What will be solubility of `A_(2)X_(3)` if its solubility product `(K_(sp))` is
equal to `1.08 xx 10^(-23)` ?

Options

A. `10^(-5) mol//L`
B. `3.7 xx 10^(-4) mol//L`
C. `1.2 xx 10^(-3) mol//L`
D. `7.5 xx 10^(-4) mol//L`

Explanation

To find the solubility of \( A_2X_3 \) given its solubility product \( K_{sp} = 1.08 \times 10^{-23} \), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of \( A_2X_3 \) in water can be represented as:
\[ A_2X_3 (s) \rightleftharpoons 2 A^{3+} (aq) + 3 X^{2-} (aq) \]

### Step 2: Define the Solubility
Let the solubility of \( A_2X_3 \) be \( S \) moles per liter. According to the dissociation:
- The concentration of \( A^{3+} \) ions will be \( 2S \).
- The concentration of \( X^{2-} \) ions will be \( 3S \).

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [A^{3+}]^2 [X^{2-}]^3 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (2S)^2 (3S)^3 \]

### Step 4: Simplify the Expression
Now, simplify the expression:
\[ K_{sp} = 4S^2 \cdot 27S^3 = 108S^5 \]

### Step 5: Set Up the Equation
Now, we can set up the equation using the given \( K_{sp} \):
\[ 1.08 \times 10^{-23} = 108S^5 \]

### Step 6: Solve for \( S^5 \)
Rearranging the equation gives:
\[ S^5 = \frac{1.08 \times 10^{-23}}{108} \]
Calculating the right side:
\[ S^5 = 1.0 \times 10^{-25} \]

### Step 7: Solve for \( S \)
Now, take the fifth root to find \( S \):
\[ S = (1.0 \times 10^{-25})^{1/5} = 10^{-5} \text{ moles per liter} \]

### Final Answer
The solubility of \( A_2X_3 \) is:
\[ S = 10^{-5} \text{ moles per liter} \]

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