A sample of hard water is found to contain 40 mg of `Ca^(2+)` ions per litre. The amount of washing shoda `(Na_(2)CO_(3))` required to soften `5*0`L of the sample would be

Options

A. 1.06 g
B. 5.3 g
C. 53 mg
D. 530 mg

Answer

To solve the problem, we need to determine the amount of washing soda (Na₂CO₃) required to soften 5 liters of hard water that contains 40 mg of Ca²⁺ ions per liter.

### Step-by-Step Solution:

1. **Determine the total amount of Ca²⁺ in 5 liters:**
- Given that there are 40 mg of Ca²⁺ ions per liter, for 5 liters:
\[ \text{Total Ca}^{2+} = 40 \, \text{mg/L} \times 5 \, \text{L} = 200 \, \text{mg} \]

2. **Convert the mass of Ca²⁺ from milligrams to grams:**
- Since 1 mg = \(10^{-3}\) g:
\[ 200 \, \text{mg} = 200 \times 10^{-3} \, \text{g} = 0.200 \, \text{g} \]

3. **Determine the moles of Ca²⁺:**
- The molar mass of Ca²⁺ is approximately 40 g/mol. Therefore, the number of moles of Ca²⁺ is calculated as follows:
\[ \text{Moles of Ca}^{2+} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.200 \, \text{g}}{40 \, \text{g/mol}} = 0.005 \, \text{mol} \]

4. **Use the stoichiometry of the reaction:**
- The balanced reaction between Ca²⁺ and Na₂CO₃ is:
\[ \text{Ca}^{2+} + \text{Na}_2\text{CO}_3 \rightarrow \text{CaCO}_3 + 2 \text{Na}^+ \]
- From the reaction, 1 mole of Ca²⁺ reacts with 1 mole of Na₂CO₃. Therefore, 0.005 moles of Ca²⁺ will require 0.005 moles of Na₂CO₃.

5. **Calculate the mass of Na₂CO₃ needed:**
- The molar mass of Na₂CO₃ is approximately 106 g/mol. Thus, the mass of Na₂CO₃ required is:
\[ \text{Mass of Na}_2\text{CO}_3 = \text{moles} \times \text{molar mass} = 0.005 \, \text{mol} \times 106 \, \text{g/mol} = 0.530 \, \text{g} \]

6. **Convert the mass of Na₂CO₃ to milligrams:**
- Since 1 g = 1000 mg:
\[ 0.530 \, \text{g} = 530 \, \text{mg} \]

### Final Answer:
The amount of washing soda (Na₂CO₃) required to soften 5 liters of the sample is **530 mg**.