A sample of hard water contains of Calcium Bicarbonate () and of Magnesium Bicarbonate (). Assuming — Hydrogen Chemistry Question
Question
A $1\text{ Litre}$ sample of hard water contains $16.2\text{ mg}$ of Calcium Bicarbonate ($Ca(HCO_3)_2$) and $7.3\text{ mg}$ of Magnesium Bicarbonate ($Mg(HCO_3)_2$). Assuming no other hardness-causing salts are present, what is the total temporary hardness of this sample expressed in parts per million (ppm) of $CaCO_3$ equivalent? (Molar masses: $Ca(HCO_3)_2 = 162$, $Mg(HCO_3)_2 = 146$, $CaCO_3 = 100$).
Answer: B
💡 Solution & Explanation
Equivalents of $CaCO_3$ from Calcium Bicarbonate = $(16.2 / 162) \times 100 = 10\text{ mg}$. Equivalents of $CaCO_3$ from Magnesium Bicarbonate = $(7.3 / 146) \times 100 = 5\text{ mg}$. Total hardness = $10 + 5 = 15\text{ mg}$ per Litre, which equals $15\text{ ppm}$.
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