The hardness of a water sample containing 10 –3 M MgSO₄ expressed as CaCO₃ equivalents (in ppm) is ______.
(molar mass of MgSO₄ is 120.37 g/mol)

Solution

Chemistry ma’am · 3 steps

Hardness counts moles, not grams: one mole of Mg²⁺ stands in for one mole of CaCO₃.

1. Match the moles. Each MgSO₄ gives one Mg²⁺, and each CaCO₃ gives one Ca²⁺. Both ions carry a 2+ charge, so 10⁻³ mol of MgSO₄ per litre counts as 10⁻³ mol of CaCO₃ per litre.
2. Convert to mass. The molar mass of CaCO₃ is 100 g mol⁻¹, so 10⁻³ mol of it is 0.1 g = 100 mg in each litre.

   ppm = (100 mg)/(1 L water) = (100 mg)/(10⁶ mg) × 10⁶ = 100

3. So. One litre of dilute water has a mass of about 10⁶ mg, so 100 mg of CaCO₃ in it is 100 ppm. The molar mass of MgSO₄ is never needed.

The common slip: 120.37. Using the mass of MgSO₄ instead of CaCO₃ gives 10⁻³ × 120.37 g per litre, which is 120.37 ppm and not 100.

The idea to keep. For hardness, turn the salt into moles of CaCO₃ first, then multiply by 100. For dilute water, mg per litre is the same as ppm.

Answer from the NTA final answer key. Solution by PrepWiser.