The solubility product of `BaSO_(4)" is " 1.5 xx 10^(-9)`. Find the solubility of `BaSO_(4)` in
pure water

Answer

To find the solubility of barium sulfate (BaSO₄) in pure water 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: Define the molar solubility
Let the molar solubility of BaSO₄ in pure water be \( S \) (in moles per liter). When BaSO₄ dissolves, it produces \( S \) moles of Ba²⁺ ions and \( S \) moles of SO₄²⁻ ions.

### Step 3: Write the expression for Ksp
The solubility product (Ksp) expression for BaSO₄ is given by:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = S \cdot S = S^2 \]

### Step 4: Substitute the value of Ksp
We are given that \( K_{sp} = 1.5 \times 10^{-9} \). Therefore, we can write:
\[ S^2 = 1.5 \times 10^{-9} \]

### Step 5: Solve for S
To find \( S \), take the square root of both sides:
\[ S = \sqrt{1.5 \times 10^{-9}} \]
Calculating this gives:
\[ S \approx 3.87 \times 10^{-5} \, \text{mol/L} \]

### Step 6: Convert molar solubility to grams per liter
To convert the solubility from moles per liter to grams per liter, we need the molar mass of BaSO₄. The molar mass can be calculated as follows:
- Molar mass of Ba = 137 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol (4 atoms of O)

Calculating the total:
\[ \text{Molar mass of BaSO}_4 = 137 + 32 + (4 \times 16) = 233 \, \text{g/mol} \]

Now, to find the solubility in grams per liter:
\[ \text{Solubility (g/L)} = S \times \text{Molar mass} \]
\[ \text{Solubility (g/L)} = 3.87 \times 10^{-5} \, \text{mol/L} \times 233 \, \text{g/mol} \]
Calculating this gives:
\[ \text{Solubility (g/L)} \approx 9.01 \times 10^{-3} \, \text{g/L} \]

### Final Answer:
The solubility of BaSO₄ in pure water is approximately \( 3.87 \times 10^{-5} \, \text{mol/L} \) or \( 9.01 \times 10^{-3} \, \text{g/L} \).

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