5 gm of `K_2 SO_4` solution were dissolved in 250 ml of water. How many ml of this solution are needed to precipitate out 1.2 gm of `BaSO_4` on addition of `BaCI_2`?

Options

A. 44.8 ml
B. 22.4 ml
C. 11.8 ml
D. 22400 ml

Answer

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

### Step 1: Write the balanced chemical equation
The reaction between potassium sulfate (K₂SO₄) and barium chloride (BaCl₂) can be represented as:
\[ \text{BaCl}_2 + \text{K}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2 \text{KCl} \]

### Step 2: Calculate the molar mass of BaSO₄ and K₂SO₄
- Molar mass of BaSO₄:
- Barium (Ba) = 137 g/mol
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol × 4 = 64 g/mol
- Total = 137 + 32 + 64 = 233 g/mol

- Molar mass of K₂SO₄:
- Potassium (K) = 39 g/mol × 2 = 78 g/mol
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol × 4 = 64 g/mol
- Total = 78 + 32 + 64 = 174 g/mol

### Step 3: Determine the amount of K₂SO₄ needed to precipitate 1.2 g of BaSO₄
Using the stoichiometry of the reaction:
- From the balanced equation, 1 mole of K₂SO₄ produces 1 mole of BaSO₄.
- Therefore, 233 g of BaSO₄ is produced from 174 g of K₂SO₄.

Using a unitary method:
\[ \text{Mass of K}_2\text{SO}_4 = \left(\frac{174 \text{ g}}{233 \text{ g}}\right) \times 1.2 \text{ g} = 0.8961 \text{ g} \]

### Step 4: Calculate the concentration of K₂SO₄ in the solution
We have 5 g of K₂SO₄ in 250 mL of solution. To find the concentration:
\[ \text{Concentration} = \frac{\text{mass of K}_2\text{SO}_4}{\text{volume of solution}} = \frac{5 \text{ g}}{250 \text{ mL}} = 0.02 \text{ g/mL} \]

### Step 5: Calculate the volume of solution needed for 0.8961 g of K₂SO₄
Using the concentration calculated:
\[ \text{Volume} = \frac{\text{mass of K}_2\text{SO}_4}{\text{concentration}} = \frac{0.8961 \text{ g}}{0.02 \text{ g/mL}} = 44.805 \text{ mL} \]

### Step 6: Round the volume to appropriate significant figures
The volume needed is approximately 44.8 mL.

### Final Answer
The volume of the K₂SO₄ solution needed to precipitate out 1.2 g of BaSO₄ is **approximately 44.8 mL**.

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