1L of pond water contains `20mg of Ca^(2+) and 12mg "of" Mg^(2+) ions`. What is the volume of a `2N Na_(2)CO_(3)` solution required to soften 5000L of pond water?

Options

A. 500L
B. 50L
C. 5L
D. none of these

Answer

To solve the problem, we need to calculate the volume of a 2N Na₂CO₃ solution required to soften 5000 liters of pond water, which contains calcium (Ca²⁺) and magnesium (Mg²⁺) ions.

### Step-by-Step Solution:

1. **Determine the amount of Ca²⁺ and Mg²⁺ in 5000 liters of pond water:**
- The concentration of Ca²⁺ in 1L of pond water = 20 mg
- The concentration of Mg²⁺ in 1L of pond water = 12 mg
- For 5000 liters:
- Amount of Ca²⁺ = 20 mg/L × 5000 L = 100,000 mg = 100 g
- Amount of Mg²⁺ = 12 mg/L × 5000 L = 60,000 mg = 60 g

2. **Calculate the milliequivalents (mEq) of Ca²⁺ and Mg²⁺:**
- The equivalent weight of Ca²⁺ (atomic weight = 40 g/mol, valency = 2):
\[ \text{mEq of Ca}^{2+} = \frac{\text{mass (g)}}{\text{equivalent weight (g/equiv)}} = \frac{100 \text{ g}}{40 \text{ g/mol} / 2} = \frac{100}{20} = 5 \text{ mEq} \]
- The equivalent weight of Mg²⁺ (atomic weight = 24 g/mol, valency = 2):
\[ \text{mEq of Mg}^{2+} = \frac{60 \text{ g}}{24 \text{ g/mol} / 2} = \frac{60}{12} = 5 \text{ mEq} \]

3. **Calculate the total milliequivalents of hardness:**
- Total mEq = mEq of Ca²⁺ + mEq of Mg²⁺
\[ \text{Total mEq} = 5 \text{ mEq} + 5 \text{ mEq} = 10 \text{ mEq} \]

4. **Relate the milliequivalents of Na₂CO₃ to the required volume:**
- The concentration of Na₂CO₃ solution is 2N (2 equivalents/L).
- To find the volume of Na₂CO₃ needed:
\[ \text{Volume (L)} = \frac{\text{Total mEq}}{\text{Normality}} = \frac{10 \text{ mEq}}{2 \text{ Eq/L}} = 5 \text{ L} \]

5. **Convert the volume to milliliters (if needed):**
- Since 1 L = 1000 mL, the volume in mL is:
\[ 5 \text{ L} = 5 \times 1000 \text{ mL} = 5000 \text{ mL} \]

### Final Answer:
The volume of a 2N Na₂CO₃ solution required to soften 5000 liters of pond water is **5 liters**.

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