Answer :
The molar solubility of PbI₂ in a solution of Pb(NO₃)₂ is C. 4.2×10−5M.
To determine the solubility of PbI₂ (lead(II) iodide) in a 2.00 M solution of Pb(NO₃)₂ (lead(II) nitrate), we need to consider the common ion effect and the solubility product constant (Ksp).
Step 1: Understanding the Ksp
The K(sp) for PbI₂ is given as 1.4 × 10⁻⁸. The dissociation of PbI₂ in water can be represented as:
PbI₂(s)⇌Pb²⁺(aq)+2I⁻(aq)
This means that for every mole of PbI₂ that dissolves, one mole of Pb²⁺ and two moles of I⁻ are produced.
Step 2: Setting Up the Expression for K(sp)
The relationship for K(sp) is:
K(sp)=[Pb²⁺][I⁻]²
Step 3: Calculating Ion Concentrations in the Solution
In a 2.00 M Pb(NO₃)₂ solution, the concentration of Pb²⁺ ions is already 2.00 M due to the dissociation of Pb(NO₃)₂:
[Pb²⁺]=2.00M
Let s be the solubility of PbI₂ in this solution. The concentration of iodide ions produced from PbI₂ would be:
[I⁻]=2s
Step 4: Plugging into the K(sp) Expression
Now, substituting into the K(sp) expression gives:
K(sp)=(2.00)(2s)²
Step 5: Substitute K(sp) into the Equation
Substituting the value for K(sp):
1.4×10⁻⁸ = (2.00)(2s)²
This simplifies to:
1.4×10⁻⁸ = (2.00)(4s)²
1.4×10⁻⁸ = 8.00s²
Step 6: Solving for s
Now, solve for s:
[tex]s^{2} = \frac{1.4*10^{-8} }{8.00}[/tex]
s² = 1.75×10⁻⁹
s = √1.75×10⁻⁹
s ≈ 4.18×10⁻⁵ M.
s ≈ 4.2×10⁻⁵ M.
