardness = 103.00 mg/l (as CaCO$_3$).