The solubility product of `BaSO_(4)` at `18^(0)` C is `1.5xx10^(-9)` . It solubility (mole `lit^(-1))` at the same temperature is

Options

A. `1.5xx10^(-9)`
B. `1.5xx10^(-5)`
C. `3.9xx10^(-9)`
D. `3.9xx10^(-5)`

Answer

To find the solubility of barium sulfate (BaSO₄) at 18°C given its solubility product (Ksp), we can follow these steps:

### Step 1: Write the dissociation equation
Barium sulfate dissociates in water as follows:
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

### Step 2: Write the expression for the solubility product (Ksp)
The solubility product constant (Ksp) for barium sulfate can be expressed as:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]

### Step 3: Define the solubility (s)
Let the solubility of BaSO₄ in moles per liter be \( s \). At equilibrium, the concentrations of the ions will be:
- \([\text{Ba}^{2+}] = s\)
- \([\text{SO}_4^{2-}] = s\)

### Step 4: Substitute into the Ksp expression
Substituting these values into the Ksp expression gives:
\[ K_{sp} = s \cdot s = s^2 \]

### Step 5: Substitute the given Ksp value
We know that the Ksp of BaSO₄ at 18°C is \( 1.5 \times 10^{-9} \):
\[ s^2 = 1.5 \times 10^{-9} \]

### Step 6: Solve for s
To find the solubility \( s \), take the square root of both sides:
\[ s = \sqrt{1.5 \times 10^{-9}} \]

Calculating this gives:
\[ s = \sqrt{1.5} \times 10^{-4.5} \]
\[ s \approx 1.22 \times 10^{-5} \, \text{mol/L} \]

### Final Answer
The solubility of barium sulfate (BaSO₄) at 18°C is approximately:
\[ s \approx 1.22 \times 10^{-5} \, \text{mol/L} \]

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