Water quality — A well-water sample contains 140 g/m³ of Ca²⁺ ions and 345 g/m³ of Na⁺ ions. What is the hardness of this water expressed as equivalent CaCO₃ (g/m³)? (Atomic masses: Ca = 40, Na = 23, C = 12, O = 16.)

Introduction:Hardness expresses the concentration of multivalent cations (mainly Ca²⁺ and Mg²⁺) as an equivalent amount of CaCO3. It is widely used in water-treatment design and reporting.

Given Data / Assumptions:

• [Ca²⁺] = 140 g/m³ = 140 mg/L.
• [Na⁺] = 345 g/m³ = 345 mg/L (does not contribute to hardness).
• Equivalent weight: Ca²⁺ → 40/2 = 20 g/eq; CaCO3 → 50 g/eq.

Concept / Approach:Hardness as CaCO3 is computed by converting the Ca²⁺ concentration to equivalents, then to CaCO3 using 50 g/eq. Sodium does not contribute to hardness because it is monovalent.

Step-by-Step Solution:Convert Ca²⁺ to mg/L: 140 g/m³ = 140 mg/L.Compute hardness as CaCO3: Hardness = (mg/L of Ca²⁺) * (50 / equivalent weight of Ca²⁺).Equivalent weight of Ca²⁺ = 40/2 = 20 g/eq.Hardness = 140 * (50/20) = 140 * 2.5 = 350 mg/L.In g/m³, the numeric value is the same: 350 g/m³.

Verification / Alternative check:Dimensional consistency: mg/L × (dimensionless 50/20) gives mg/L as CaCO3; converting to g/m³ preserves the number.

Why Other Options Are Wrong:

• 485 assumes Na⁺ contributes like Ca²⁺; it does not.
• 140 uses Ca²⁺ as if already CaCO3 equivalents (missing the 50/20 factor).
• 345 incorrectly uses Na⁺ concentration as hardness.
• 700 would double-count or misapply equivalent weights.

Common Pitfalls:Including Na⁺/K⁺ in hardness; using molar instead of equivalent weights; unit confusion between mg/L and g/m³.

Final Answer:350