The solubility of `BaSO_(4)` in water is `2.42xx10^(-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. `1.08xx10^(-10) mol^(2)L^(-2)`
B. `1.08xx10^(-12) mol^(2)L^(-2)`
C. `1.08xx10^(-14) mol^(2)L^(-2)`
D. `1.08xx10^(-8) mol^(2)L^(-2)`

Explanation

To find the solubility product (K_sp) of BaSO₄, we will follow these steps:

### Step 1: Convert the solubility from grams per liter to moles per liter.

The solubility of BaSO₄ is given as \(2.42 \times 10^{-3} \, \text{g/L}\). We need to convert this to moles per liter (mol/L).

**Calculation:**
\[ \text{Moles of BaSO₄} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Given:
- Mass = \(2.42 \times 10^{-3} \, \text{g}\)
- Molar mass of BaSO₄ = 233 g/mol

\[ \text{Moles of BaSO₄} = \frac{2.42 \times 10^{-3} \, \text{g}}{233 \, \text{g/mol}} \approx 1.04 \times 10^{-5} \, \text{mol/L} \]

### Step 2: Write the dissociation equation for BaSO₄.

When BaSO₄ dissolves in water, it dissociates into barium ions (Ba²⁺) and sulfate ions (SO₄²⁻):

\[ \text{BaSO₄ (s)} \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

### Step 3: Determine the concentrations of the ions at equilibrium.

From the dissociation equation, we see that for every mole of BaSO₄ that dissolves, we get:
- 1 mole of Ba²⁺
- 1 mole of SO₄²⁻

Thus, the concentration of Ba²⁺ and SO₄²⁻ will both be equal to the solubility of BaSO₄:

\[ [\text{Ba}^{2+}] = 1.04 \times 10^{-5} \, \text{mol/L} \]
\[ [\text{SO}_4^{2-}] = 1.04 \times 10^{-5} \, \text{mol/L} \]

### Step 4: Write the expression for the solubility product (K_sp).

The solubility product \(K_{sp}\) for BaSO₄ is given by the equation:

\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]

### Step 5: Substitute the concentrations into the K_sp expression.

Substituting the values we found:

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

Calculating this gives:

\[ K_{sp} = 1.0816 \times 10^{-10} \]

### Final Answer:
\[ K_{sp} \approx 1.08 \times 10^{-10} \]

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