`RH_(2)` ( ion exchange resin) can replace `Ca^(2+)`d in hard water as.
`RH_(2)+Ca^(2+)rarrRCa+2H^(+)`
`1 "litre"` of hard water passing through `RH_(2)` has `pH2`. Hence hardness in `pp m "of" Ca^(2+)` is:

Options

A. 200
B. 100
C. 50
D. 125

Explanation

To solve the problem, we need to determine the hardness of hard water in terms of ppm (parts per million) of \( \text{Ca}^{2+} \) ions based on the given pH value. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Ion Exchange Reaction
The ion exchange reaction given is:
\[ \text{RH}_2 + \text{Ca}^{2+} \rightarrow \text{RCa} + 2\text{H}^+ \]

This indicates that for every mole of \( \text{Ca}^{2+} \) replaced, two moles of \( \text{H}^+ \) ions are released.

### Step 2: Calculate the Concentration of \( \text{H}^+ \) Ions
The pH of the hard water is given as 2. The concentration of \( \text{H}^+ \) ions can be calculated using the formula:
\[ [\text{H}^+] = 10^{-\text{pH}} \]
Substituting the given pH:
\[ [\text{H}^+] = 10^{-2} = 0.01 \, \text{mol/L} \]

### Step 3: Relate \( \text{H}^+ \) Concentration to \( \text{Ca}^{2+} \) Concentration
From the reaction, we know that:
- 2 moles of \( \text{H}^+ \) correspond to 1 mole of \( \text{Ca}^{2+} \).

Thus, the concentration of \( \text{Ca}^{2+} \) ions can be calculated as:
\[ [\text{Ca}^{2+}] = \frac{1}{2} [\text{H}^+] = \frac{1}{2} \times 0.01 = 0.005 \, \text{mol/L} \]

### Step 4: Calculate the Mass of \( \text{Ca}^{2+} \) Ions
To find the mass of \( \text{Ca}^{2+} \) ions in 1 liter of water, we use the molar mass of calcium, which is approximately 40 g/mol:
\[ \text{Mass of } \text{Ca}^{2+} = \text{moles} \times \text{molar mass} \]
\[ \text{Mass of } \text{Ca}^{2+} = 0.005 \, \text{mol} \times 40 \, \text{g/mol} = 0.2 \, \text{g} \]

### Step 5: Convert Mass to ppm
To convert the mass of \( \text{Ca}^{2+} \) ions to ppm, we use the formula:
\[ \text{ppm} = \left( \frac{\text{mass of solute (g)}}{\text{volume of solution (L)}} \right) \times 10^6 \]
Since we have 0.2 g of \( \text{Ca}^{2+} \) in 1 liter of water:
\[ \text{ppm} = \left( \frac{0.2 \, \text{g}}{1 \, \text{L}} \right) \times 10^6 = 200 \, \text{ppm} \]

### Conclusion
The hardness of the hard water in terms of \( \text{Ca}^{2+} \) ions is:
**200 ppm**