What is the molartiy of `SO_4^(2-)` ion in aqueous solution that contain `34.2 ppm` of `AI_2(SO_4)_3`? (Assume complete dissociation and density of solution `1g/mL`)

Options

A. `3xx10^(-4) M`
B. `2xx10^(-4)`
C. `10^(-4)M`
D. None of these

Answer

To find the molarity of the `SO_4^(2-)` ion in an aqueous solution containing `34.2 ppm` of `Al_2(SO_4)_3`, we can follow these steps:

### Step 1: Understand the definition of ppm
PPM (parts per million) is defined as the mass of solute (in grams) per million parts of solution (in milliliters). For our calculation, we will use the formula:
\[ \text{ppm} = \frac{\text{mass of solute (g)}}{\text{volume of solution (L)}} \times 10^6 \]
Given that the solution has a density of `1 g/mL`, we can assume that `1 L` of solution weighs `1000 g`.

### Step 2: Calculate the mass of `Al_2(SO_4)_3`
Since we are given `34.2 ppm` of `Al_2(SO_4)_3`, we can rearrange the ppm formula to find the mass of `Al_2(SO_4)_3` in `1 L` of solution:
\[ \text{mass of } Al_2(SO_4)_3 = \frac{34.2 \, \text{ppm} \times 1 \, \text{L} \times 10^6 \, \text{mL}}{10^6} = 34.2 \times 10^{-3} \, \text{g} \]

### Step 3: Calculate the moles of `Al_2(SO_4)_3`
Next, we need to calculate the number of moles of `Al_2(SO_4)_3` using its molar mass. The molar mass of `Al_2(SO_4)_3` can be calculated as follows:
- Aluminum (Al): 26.98 g/mol (2 Al)
- Sulfur (S): 32.07 g/mol (3 S)
- Oxygen (O): 16.00 g/mol (12 O)

Calculating the molar mass:
\[ \text{Molar mass of } Al_2(SO_4)_3 = (2 \times 26.98) + (3 \times 32.07) + (12 \times 16.00) = 342.17 \, \text{g/mol} \]

Now, we can calculate the moles of `Al_2(SO_4)_3`:
\[ \text{moles of } Al_2(SO_4)_3 = \frac{34.2 \times 10^{-3} \, \text{g}}{342.17 \, \text{g/mol}} \approx 1.00 \times 10^{-4} \, \text{mol} \]

### Step 4: Calculate the molarity of `Al_2(SO_4)_3`
Molarity (M) is defined as the number of moles of solute per liter of solution. Since we assumed the volume of the solution is `1 L`, the molarity of `Al_2(SO_4)_3` is:
\[ \text{Molarity of } Al_2(SO_4)_3 = \frac{1.00 \times 10^{-4} \, \text{mol}}{1 \, \text{L}} = 1.00 \times 10^{-4} \, \text{M} \]

### Step 5: Determine the molarity of `SO_4^(2-)` ions
The dissociation of `Al_2(SO_4)_3` in solution can be represented as:
\[ Al_2(SO_4)_3 \rightarrow 2 Al^{3+} + 3 SO_4^{2-} \]
From the equation, we see that `1 mol` of `Al_2(SO_4)_3` produces `3 mol` of `SO_4^{2-}` ions. Therefore, the molarity of `SO_4^{2-}` ions will be:
\[ \text{Molarity of } SO_4^{2-} = 3 \times \text{Molarity of } Al_2(SO_4)_3 = 3 \times 1.00 \times 10^{-4} \, \text{M} = 3.00 \times 10^{-4} \, \text{M} \]

### Final Answer
The molarity of the `SO_4^(2-)` ion in the solution is:
\[ \text{Molarity of } SO_4^{2-} = 3.00 \times 10^{-4} \, \text{M} \]