What concentration of the lead ion, Pb^{2+} , must be exceeded to precipitate PbCl2 from a solution that is 1.00 x 10^{-2} M in the chloride ion, Cl^- ? Ksp for lead chloride is 1.17×10−5 .
To determine the concentration of lead ion, Pb^{2+}, that must be exceeded to precipitate PbCl2 from a solution that is 1.00 x 10^{-2} M in chloride ion, Cl^-, we need to use the solubility product constant (Ksp) for lead chloride (PbCl2), which is given as 1.17×10^−5.
The solubility product constant (Ksp) expression for the dissolution of PbCl2 can be written as:
Ksp = [Pb^{2+}][Cl^-]^2
Since the stoichiometry of the reaction is 1:2 (Pb^{2+}: Cl^-), we can write the equilibrium expression as follows:
Ksp = [Pb^{2+}][Cl^-]^2
Given that the concentration of chloride ion (Cl^-) is 1.00 x 10^{-2} M, we can substitute this value into the equation:
Ksp = [Pb^{2+}](1.00 x 10^{-2})^2
Simplifying the equation, we have:
1.17×10^−5 = [Pb^{2+}](1.00 x 10^{-2})^2
Solving for [Pb^{2+}], we can rearrange the equation as:
[Pb^{2+}] = (1.17×10^−5) / (1.00 x 10^{-2})^2
[Pb^{2+}] = (1.17×10^−5) / (1.00 x 10^{-2} x 1.00 x 10^{-2})
[Pb^{2+}] = 1.17×10^−5 / 1.00 x 10^{-4}
[Pb^{2+}] = 1.17×10^1
Therefore, the concentration of lead ion (Pb^{2+}) that must be exceeded to precipitate PbCl2 from a solution that is 1.00 x 10^{-2} M in chloride ion (Cl^-) is 11.7 M.
To find the concentration of the lead ion, Pb²⁺, at which PbCl₂ will start to precipitate, we need to calculate the ion product (IP) for lead chloride (PbCl₂) and compare it to the solubility product constant (Ksp).
The ion product (IP) is calculated by multiplying the concentrations of the ions present in the solution at any given moment. For the dissociation of PbCl₂, the balanced equation is:
PbCl₂ ⇌ Pb²⁺ + 2Cl⁻
The IP expression for PbCl₂ is: IP = [Pb²⁺] * [Cl⁻]²
Given that the concentration of chloride ion (Cl⁻) is 1.00 x 10⁻² M, we can substitute this value into the IP expression:
IP = [Pb²⁺] * (1.00 x 10⁻²)² = [Pb²⁺] * 1.00 x 10⁻⁴
To determine the concentration of Pb²⁺ at which PbCl₂ will start to precipitate, we compare the IP to the solubility product constant (Ksp). The Ksp for lead chloride is given as 1.17 x 10⁻⁵.
If IP > Ksp, then precipitation will occur. Therefore, we need to find the concentration of Pb²⁺ at which IP is equal to Ksp.
Thus: IP = Ksp
[Pb²⁺] * 1.00 x 10⁻⁴ = 1.17 x 10⁻⁵
Solving for [Pb²⁺]:
[Pb²⁺] = (1.17 x 10⁻⁵) / (1.00 x 10⁻⁴)
[Pb²⁺] ≈ 0.117 M
Therefore, the concentration of lead ion (Pb²⁺) must exceed 0.117 M to precipitate PbCl₂ from a solution that is 1.00 x 10⁻² M in chloride ion (Cl⁻).
Ksp = (Pb^2+)(Cl^-)^2
Substitute for Ksp and (Cl^-) and solve for (Pb^2+)