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: Understanding the Concept:
Hardness is expressed in terms of $CaCO_3$ equivalents. Each salt contribution is calculated by converting its mass to the equivalent mass of $CaCO_3$.
Step 2: Key Formula or Approach:
\[ \text{Hardness (as } CaCO_3) = \frac{\text{Mass of salt}}{\text{Molar mass of salt}} \times \text{Molar mass of } CaCO_3 \] (Note: This works because all these salts have the same n-factor as $CaCO_3$, which is 2).
Step 3: Detailed Explanation:
1. Molar Masses: $Mg(HCO_3)_2 = 146$, $Ca(HCO_3)_2 = 162$, $CaSO_4 = 136$, $CaCO_3 = 100$.
2. Calculation:
Due to $Mg(HCO_3)_2 = \frac{7.3}{146} \times 100 = 0.05 \times 100 = 5 \text{ ppm}$.
Due to $Ca(HCO_3)_2 = \frac{8.1}{162} \times 100 = 0.05 \times 100 = 5 \text{ ppm}$.
Due to $CaSO_4 = \frac{27.2}{136} \times 100 = 0.2 \times 100 = 20 \text{ ppm}$.
3. Total Hardness = $5 + 5 + 20 = 30 \text{ ppm}$.
Step 4: Final Answer:
The total hardness is 30 ppm.