The balanced chemical equation for the dissociation of MgCl2 is:
MgCl2(s) ⇌ Mg2+(aq) + 2Cl-(aq)
The equilibrium expression for the dissociation reaction is:
Ksp = [Mg2+][Cl-]^2
where Ksp is the solubility product constant, [Mg2+] is the concentration of Mg2+ ions in solution, and [Cl-] is the concentration of Cl- ions in solution.
To calculate the Ksp of MgCl2, we need to first determine the concentration of Mg2+ and Cl- ions in the saturated solution. We can do this by using the given information that 200 g of MgCl2 is required to saturate 1.5 L of solution at 20°C.
The molar mass of MgCl2 is:
MgCl2 = 24.31 + 2(35.45) = 95.21 g/mol
So, the number of moles of MgCl2 in 200 g is:
n = mass / molar mass = 200 g / 95.21 g/mol = 2.10 mol
Since MgCl2 dissociates into one Mg2+ ion and two Cl- ions, the number of moles of Mg2+ ions in the solution is equal to the number of moles of MgCl2:
[Mg2+] = 2.10 mol / 1.5 L = 1.40 M
[Cl-] is twice the concentration of Mg2+ ions:
[Cl-] = 2 × [Mg2+] = 2.80 M
Now we can substitute these values into the Ksp expression to calculate the Ksp:
Ksp = [Mg2+][Cl-]^2 = (1.40 M)(2.80 M)^2 = 11.4
Therefore, the Ksp of MgCl2 at 20°C is 11.4. The units for Ksp depend on the units used for the concentrations. In this case, the units for Ksp are (M) x (M)^2 = M^3.
*IG:whis.sama_ent*