`RH_(2)` (ion exchange resin) can replace `Ca^(2+)` ions in hard water as `RH_(2)+Ca^(2+) to RCa+2H^(+)`. If L of hard water after passing through `RH_(2)` has pH=3 then hardness in parts per million of `Ca^(2+)` is :

Options

A. 20
B. 10
C. 40
D. 100

Answer

To solve the problem, we need to determine the hardness of water in parts per million (ppm) of \( Ca^{2+} \) ions after passing through the ion exchange resin \( RH_2 \). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
The reaction given is:
\[ RH_2 + Ca^{2+} \rightarrow RCa + 2H^+ \]
This indicates that for every mole of \( Ca^{2+} \) replaced, 2 moles of \( H^+ \) ions are produced.

### Step 2: Determine the Concentration of \( H^+ \) Ions
The pH of the water after treatment is given as 3. The concentration of \( H^+ \) ions can be calculated using the formula:
\[ [H^+] = 10^{-\text{pH}} \]
Substituting the pH value:
\[ [H^+] = 10^{-3} \, \text{mol/L} \]

### Step 3: Calculate the Moles of \( H^+ \) Ions
Since we have 1 liter of hard water, the number of moles of \( H^+ \) ions is:
\[ \text{Moles of } H^+ = [H^+] \times \text{Volume} = 10^{-3} \, \text{mol/L} \times 1 \, \text{L} = 10^{-3} \, \text{mol} \]

### Step 4: Relate Moles of \( Ca^{2+} \) to Moles of \( H^+ \)
From the reaction stoichiometry, we know that 1 mole of \( Ca^{2+} \) produces 2 moles of \( H^+ \). Therefore, the moles of \( Ca^{2+} \) can be calculated as:
\[ \text{Moles of } Ca^{2+} = \frac{1}{2} \times \text{Moles of } H^+ = \frac{1}{2} \times 10^{-3} \, \text{mol} = 5 \times 10^{-4} \, \text{mol} \]

### Step 5: Calculate the Mass of \( Ca^{2+} \)
The molar mass of \( Ca^{2+} \) is approximately 40 g/mol. Thus, the mass of \( Ca^{2+} \) in grams is:
\[ \text{Mass of } Ca^{2+} = \text{Moles} \times \text{Molar Mass} = 5 \times 10^{-4} \, \text{mol} \times 40 \, \text{g/mol} = 0.02 \, \text{g} \]

### Step 6: Convert Mass to Parts Per Million (ppm)
Parts per million (ppm) is defined as the mass of solute (in grams) per million grams of solution (or in this case, per million milliliters of water). Since we have 1 liter (1000 mL) of water, we need to scale up to 1 million mL:
\[ \text{Hardness in ppm} = \left( \frac{0.02 \, \text{g}}{1000 \, \text{mL}} \right) \times 10^6 \, \text{mL} = 20 \, \text{ppm} \]

### Final Answer
The hardness of the water in parts per million of \( Ca^{2+} \) is:
\[ \text{Hardness} = 20 \, \text{ppm} \]

---