The solubility of `BaSO_4` in water is ` 2.42 xx 10^(-3) gL^(-1) ` at 298 K. The value of its solubility product `(K_(sp)) ` will be (Given molar mass of `BaSO_4= 233 g " mol"^(-1))`

Options

A. ` 1.08 xx 10^(-10) mol^(2) L^(-2)`
B. ` 1.08 xx 10^(-12) mol ^(2) L^(-2)`
C. ` 1.08 xx 10 ^(-14) mol^(2) L^(-2)`
D. ` 1.08 xx 10^(-8) mol^(2) L^(-2)`

Answer

To find the solubility product \( K_{sp} \) of \( BaSO_4 \), we will follow these steps:

### Step 1: Calculate the solubility in moles per liter
Given the solubility of \( BaSO_4 \) in water is \( 2.42 \times 10^{-3} \, g/L \) and the molar mass of \( BaSO_4 \) is \( 233 \, g/mol \), we can convert the solubility from grams per liter to moles per liter using the formula:

\[ \text{Solubility (S)} = \frac{\text{mass (g/L)}}{\text{molar mass (g/mol)}} \]

Substituting the values:

\[ S = \frac{2.42 \times 10^{-3} \, g/L}{233 \, g/mol} = 1.04 \times 10^{-5} \, mol/L \]

### Step 2: Write the dissociation equation
The dissociation of \( BaSO_4 \) in water can be represented as:

\[ BaSO_4 (s) \rightleftharpoons Ba^{2+} (aq) + SO_4^{2-} (aq) \]

From the equation, we see that for every mole of \( BaSO_4 \) that dissolves, it produces 1 mole of \( Ba^{2+} \) ions and 1 mole of \( SO_4^{2-} \) ions.

### Step 3: Set up the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is defined as:

\[ K_{sp} = [Ba^{2+}][SO_4^{2-}] \]

Since the concentration of \( Ba^{2+} \) and \( SO_4^{2-} \) ions at equilibrium is equal to the solubility \( S \):

\[ K_{sp} = S \times S = S^2 \]

### Step 4: Calculate \( K_{sp} \)
Now we can substitute the value of \( S \):

\[ K_{sp} = (1.04 \times 10^{-5})^2 \]

Calculating this gives:

\[ K_{sp} = 1.0816 \times 10^{-10} \, mol^2/L^2 \]

### Step 5: Round the answer
Rounding to two significant figures, we find:

\[ K_{sp} \approx 1.08 \times 10^{-10} \, mol^2/L^2 \]

### Final Answer
The solubility product \( K_{sp} \) of \( BaSO_4 \) at 298 K is approximately:

\[ K_{sp} = 1.08 \times 10^{-10} \, mol^2/L^2 \]

---