Question
How to calculate the solubility product constant of PbI₂
Answers
GPT-4o mini
To calculate the solubility product constant (Ksp) of lead(II) iodide (PbI₂), you will need to understand how to set up the equilibrium expression based on its dissociation in water.
### Step 1: Write the Dissociation Equation
When lead(II) iodide (PbI₂) dissolves in water, it dissociates into its ions:
\[
\text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq)
\]
### Step 2: Define Solubility (s)
Let the solubility of PbI₂ in water be \( s \) mol/L. From the dissociation equation, you can see that:
- 1 mole of PbI₂ produces 1 mole of Pb²⁺ ions.
- 1 mole of PbI₂ produces 2 moles of I⁻ ions.
So at equilibrium:
- The concentration of Pb²⁺ (c) = \( s \)
- The concentration of I⁻ (b) = \( 2s \)
### Step 3: Write the Ksp Expression
The solubility product constant (Ksp) expression for the dissolution of PbI₂ is given by:
\[
K_{sp} = [\text{Pb}^{2+}] \cdot [\text{I}^-]^2
\]
Substituting the concentrations in terms of \( s \):
\[
K_{sp} = [s] \cdot [2s]^2
\]
### Step 4: Simplify the Expression
Calculating the expression:
\[
K_{sp} = s \cdot (2s)^2 = s \cdot 4s^2 = 4s^3
\]
### Step 5: Find the Value of Solubility (s)
To calculate a numerical value for \( Ksp \), you must determine the value of \( s \) experimentally, often through solubility measurements in a saturated solution. Once you have \( s \) (in moles per liter), you can substitute it back into the equation:
\[
K_{sp} = 4s^3
\]
### Example Calculation
For example, if the solubility (s) of PbI₂ is found to be \( 0.0002 \, \text{mol/L} \):
\[
K_{sp} = 4(0.0002)^3 = 4(8 \times 10^{-12}) = 3.2 \times 10^{-11}
\]
Make sure to replace \( 0.0002 \) with the actual solubility value you obtained experimentally to find the corresponding Ksp.
### Step 1: Write the Dissociation Equation
When lead(II) iodide (PbI₂) dissolves in water, it dissociates into its ions:
\[
\text{PbI}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{I}^- (aq)
\]
### Step 2: Define Solubility (s)
Let the solubility of PbI₂ in water be \( s \) mol/L. From the dissociation equation, you can see that:
- 1 mole of PbI₂ produces 1 mole of Pb²⁺ ions.
- 1 mole of PbI₂ produces 2 moles of I⁻ ions.
So at equilibrium:
- The concentration of Pb²⁺ (c) = \( s \)
- The concentration of I⁻ (b) = \( 2s \)
### Step 3: Write the Ksp Expression
The solubility product constant (Ksp) expression for the dissolution of PbI₂ is given by:
\[
K_{sp} = [\text{Pb}^{2+}] \cdot [\text{I}^-]^2
\]
Substituting the concentrations in terms of \( s \):
\[
K_{sp} = [s] \cdot [2s]^2
\]
### Step 4: Simplify the Expression
Calculating the expression:
\[
K_{sp} = s \cdot (2s)^2 = s \cdot 4s^2 = 4s^3
\]
### Step 5: Find the Value of Solubility (s)
To calculate a numerical value for \( Ksp \), you must determine the value of \( s \) experimentally, often through solubility measurements in a saturated solution. Once you have \( s \) (in moles per liter), you can substitute it back into the equation:
\[
K_{sp} = 4s^3
\]
### Example Calculation
For example, if the solubility (s) of PbI₂ is found to be \( 0.0002 \, \text{mol/L} \):
\[
K_{sp} = 4(0.0002)^3 = 4(8 \times 10^{-12}) = 3.2 \times 10^{-11}
\]
Make sure to replace \( 0.0002 \) with the actual solubility value you obtained experimentally to find the corresponding Ksp.