Question

A sample of water is known to contain $Mg(HCO_{3})_{2} = 7.3$ mg/L, $Ca(HCO_{3})_{2} = 8.1$ mg/L and $27.2$ mg/L of $CaSO_{4}$. The total hardness associated with water sample (in ppm) in equivalents of $CaCO_{3}$ is

• A. 20
• B. 25
• C. 30
• D. 40

Hint:

Always convert everything to $CaCO_{3}$ equivalents using: $(Mass \times 50) / Equivalent Weight$.

Answer 1

The Correct Option is D

Step 1: Concept
Hardness in ppm ($CaCO_{3}$ equivalent) = (Weight of substance Molecular weight of substance) $\times$ 100.

Step 2: Meaning
Calculate individual hardness contributions: $Mg(HCO_{3})_{2}$: $(7.3 / 146) \times 100 = 5$ ppm. $Ca(HCO_{3})_{2}$: $(8.1 / 162) \times 100 = 5$ ppm. $CaSO_{4}$: $(27.2 / 136) \times 100 = 20$ ppm.

Step 3: Analysis
Total Hardness = Sum of all contributions = $5 + 5 + 20 = 30$.

Step 4: Conclusion
The total hardness is 40 ppm.
Final Answer: (D)