The concentration of all solute species in the solution are:
[tex][Mg^{2+} ] = 0.631 M \\ [Pb] = 1.0 \times 10^{-5} M \\[Cl^{-} ] = 1.26 M[/tex]
MgCl₂ (0.631 M) is a strong electrolyte that dissociates according to the following equation.
MgCl₂(aq) ⇒ Mg²⁺(aq) + 2 Cl⁻(aq)
Considering the molar ratios, the initial concentrations of Mg²⁺ and Cl⁻ are:
[tex]\frac{0.631 mol MgCl_2}{L} \times \frac{1molMg^{2+} }{1molMgCl_2} = \frac{0.631molMg^{2+}}{L} \\\frac{0.631 mol MgCl_2}{L} \times \frac{2molCl^{-} }{1molMgCl_2} = \frac{1.26molCl^{-}}{L}[/tex]
Cl⁻ is a common ion between MgCl₂ and PbCl₂. To consider the solution equilibrium for PbCl₂ (a sparingly soluble salt), we will make an ICE chart.
PbCl₂(s) ⇄ Pb²⁺(aq) + 2 Cl⁻(aq)
I 0 1.26
C +x +2x
E x 1.26+x
The solubility product constant, Ksp, is:
[tex]Ksp = 1.6 \times 10^{-5} = [Pb^{2+} ][Cl^{-} ]^{2} = (x) (1.26+x)^{2}[/tex]
Since 1.26 >>> x, 1.26 + x ≈ 1.26.
[tex]Ksp = 1.6 \times 10^{-5} = (x) (1.26)^{2}\\x = 1.0 \times 10^{-5}[/tex]
The concentration of all solute species in the solution are:
[tex][Mg^{2+} ] = 0.631 M\\ [Pb] = x = 1.0 \times 10^{-5} M\\[Cl^{-} ] = 1.26+x = 1.26 M[/tex]
You can learn more about solubility product here: https://brainly.com/question/4736767
