What will the concentration of `[Ca^(+2)]` in a sample of 1 litre hard water if after treatment with washing soda 10 g insoluable `CaCO_(3)` is precipitated.

Options

A. 0.2 M
B. 0.1 M
C. 0.3 M
D. 0.4 M

Answer

To find the concentration of calcium ions \([Ca^{2+}]\) in a 1-litre sample of hard water after treatment with washing soda, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
When hard water, which contains calcium ions \([Ca^{2+}]\), is treated with washing soda (sodium carbonate, \(Na_2CO_3\)), a precipitation reaction occurs:
\[ Ca^{2+} + Na_2CO_3 \rightarrow CaCO_3 (s) + 2Na^+ \]
In this reaction, calcium ions react with carbonate ions to form insoluble calcium carbonate \((CaCO_3)\).

2. **Determine the Mass of Precipitate**:
The question states that 10 grams of insoluble \(CaCO_3\) is precipitated.

3. **Calculate the Molar Mass of \(CaCO_3\)**:
The molar mass of \(CaCO_3\) can be calculated as follows:
- Calcium (Ca): 40 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol
\[ \text{Molar mass of } CaCO_3 = 40 + 12 + 48 = 100 \text{ g/mol} \]

4. **Calculate the Number of Moles of \(CaCO_3\)**:
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \text{ g}}{100 \text{ g/mol}} = 0.1 \text{ mol} \]

5. **Relate Moles of \(CaCO_3\) to Moles of \(Ca^{2+}\)**:
From the stoichiometry of the reaction, 1 mole of \(Ca^{2+}\) produces 1 mole of \(CaCO_3\). Therefore, the moles of \(Ca^{2+}\) in the original hard water sample is also 0.1 mol.

6. **Calculate the Concentration of \(Ca^{2+}\)**:
The concentration (molarity) of \(Ca^{2+}\) can be calculated using the formula:
\[ \text{Molarity} = \frac{\text{number of moles}}{\text{volume in litres}} = \frac{0.1 \text{ mol}}{1 \text{ L}} = 0.1 \text{ M} \]

### Final Answer:
The concentration of \([Ca^{2+}]\) in the hard water sample is **0.1 M**.