Question

The analysis of a water sample is shown in the table. The hardness of the water sample is ______ mg/L as CaCO₃.

Atomic weight (g/mol): Ca = 40; Mg = 24; Na = 23; Cl = 35.5; S = 32; O = 16; N = 14

Species	Concentration (mg/L)
Na+	30
Ca2+	10
Mg2+	15
Cl-	30
SO42-	60
NO3-	5

• A. 87.5
• B. 62.5
• C. 25.5
• D. 47.5

Hint:

Only Ca²⁺ and Mg²⁺ cause hardness. Convert each using \( \text{mg/L} \times \dfrac{50}{\text{equivalent weight of ion}} \), then add the two results together.

Answer 1

The Correct Option is A

Step 1: Identify the data that actually matters.
Total hardness of water is caused only by the divalent metal ions present, mainly Ca²⁺ and Mg²⁺. The other species listed in the table, Na⁺, Cl^-, SO₄^2- and NO₃^-, do not contribute to hardness at all, so they are not used anywhere in the calculation. From the table: Ca²⁺ = 10 mg/L and Mg²⁺ = 15 mg/L.

Step 2: Work out the equivalent weights.
Equivalent weight = atomic (or molecular) weight divided by valence.
Equivalent weight of Ca = 40 / 2 = 20
Equivalent weight of Mg = 24 / 2 = 12
Equivalent weight of CaCO₃ = 100 / 2 = 50 (molecular weight of CaCO₃ is 100 g/mol, valence 2)

Step 3: Convert the calcium concentration to CaCO3 equivalent.
\( \text{Hardness}_{Ca} = \text{concentration} \times \dfrac{\text{equivalent weight of CaCO}_3}{\text{equivalent weight of Ca}} \)
\[ \text{Hardness}_{Ca} = 10 \times \dfrac{50}{20} = 10 \times 2.5 = 25 \text{ mg/L as CaCO}_3 \]

Step 4: Convert the magnesium concentration to CaCO3 equivalent.
\( \text{Hardness}_{Mg} = \text{concentration} \times \dfrac{\text{equivalent weight of CaCO}_3}{\text{equivalent weight of Mg}} \)
\[ \text{Hardness}_{Mg} = 15 \times \dfrac{50}{12} = 15 \times 4.1\overline{6} = 62.5 \text{ mg/L as CaCO}_3 \]

Step 5: Add both contributions to get total hardness.
\[ \text{Total hardness} = 25 + 62.5 = 87.5 \text{ mg/L as CaCO}_3 \]

Step 6: Match with the options.
The computed value of 87.5 mg/L matches option (A), confirming it as the correct choice.