Question

Calculate in detail total temporary hardness, total permanent hardness, in terms of calcium
carbonate equivalents in a water sample containing Magnesium bicarbonate (15.5 mg), Calcium
Sulphate (8.5 mg), Magnesium Chloride (9.1 mg), calcium sulphate (13 mg)

Answer 1

$\begin{aligned} Mg(HCO_3)_2 =& \ 15.5mg\\ CaSO_4 =&\ 8.5mg\\ MgCl_2 =&\ 9.1g\\ Ca(HCO_3)_2 =&\ 13mg \end{aligned}$

Temporary hardness is present in bicarbonates only;

$\begin{aligned} \textsf{mole of }Mg(HCO_3)_2 &= \dfrac{15.5×10^{-3}}{146g/mol} \\ \\ &= 1.06× 10^{-4} moles\\ \\ \textsf{mole of }Ca(HCO_3)_2 &= \dfrac{13×10^{-3}}{162g/mol} \\ \\ &= 8.02× 10^{-5} moles \end{aligned}$

$\begin{aligned} \\ \textsf{Total moles of }Ca^+\textsf{and }Mg ^+ &= (1.06 × 10^{-4})+(8.02×10^{-5})\\ &=1.86×10^{-4} moles \end{aligned}$

$\begin{aligned} \\ \therefore \textsf{Total moles of }CaCO_3 = 1.86×10^{-4} moles \end{aligned}$

$\begin{aligned} \textsf{Total mass of } CaCO_3 =1.86×1&0^{-4} moles × 100g/mol\\ =0.0186g&\\ \\ \therefore \textsf{ppm (temporary hardness)} &= \dfrac{0.0186g}{1000} × 10^6\\ \\ &= 18.6ppm \end{aligned}$

Permanent hardness is present in Chlorides and Sulphates;

$\begin{aligned} \\ \textsf{mole of }MgCl_2 &= \dfrac{9.1×10^{-3}g}{95g/mol} \\ \\ &= 9.58× 10^{-5} moles\\ \\ \textsf{mole of }CaSO_4 &= \dfrac{8.5×10^{-3}g}{136g/mol} \\ \\ &= 6.25× 10^{-5} moles \end{aligned}$

$\begin{aligned} \\ \textsf{Total moles of }Ca^+\textsf{and }Mg ^+ &= (9.58 × 10^{-5})+(6.25×10^{-5})\\ &=1.58×10^{-4} moles \end{aligned}$

$\begin{aligned} \\ \therefore \textsf{Total moles of }CaCO_3 = 1.58×10^{-4} moles \end{aligned}$

$\begin{aligned} \textsf{Total mass of } CaCO_3 =1.58×1&0^{-4} moles × 100g/mol\\ =0.0158g&\\ \\ \therefore \textsf{ppm (temporary hardness)} &= \dfrac{0.0158g}{1000} × 10^6\\ \\ &= 15.8ppm \end{aligned}$

$\therefore$ The total temporary hardness and the total permanent hardness in terms of Calcium Carbonate equivalents is 18.6ppm and 15.8ppm respectively.