The hardness of water due to `HCO_3` is `122pm`. Select the correct statement(s).

Options

A. The hardness of water in terms of `CaCO_(3)` is `200 p p m`
B. The hardness of water in terms of `CaCO_(3)` is `100 p p m`
C. The hardness of water in terms of `CaCO_(3)` is `22 p p m`
D. The hardness of water in terms of `CaCO_(3)` is `95 p p m`

Explanation

To solve the problem regarding the hardness of water due to bicarbonate ions (HCO₃⁻), we need to convert the given hardness from HCO₃⁻ to the equivalent hardness in terms of CaCO₃ (calcium carbonate). Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
The hardness of water due to HCO₃⁻ is given as 122 ppm. This means there are 122 grams of HCO₃⁻ in 1 million grams of water (or 1 million mL of water).

### Step 2: Calculate the Moles of HCO₃⁻
The molecular weight of HCO₃⁻ (bicarbonate ion) is approximately 61 g/mol. To find the number of moles of HCO₃⁻ present in 122 grams, we use the formula:

\[ \text{Moles of HCO₃⁻} = \frac{\text{mass}}{\text{molecular weight}} = \frac{122 \, \text{g}}{61 \, \text{g/mol}} = 2 \, \text{moles} \]

### Step 3: Relate HCO₃⁻ to CaCO₃
For every 2 moles of HCO₃⁻, there is 1 mole of CaCO₃ formed. Thus, from 2 moles of HCO₃⁻, we can form 1 mole of CaCO₃.

### Step 4: Calculate the Equivalent Hardness in CaCO₃
The molecular weight of CaCO₃ is approximately 100 g/mol. Therefore, the equivalent hardness in terms of CaCO₃ can be calculated as follows:

\[ \text{Hardness in ppm of CaCO₃} = \text{Moles of CaCO₃} \times \text{Molecular weight of CaCO₃} \]
Since we have 1 mole of CaCO₃:

\[ \text{Hardness in ppm of CaCO₃} = 1 \, \text{mole} \times 100 \, \text{g/mol} = 100 \, \text{ppm} \]

### Step 5: Conclusion
Thus, the hardness of water in terms of CaCO₃ is 100 ppm.

### Answer
The correct statement regarding the hardness of water due to HCO₃⁻ is that it is **100 ppm** when expressed as CaCO₃.