The solubility of `BaSO_(4)` in water `2.42 xx 10^(-3) gL^(-1)` at 298 K. The value of 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^(14) mol^(2) L^(-2)`

Explanation

To find the solubility product \( K_{sp} \) of \( BaSO_4 \) given its solubility in water, we can follow these steps:

### Step 1: Convert solubility from g/L to mol/L
The solubility of \( BaSO_4 \) is given as \( 2.42 \times 10^{-3} \) g/L. To convert this to moles per liter (mol/L), we use the molar mass of \( BaSO_4 \).

\[ \text{Molar mass of } BaSO_4 = 233 \, \text{g/mol} \]

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

### Step 2: Write the dissociation equation
When \( BaSO_4 \) dissolves in water, it dissociates into its ions:

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

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

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

Since the solubility of \( BaSO_4 \) is \( S \) (in mol/L), at equilibrium:

\[ [Ba^{2+}] = S \quad \text{and} \quad [SO_4^{2-}] = S \]

Thus, we can express \( K_{sp} \) as:

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

### Step 4: Calculate \( K_{sp} \)
Substituting the value of \( S \):

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

Calculating this gives:

\[ K_{sp} = 1.08 \times 10^{-10} \]

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

\[ K_{sp} = 1.08 \times 10^{-10} \]

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