Question

The hardness of a water sample containing ${10}^{-3}\mathrm{M}{\mathrm{MgSO}}_4$ expressed as ${\mathrm{CaCO}}_3$ equivalents (in ppm)is (molar mass of ${\mathrm{MgSO}}_4$is $120.37$$\mathrm{g}/\mathrm{mol}$)

Answer 1

The hardness of water:

• Compounds of calcium and magnesium, as well as a number of other metals, contribute to hardness. According to general recommendations, fluids should be categorized as soft up to 60 mg/L (milligrams per liter) of calcium carbonate, moderately hard between 61 and 120 mg/L, hard between 121 and 180 mg/L, and very hard over 180 mg/L.

Solution:

$$\begin{gathered} \mathrm{Hardness}\mathrm{of}\mathrm{water}\mathrm{is}\mathrm{measured}\mathrm{in}\mathrm{ppm}\mathrm{in}\mathrm{terms}{\mathrm{CaCO}}_{3.} \\ {\mathrm{nCaCO}}_3={\mathrm{nMgSO}}_4 \\ \mathrm{ppm}\mathrm{is}\mathrm{the}\mathrm{parts}(\mathrm{in}\mathrm{grams})\mathrm{present}\mathrm{per}\mathrm{million}\mathrm{i}.\mathrm{e},{10}^6 \\ 1000\mathrm{mL}\mathrm{has}{10}^{-3}\mathrm{moles}\mathrm{of}{\mathrm{MgSO}}_{4.} \\ \mathrm{Grams}\mathrm{of}{\mathrm{CaCO}}_3\mathrm{in}1000\mathrm{mL}={10}^{-3}\times 100\mathrm{grams} \\ \mathrm{Grams}\mathrm{of}{\mathrm{CaCO}}_3\mathrm{in}1\mathrm{mL}={10}^{-3}\times 100/1000\mathrm{grams} \\ \mathrm{Hardness}=({10}^{-3}\times 100/1000)\times {10}^6 \\ =100\mathrm{ppm}. \end{gathered}$$