The solubility of `Ba_(3)(AsO_(4))_(2)`(formula mass=690) is `6.9xx10^(-2)` g//100 mL. What is the `K_(sp)`?

Options

A. `1.08xx10^(-11)`
B. `1.08xx10^(-13)`
C. `1.0xx10^(-15)`
D. `6.0xx10^(-13)`

Answer

To find the solubility product constant \( K_{sp} \) for \( Ba_3(AsO_4)_2 \), we will follow these steps:

### Step 1: Convert the solubility from grams to moles
The solubility of \( Ba_3(AsO_4)_2 \) is given as \( 6.9 \times 10^{-2} \) g per 100 mL. First, we need to convert this to moles.

1. **Convert grams to moles:**
\[ \text{Moles of } Ba_3(AsO_4)_2 = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Given:
- Mass = \( 6.9 \times 10^{-2} \) g
- Molar mass = 690 g/mol

\[ \text{Moles} = \frac{6.9 \times 10^{-2}}{690} = 1 \times 10^{-4} \text{ moles} \]

### Step 2: Calculate the molarity
Since the solubility is given for 100 mL, we need to convert this to liters to find the molarity.

\[ \text{Molarity (M)} = \frac{\text{moles}}{\text{volume (L)}} \]
\[ \text{Molarity} = \frac{1 \times 10^{-4}}{0.1} = 1 \times 10^{-3} \text{ M} \]

### Step 3: Write the dissociation equation
The dissociation of \( Ba_3(AsO_4)_2 \) in water can be represented as:

\[ Ba_3(AsO_4)_2 (s) \rightleftharpoons 3 Ba^{2+} (aq) + 2 AsO_4^{3-} (aq) \]

### Step 4: Determine the concentration of ions
From the dissociation equation, we can see that:
- For every 1 mole of \( Ba_3(AsO_4)_2 \) that dissolves, it produces 3 moles of \( Ba^{2+} \) and 2 moles of \( AsO_4^{3-} \).

If the solubility (S) is \( 1 \times 10^{-3} \) M:
- Concentration of \( Ba^{2+} \) = \( 3S = 3 \times 1 \times 10^{-3} = 3 \times 10^{-3} \) M
- Concentration of \( AsO_4^{3-} \) = \( 2S = 2 \times 1 \times 10^{-3} = 2 \times 10^{-3} \) M

### Step 5: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) is given by the expression:

\[ K_{sp} = [Ba^{2+}]^3 \times [AsO_4^{3-}]^2 \]

Substituting the concentrations:

\[ K_{sp} = (3 \times 10^{-3})^3 \times (2 \times 10^{-3})^2 \]

### Step 6: Calculate \( K_{sp} \)
Calculating each part:

\[ (3 \times 10^{-3})^3 = 27 \times 10^{-9} \]
\[ (2 \times 10^{-3})^2 = 4 \times 10^{-6} \]

Now multiplying these together:

\[ K_{sp} = 27 \times 10^{-9} \times 4 \times 10^{-6} = 108 \times 10^{-15} = 1.08 \times 10^{-13} \]

### Final Answer
\[ K_{sp} = 1.08 \times 10^{-13} \]