100 mL of a water sample contains 0.81g of calcium bicarbonate and 0.73g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalent of `CaCo_3` is `10^x` ppm. What is numerical value of x? (molar mass of calcium bicarbonate is `162 gmol^(-1)` and magnesium bicarbonate is `146g mol^(-1))`

Options

A. `5,000 ppm`
B. `1,000 ppm`
C. `100 ppm`
D. `10,000ppm`

Answer

To solve the problem, we need to calculate the hardness of the water sample in terms of equivalent calcium carbonate (CaCO₃) and express it in parts per million (ppm). Here’s the step-by-step solution:

### Step 1: Identify the Given Data
- Mass of calcium bicarbonate (Ca(HCO₃)₂) = 0.81 g
- Mass of magnesium bicarbonate (Mg(HCO₃)₂) = 0.73 g
- Molar mass of calcium bicarbonate = 162 g/mol
- Molar mass of magnesium bicarbonate = 146 g/mol
- Volume of water sample = 100 mL

### Step 2: Calculate the Equivalent Weight of Calcium and Magnesium Bicarbonate
The equivalent weight is calculated using the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{Valency Factor}} \]
For both calcium bicarbonate and magnesium bicarbonate, the valency factor is 2 (since they can release 2 bicarbonate ions).

- Equivalent weight of calcium bicarbonate:
\[ \text{Equivalent Weight of Ca(HCO₃)₂} = \frac{162 \text{ g/mol}}{2} = 81 \text{ g/equiv} \]

- Equivalent weight of magnesium bicarbonate:
\[ \text{Equivalent Weight of Mg(HCO₃)₂} = \frac{146 \text{ g/mol}}{2} = 73 \text{ g/equiv} \]

### Step 3: Calculate the Number of Equivalents for Each Bicarbonate
Using the formula:
\[ \text{Number of Equivalents} = \frac{\text{Mass}}{\text{Equivalent Weight}} \]

- For calcium bicarbonate:
\[ \text{Number of Equivalents of Ca(HCO₃)₂} = \frac{0.81 \text{ g}}{81 \text{ g/equiv}} = 0.01 \text{ equiv} \]

- For magnesium bicarbonate:
\[ \text{Number of Equivalents of Mg(HCO₃)₂} = \frac{0.73 \text{ g}}{73 \text{ g/equiv}} = 0.01 \text{ equiv} \]

### Step 4: Calculate Total Equivalents of Calcium Carbonate
The total equivalents of hardness in terms of calcium carbonate is the sum of the equivalents from both bicarbonates:
\[ \text{Total Equivalents} = 0.01 + 0.01 = 0.02 \text{ equiv} \]

### Step 5: Convert Equivalents to Grams of Calcium Carbonate
To find the weight of calcium carbonate equivalent to the total equivalents:
\[ \text{Weight of CaCO₃} = \text{Total Equivalents} \times \text{Equivalent Weight of CaCO₃} \]
The equivalent weight of CaCO₃ is:
\[ \text{Equivalent Weight of CaCO₃} = \frac{100 \text{ g/mol}}{2} = 50 \text{ g/equiv} \]
Thus,
\[ \text{Weight of CaCO₃} = 0.02 \text{ equiv} \times 50 \text{ g/equiv} = 1 \text{ g} \]

### Step 6: Convert Weight to ppm
Since the sample volume is 100 mL, we can convert the weight of CaCO₃ to ppm:
\[ \text{ppm} = \frac{\text{Weight of CaCO₃ in grams}}{\text{Volume in liters}} \times 10^6 \]
\[ \text{ppm} = \frac{1 \text{ g}}{0.1 \text{ L}} \times 10^6 = 10,000 \text{ ppm} \]

### Step 7: Express in Terms of \(10^x\)
The hardness of the water sample is \(10,000\) ppm, which can be expressed as:
\[ 10^4 \text{ ppm} \]
Thus, \(x = 4\).

### Final Answer
The numerical value of \(x\) is **4**.

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