Washing soda (`Na_2CO_(3).10H_2O`) is widely used in softening of hard water. If `1L` of hard water requires 0.0143g of washing soda, what is hardness of water in terms of `p p m` of `CaCO_3`?

Explanation

To determine the hardness of water in terms of parts per million (ppm) of CaCO₃ when 1 liter of hard water requires 0.0143 g of washing soda (Na₂CO₃·10H₂O), we can follow these steps:

### Step 1: Calculate the moles of washing soda used
First, we need to find the molar mass of washing soda (Na₂CO₃·10H₂O).

- The molar mass of Na₂CO₃·10H₂O is calculated as follows:
- Na: 22.99 g/mol × 2 = 45.98 g/mol
- C: 12.01 g/mol × 1 = 12.01 g/mol
- O: 16.00 g/mol × 3 = 48.00 g/mol
- H₂O: 18.02 g/mol × 10 = 180.20 g/mol
- Total = 45.98 + 12.01 + 48.00 + 180.20 = 286.19 g/mol (approximately 286 g/mol)

Now, we can calculate the moles of washing soda:

\[ \text{Moles of washing soda} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.0143 \text{ g}}{286 \text{ g/mol}} \approx 5 \times 10^{-5} \text{ moles} \]

### Step 2: Relate moles of washing soda to moles of CaCO₃
The reaction between washing soda and calcium ions can be represented as:

\[ \text{Ca}^{2+} + \text{Na}_2\text{CO}_3 \rightarrow \text{CaCO}_3 + 2\text{Na}^+ \]

From the stoichiometry of the reaction, we see that 1 mole of washing soda reacts with 1 mole of Ca²⁺ to produce 1 mole of CaCO₃. Therefore, the moles of CaCO₃ produced will be equal to the moles of washing soda used:

\[ \text{Moles of CaCO}_3 = 5 \times 10^{-5} \text{ moles} \]

### Step 3: Calculate the mass of CaCO₃ produced
Now, we can calculate the mass of CaCO₃ produced using its molar mass (100 g/mol):

\[ \text{Mass of CaCO}_3 = \text{moles} \times \text{molar mass} = 5 \times 10^{-5} \text{ moles} \times 100 \text{ g/mol} = 5 \times 10^{-3} \text{ g} \]

### Step 4: Calculate the hardness in ppm
To find the hardness in ppm, we use the formula:

\[ \text{ppm} = \left( \frac{\text{mass of CaCO}_3}{\text{mass of water}} \right) \times 10^6 \]

Assuming the density of water is 1 g/mL, the mass of 1 L of water is 1000 g:

\[ \text{ppm} = \left( \frac{5 \times 10^{-3} \text{ g}}{1000 \text{ g}} \right) \times 10^6 = 5 \text{ ppm} \]

### Final Answer
The hardness of the water in terms of ppm of CaCO₃ is **5 ppm**.

---