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 \( \text{BaSO}_4 \) in pure water given its solubility product (\( K_{sp} \)), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of barium sulfate in water can be represented as:
\[ \text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( \text{BaSO}_4 \) in pure water be \( S \) mol/L. This means:
- The concentration of \( \text{Ba}^{2+} \) ions will be \( S \).
- The concentration of \( \text{SO}_4^{2-} \) ions will also be \( S \).

### Step 3: Write the expression for \( K_{sp} \)
The solubility product (\( K_{sp} \)) is given by the equation:
\[ K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \]
Substituting the concentrations, we have:
\[ K_{sp} = S \times S = S^2 \]

### Step 4: Substitute the given \( K_{sp} \) value
We know that \( K_{sp} \) for \( \text{BaSO}_4 \) is \( 1.5 \times 10^{-9} \). Therefore, we can write:
\[ S^2 = 1.5 \times 10^{-9} \]

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

### Step 6: Convert to grams per liter (if required)
To convert the solubility from mol/L to grams/L, we need the molar mass of \( \text{BaSO}_4 \):
- Molar mass of \( \text{Ba} = 137 \, \text{g/mol} \)
- Molar mass of \( \text{S} = 32 \, \text{g/mol} \)
- Molar mass of \( \text{O} = 16 \, \text{g/mol} \times 4 = 64 \, \text{g/mol} \)

Calculating the total molar mass:
\[ \text{Molar mass of } \text{BaSO}_4 = 137 + 32 + 64 = 233 \, \text{g/mol} \]

Now, multiply the solubility in mol/L by the molar mass to convert to grams/L:
\[ \text{Solubility in g/L} = S \times \text{Molar mass} = (3.87 \times 10^{-5} \, \text{mol/L}) \times (233 \, \text{g/mol}) \approx 9.01 \times 10^{-3} \, \text{g/L} \]

### Final Answer
The solubility of \( \text{BaSO}_4 \) in pure water is approximately \( 9.01 \times 10^{-3} \, \text{g/L} \).

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