"solubilities of the following compounds in grams per 100 mL of water. Because these compounds are only slightly soluble, assume that the volume does not change on dissolution and calculate the solubility product for each.
BaSiF6, 0.026 g/100 mL (contains SiF6 2− ions)"
To calculate the solubility product (Ksp) of BaSiF₆ from its solubility in grams per 100 mL, we will follow these steps:

### Step 1: **Convert Solubility to Moles per Liter**
The given solubility of BaSiF₆ is 0.026 g/100 mL, which is equivalent to 0.260 g/L.

Next, we convert grams to moles using the molar mass of BaSiF₆.

#### Molar Mass of BaSiF₆:
- Ba: 137.33 g/mol
- Si: 28.09 g/mol
- F₆: 6 × 19.00 = 114.00 g/mol

Molar mass of BaSiF₆ = 137.33 + 28.09 + 114.00 = 279.42 g/mol

Now, convert the solubility to moles per liter (mol/L):

Molar solubility = 0.260 g/L ÷ 279.42 g/mol
Molar solubility ≈ 9.31 × 10⁻⁴ mol/L

### Step 2: **Write the Dissolution Equation for BaSiF₆**
The dissolution equation for BaSiF₆ is:
BaSiF₆ (s) ⇌ Ba²⁺ (aq) + SiF₆²⁻ (aq)

### Step 3: **Set Up the Ksp Expression**
The Ksp expression for BaSiF₆ is:
Ksp = [Ba²⁺][SiF₆²⁻]

Let the molar solubility of BaSiF₆ be 's'. At equilibrium:
- [Ba²⁺] = s
- [SiF₆²⁻] = s

The Ksp expression becomes:
Ksp = s × s = s²

### Step 4: **Substitute the Molar Solubility into the Ksp Expression**
Substitute the molar solubility (s ≈ 9.31 × 10⁻⁴ mol/L) into the Ksp expression:
Ksp = (9.31 × 10⁻⁴)²

### Step 5: **Calculate Ksp**
Ksp ≈ 8.68 × 10⁻⁷

Thus, the solubility product (Ksp) of BaSiF₆ is approximately 8.68 × 10⁻⁷.

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