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What is the solubility of PbI2 in a 2.00 M solution of Pb(NO3)2? [Ksp(PbI2) =1.4 × 10–8]

Question options:



7.0 × 10–9 M



1.4 × 10–8 M



4.2 × 10–5 M



5.9 × 10–5 M



8.4 × 10–5 M

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.

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