Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solution of salt in contact with a solution containing a common ion. Show that changes in the initial concentrations: PbCl2(s) in 0.631 M MgCl2 (MgCl2 is strong electrolyte)
Ksp PbCl2 = 1.6 × 10−5

The concentration of all solute species in the solution are:

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:

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:

Since 1.26 >>> x, 1.26 + x ≈ 1.26.

The concentration of all solute species in the solution are:

You can learn more about solubility product here:

i think the answer is terrae but not 100 percent sure. ▓▬▬ §

the answer is c. nacl