A 100 mL water sample contains exactly 0.81 g of calcium bicarbonate and 0.73 g of magnesium bicarbo — Redox Reactions and Volumetric Analysis Chemistry Question
Question
A 100 mL water sample contains exactly 0.81 g of calcium bicarbonate and 0.73 g of magnesium bicarbonate. The total hardness of this water sample, expressed conventionally in parts per million (ppm) of $CaCO_3$ equivalents, is $H$. What is the value of $H$? (Molar mass: $Ca(HCO_3)_2$ = 162 g/mol, $Mg(HCO_3)_2$ = 146 g/mol, $CaCO_3$ = 100 g/mol, density of water $\approx$ 1.0 g/mL)
💡 Solution & Explanation
Moles of $Ca(HCO_3)_2$ = $0.81 / 162 = 0.005$ mol. Moles of $Mg(HCO_3)_2$ = $0.73 / 146 = 0.005$ mol. Each of these compounds has an n-factor of 2 (containing 2 moles of $HCO_3^-$ or $+2$ cationic charge). Total equivalents = $(0.005 \times 2) + (0.005 \times 2) = 0.02$ eq. Equivalent mass of $CaCO_3$ = 50. Mass of $CaCO_3$ equivalent = $0.02 \times 50 = 1.0$ g. 100 mL water = 100 g. ppm = $(1.0 / 100) \times 10^6 = 10,000$ ppm.