To calculate the solubility product constant (\( K_{sp} \)) of lead(II) iodide \(\text{PbI}2\), we will first write the dissociation equation for the salt, construct an ICE table, and then derive the expression for \( K{sp} \).
Step 1: Write the Dissociation Equation
The dissociation of lead(II) iodide when it dissolves in water can be expressed as:
\[ \text{PbI}_2(s) \leftrightarrow \text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \]
Step 2: Define the ICE Table
The ICE (Initial, Change, Equilibrium) table will help us keep track of the concentrations of the species involved.
| Species | Initial (M) | Change (M) | Equilibrium (M) | |----------------|-------------|------------------|-------------------------| | \(\text{Pb}^{2+}\) | 0 | +\(s\) | \(s\) | | \(\text{I}^-\) | 0 | +\(2s\) | \(2s\) | | \(\text{PbI}_2\) | Solid | - | Solid (remains constant) |
Step 3: Write the Equilibrium Expression
The solubility product \( K_{sp} \) is defined based on the concentrations of the ions at equilibrium. According to the dissolution of \(\text{PbI}2\), the expression for \( K{sp} \) is:
\[ K_{sp} = [\text{Pb}^{2+}][\text{I}^-]^2 \]
Step 4: Substitute Equilibrium Concentrations
From the ICE table, we have:
- \([\text{Pb}^{2+}] = s\)
- \([\text{I}^-] = 2s\)
Substituting these concentrations into the \( K_{sp} \) expression, we get:
\[ K_{sp} = (s)(2s)^2 \]
This simplifies to:
\[ K_{sp} = s \cdot 4s^2 = 4s^3 \]
Final Step: Summarize
The solubility product constant for lead(II) iodide (\( K_{sp} \)) is given by the expression:
\[ K_{sp} = 4s^3 \]
However, to calculate a numerical value for \( K_{sp} \), the solubility \( s \) would need to be known or measured. This value will not match solubility tables, as mentioned, and is consistent with the equilibrium established in this specific case.