A sample of water has a hardness expressed as 80 ppm of `Ca^(2+)` This sample is passed through an ion exchange column and the `Ca^(2+)` is replaced by `H^(+)`.What is the pH of the water after it has been so treated ? [Atomic mass of Ca=40]

Options

A. 3
B. 2.7
C. 5.4
D. 2.4

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Understand the hardness in ppm
The hardness of water is given as 80 ppm of \( \text{Ca}^{2+} \). This means that in 1 million parts (or 1 million milliliters) of water, there are 80 grams of \( \text{Ca}^{2+} \).

### Step 2: Convert grams of \( \text{Ca}^{2+} \) to moles
To find the number of moles of \( \text{Ca}^{2+} \), we use the formula:
\[ \text{Moles of } \text{Ca}^{2+} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Given that the atomic mass of \( \text{Ca} \) is 40 g/mol:
\[ \text{Moles of } \text{Ca}^{2+} = \frac{80 \text{ g}}{40 \text{ g/mol}} = 2 \text{ moles} \]

### Step 3: Ion exchange process
In the ion exchange column, \( \text{Ca}^{2+} \) ions are replaced by \( \text{H}^{+} \) ions. Each \( \text{Ca}^{2+} \) ion has a charge of +2, so 2 moles of \( \text{Ca}^{2+} \) will contribute a total charge of:
\[ \text{Total charge} = 2 \text{ moles} \times 2 = 4 \text{ moles of charge} \]
This means that 4 moles of \( \text{H}^{+} \) ions will be required to replace the \( \text{Ca}^{2+} \) ions.

### Step 4: Calculate the concentration of \( \text{H}^{+} \)
Since the original volume of water is 1 million milliliters (1 L), the concentration of \( \text{H}^{+} \) ions after the ion exchange can be calculated as:
\[ \text{Concentration of } \text{H}^{+} = \frac{4 \text{ moles}}{1 \text{ L}} = 4 \text{ moles/L} \]

### Step 5: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^{+}] \]
Substituting the concentration of \( \text{H}^{+} \):
\[ \text{pH} = -\log(4 \times 10^{-3}) = -\log(4) + 3 \]
Using the approximation \( \log(4) \approx 0.6 \):
\[ \text{pH} \approx -0.6 + 3 = 2.4 \]

### Final Answer
The pH of the water after it has been treated is approximately **2.4**.

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