`K_(sp)` for lead iodate `[Pb(IO_(3))_(2) is 3.2 xx 10^(-14)` at a given temperature. The solubility in `molL^(-1)` will be

Options

A. `2.0 xx 10^(-5)`
B. `(3.2 xx 10^(-7))^(1//2)`
C. `(3.8 xx 10^(-7))`
D. `4.0 xx 10^(-6)`

Explanation

To find the solubility of lead iodate \([Pb(IO_3)_2]\) given its solubility product constant \(K_{sp} = 3.2 \times 10^{-14}\), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of lead iodate in water can be represented as:
\[ Pb(IO_3)_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2 IO_3^{-} (aq) \]

### Step 2: Define the Solubility
Let the solubility of lead iodate be \(S\) (in mol/L). From the dissociation reaction:
- The concentration of \(Pb^{2+}\) ions will be \(S\).
- The concentration of \(IO_3^{-}\) ions will be \(2S\) (since two iodate ions are produced for each formula unit of lead iodate).

### Step 3: Write the \(K_{sp}\) Expression
The \(K_{sp}\) expression for the dissociation is given by:
\[ K_{sp} = [Pb^{2+}][IO_3^{-}]^2 \]
Substituting the concentrations from the solubility:
\[ K_{sp} = S \cdot (2S)^2 \]
This simplifies to:
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the Given \(K_{sp}\)
Now we can substitute the given \(K_{sp}\) value into the equation:
\[ 3.2 \times 10^{-14} = 4S^3 \]

### Step 5: Solve for \(S^3\)
To find \(S^3\), rearrange the equation:
\[ S^3 = \frac{3.2 \times 10^{-14}}{4} \]
Calculating this gives:
\[ S^3 = 0.8 \times 10^{-14} = 8.0 \times 10^{-15} \]

### Step 6: Find \(S\)
Now, take the cube root of both sides to find \(S\):
\[ S = \sqrt[3]{8.0 \times 10^{-15}} \]
Calculating the cube root:
\[ S = 2 \times 10^{-5} \text{ mol/L} \]

### Final Answer
Thus, the solubility of lead iodate \([Pb(IO_3)_2]\) is:
\[ \text{Solubility} = 2 \times 10^{-5} \text{ mol/L} \]