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 step by step, we will follow these calculations:

### Step 1: Understand the given data
We are given:
- Concentration of HCO₃⁻ ions = 366 ppm
- Mass of tap water = 500 g

### Step 2: Calculate the mass of HCO₃⁻ ions in the water
The formula for ppm (parts per million) is:
\[ \text{ppm} = \frac{\text{mass of solute (g)}}{\text{mass of solution (g)}} \times 10^6 \]
Rearranging this formula to find the mass of HCO₃⁻:
\[ \text{mass of HCO₃⁻} = \frac{\text{ppm} \times \text{mass of solution}}{10^6} \]
Substituting the values:
\[ \text{mass of HCO₃⁻} = \frac{366 \times 500}{10^6} = 0.183 g \]

### Step 3: Calculate the number of moles of HCO₃⁻
To find the number of moles, we use the formula:
\[ \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of HCO₃⁻ (Hydrogen Carbonate):
- H = 1 g/mol
- C = 12 g/mol
- O = 16 g/mol × 3 = 48 g/mol
- Total = 1 + 12 + 48 = 61 g/mol

Now, substituting the values:
\[ \text{moles of HCO₃⁻} = \frac{0.183}{61} = 0.003 \text{ moles} \]

### Step 4: Determine the amount of Ca(OH)₂ needed
From the reaction:
\[ \text{HCO₃⁻} + \text{Ca(OH)₂} \rightarrow \text{CaCO₃} + \text{H₂O} \]
The stoichiometry shows that 1 mole of HCO₃⁻ reacts with 1 mole of Ca(OH)₂. Therefore, the moles of Ca(OH)₂ required will also be 0.003 moles.

### Step 5: Calculate the mass of Ca(OH)₂
The molar mass of Ca(OH)₂:
- Ca = 40 g/mol
- O = 16 g/mol × 2 = 32 g/mol
- H = 1 g/mol × 2 = 2 g/mol
- Total = 40 + 32 + 2 = 74 g/mol

Now, we can find the mass of Ca(OH)₂:
\[ \text{mass of Ca(OH)₂} = \text{moles} \times \text{molar mass} = 0.003 \times 74 = 0.222 g \]

### 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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