10 L of hard water required 0.56 g of lime (CaO) for removing hardness. Hence, temporary hardness in ppm (part per million 10⁶) of CaCO₃ is:

Temporary hardness is due to ${{\mathrm{HCO}}_3}^{⊝} \mathrm{of} {\mathrm{Ca}}^{2+} \mathrm{and} {\mathrm{Mg}}^{2+}$

$\begin{array}{l}\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2+\frac{\mathrm{CaO}}{56\mathrm{g}}\underset{ }{\to }\frac{2{\mathrm{CaCO}}_3}{(2\times 100)\mathrm{g}}+{\mathrm{H}}_2\mathrm{O}\\ 0.56\mathrm{g} \mathrm{CaO}\equiv 2\mathrm{g} {\mathrm{CaCO}}_3\text{ in }10 \mathrm{L} {\mathrm{H}}_2\mathrm{O}\\ =2\mathrm{g} {\mathrm{CaCO}}_3\text{ in }{10}^4\mathrm{mL} {\mathrm{H}}_2\mathrm{O}\\ =200\mathrm{g}{\mathrm{CaCO}}_3\text{ in }{10}^6\mathrm{mL} {\mathrm{H}}_2\mathrm{O}\end{array}$