A sample of tap water contains 366 ppm of `HCO_(3)^(-)`ions with `Ca^(2+)` ion. Now `Ca^(2+)` removed by Clark's method by addition of `Ca(OH)_2`. Then what minimum mass of `Ca(OH)_2` will be required to remove `HCO_(3)^(-)`ions completely from 500 g of same tap water

Options

A. 1 g
B. 0.4 g
C. 0.222 g
D. 0.111 g

Answer

To solve the problem of determining the minimum mass of `Ca(OH)₂` required to remove `HCO₃⁻` ions completely from 500 g of tap water, we can follow these steps:

### Step 1: Understand the Reaction
The reaction between bicarbonate ions (`HCO₃⁻`) and calcium hydroxide (`Ca(OH)₂`) can be represented as:
\[ HCO₃⁻ + Ca(OH)₂ \rightarrow CaCO₃ + H₂O + OH⁻ \]
This indicates that one mole of `Ca(OH)₂` reacts with one mole of `HCO₃⁻`.

### Step 2: Calculate the Mass of `HCO₃⁻` in the Water Sample
Given that the concentration of `HCO₃⁻` is 366 ppm, we can use the following formula to find the mass of `HCO₃⁻`:
\[ \text{ppm} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^6 \]
Rearranging this gives:
\[ \text{mass of solute} = \text{ppm} \times \frac{\text{mass of solution}}{10^6} \]
Substituting the values:
\[ \text{mass of } HCO₃⁻ = 366 \times \frac{500}{10^6} \]
Calculating this:
\[ \text{mass of } HCO₃⁻ = 366 \times 0.0005 = 0.183 \text{ g} \]

### Step 3: Calculate Moles of `HCO₃⁻`
Next, we need to calculate the number of moles of `HCO₃⁻`. The molar mass of `HCO₃⁻` is calculated as follows:
- Hydrogen (H): 1 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol
Thus, the molar mass of `HCO₃⁻` is:
\[ 1 + 12 + 48 = 61 \text{ g/mol} \]
Now, we can calculate the moles of `HCO₃⁻`:
\[ \text{moles of } HCO₃⁻ = \frac{\text{mass of } HCO₃⁻}{\text{molar mass of } HCO₃⁻} = \frac{0.183 \text{ g}}{61 \text{ g/mol}} \]
Calculating this gives:
\[ \text{moles of } HCO₃⁻ \approx 0.003 \text{ moles} \]

### Step 4: Determine Moles of `Ca(OH)₂` Required
From the stoichiometry of the reaction, we see that 1 mole of `Ca(OH)₂` reacts with 1 mole of `HCO₃⁻`. Therefore, the moles of `Ca(OH)₂` required will also be:
\[ \text{moles of } Ca(OH)₂ = 0.003 \text{ moles} \]

### Step 5: Calculate the Mass of `Ca(OH)₂`
The molar mass of `Ca(OH)₂` is calculated as follows:
- Calcium (Ca): 40 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol
- Hydrogen (H): 1 g/mol × 2 = 2 g/mol
Thus, the molar mass of `Ca(OH)₂` is:
\[ 40 + 32 + 2 = 74 \text{ g/mol} \]
Now, we can calculate the mass of `Ca(OH)₂` required:
\[ \text{mass of } Ca(OH)₂ = \text{moles of } Ca(OH)₂ \times \text{molar mass of } Ca(OH)₂ \]
Substituting the values:
\[ \text{mass of } Ca(OH)₂ = 0.003 \text{ moles} \times 74 \text{ g/mol} \]
Calculating this gives:
\[ \text{mass of } Ca(OH)₂ \approx 0.222 \text{ g} \]

### Final Answer
The minimum mass of `Ca(OH)₂` required to remove `HCO₃⁻` ions completely from 500 g of tap water is **0.222 g**.

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