The solubility in water of a sparingly soluble salt `AB_2 ` is
` 1.0 xx 10 ^(-5) " mol " L^(-1) `. Its solubility product number will be

Options

A. ` 4 xx 10 ^(-15) `
B. ` 4 xx 10 ^(-10)`
C. ` 1 xx 10 ^(-15)`
D. ` 1.0 xx 10 ^(-10)`

Explanation

To find the solubility product (Ksp) of the sparingly soluble salt \( AB_2 \) with a given solubility of \( 1.0 \times 10^{-5} \) mol/L, we can follow these steps:

### Step 1: Write the dissociation equation
The salt \( AB_2 \) dissociates in water as follows:
\[ AB_2 (s) \rightleftharpoons A^{2+} (aq) + 2B^{-} (aq) \]

### Step 2: Define the solubility
Let the solubility of \( AB_2 \) be \( S = 1.0 \times 10^{-5} \) mol/L. This means that at equilibrium:
- The concentration of \( A^{2+} \) ions will be \( S \).
- The concentration of \( B^{-} \) ions will be \( 2S \) because for every mole of \( AB_2 \) that dissolves, one mole of \( A^{2+} \) and two moles of \( B^{-} \) are produced.

### Step 3: Express the concentrations at equilibrium
At equilibrium:
- \([A^{2+}] = S = 1.0 \times 10^{-5}\) mol/L
- \([B^{-}] = 2S = 2 \times 1.0 \times 10^{-5} = 2.0 \times 10^{-5}\) mol/L

### Step 4: Write the expression for Ksp
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [A^{2+}][B^{-}]^2 \]
Substituting the equilibrium concentrations:
\[ K_{sp} = (S)(2S)^2 \]

### Step 5: Substitute the values
Substituting \( S = 1.0 \times 10^{-5} \):
\[ K_{sp} = (1.0 \times 10^{-5}) \times (2 \times 1.0 \times 10^{-5})^2 \]
Calculating \( (2S)^2 \):
\[ (2 \times 1.0 \times 10^{-5})^2 = 4 \times (1.0 \times 10^{-5})^2 = 4 \times 1.0 \times 10^{-10} = 4.0 \times 10^{-10} \]
Thus,
\[ K_{sp} = (1.0 \times 10^{-5}) \times (4.0 \times 10^{-10}) = 4.0 \times 10^{-15} \]

### Final Answer
The solubility product \( K_{sp} \) of the salt \( AB_2 \) is:
\[ K_{sp} = 4.0 \times 10^{-15} \]

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