To determine the solubility product constant (\( K_{sp} \)) of \( PbI_2 \) at 0°C, let's start by calculating the molarity of the dissolved lead(II) iodide in water and then set up an ICE table to find the concentrations of the ions at equilibrium.
Step 1: Calculate the Number of Moles of \( PbI_2 \)
The molar mass of \( PbI_2 \) can be calculated as follows:
- Lead (Pb): 207.2 g/mol
- Iodine (I): 126.9 g/mol (and there are 2 I atoms)
\[ \text{Molar mass of } PbI_2 = 207.2 + 2(126.9) = 207.2 + 253.8 = 461.0 \text{ g/mol} \]
Now, we can calculate the number of moles in 0.014 grams of \( PbI_2 \):
\[ \text{Moles of } PbI_2 = \frac{0.014 \text{ g}}{461.0 \text{ g/mol}} \approx 3.03 \times 10^{-5} \text{ moles} \]
Step 2: Calculate Molarity of \( PbI_2 \)
Now we can use the moles to find the molarity (M) in 4.5 L of solution:
\[ M = \frac{\text{moles}}{\text{volume (L)}} = \frac{3.03 \times 10^{-5} \text{ moles}}{4.5 \text{ L}} \approx 6.73 \times 10^{-6} \text{ M} \]
Step 3: Set Up the ICE Table
The dissolution of \( PbI_2 \) can be represented as:
\[ PbI_2 (s) \rightleftharpoons Pb^{2+} (aq) + 2I^{-} (aq) \]
ICE Table:
Let’s denote the change in concentration of \( Pb^{2+} \) and \( I^{-} \) as \( x \):
\[ \begin{array}{|c|c|c|c|} \hline & [Pb^{2+}] & [I^{-}] \ \hline \text{Initial} & 0 & 0 \ \hline \text{Change} & +x & +2x \ \hline \text{Equilibrium} & x & 2x \ \hline \end{array} \]
From the molarity we calculated, we know that the total concentration of \( PbI_2 \) at saturation before the precipitate starts to form is \( 6.73 \times 10^{-6} \) M.
At equilibrium:
- \( [Pb^{2+}] = x = 6.73 \times 10^{-6} \) M
- \( [I^{-}] = 2x = 2(6.73 \times 10^{-6}) = 1.346 \times 10^{-5} \) M
Step 4: Calculate the Solubility Product Constant \( K_{sp} \)
The expression for the \( K_{sp} \) is given by:
\[ K_{sp} = [Pb^{2+}][I^{-}]^2 \]
Now substituting in our equilibrium concentrations:
\[ K_{sp} = (6.73 \times 10^{-6}) \cdot (1.346 \times 10^{-5})^2 \]
Calculating it step by step:
- Calculate \( (1.346 \times 10^{-5})^2 \):
\[ (1.346 \times 10^{-5})^2 \approx 1.81 \times 10^{-10} \]
- Now substitute that back into the \( K_{sp} \) expression:
\[ K_{sp} = 6.73 \times 10^{-6} \cdot 1.81 \times 10^{-10} \approx 1.22 \times 10^{-15} \]
Final Result
The solubility product constant \( K_{sp} \) of \( PbI_2 \) at 0°C is approximately:
\[ \boxed{1.22 \times 10^{-15}} \]