The molar solubility of magnesium chloride in water is 0.23 M.

What is the numerical value for Ksp for MgCl2?
Suppose you wanted to reduce the concentration of dissolved Mg2+ to 0.00537 M by adding NaCl. What concentration of NaCl would you need?

MgCl2 ==> Mg^2+ + 2Cl^-

.........0.23....0.46
Substitute those number into the Ksp expression and solve for Ksp.

part 2.
Then Ksp = (Mg^2)(Cl^-)^2
Plug in 0.00537 for Mg^2+ and solve for Cl^-.

To find the numerical value for Ksp for MgCl2, we need to first understand the dissociation equation for the compound. Magnesium chloride (MgCl2) dissociates in water to form one magnesium ion (Mg2+) and two chloride ions (2Cl-).

The balanced dissociation equation is as follows:
MgCl2 (s) ⇌ Mg2+ (aq) + 2Cl- (aq)

The molar solubility of magnesium chloride (MgCl2) in water is given as 0.23 M. This means that at equilibrium, the concentration of Mg2+ ions in the solution will also be 0.23 M. Since there is a 1:1 stoichiometric ratio between MgCl2 and Mg2+, the concentration of Mg2+ is equal to the molar solubility.

Therefore, the numerical value for Ksp for MgCl2 can be calculated by using the molar solubility value:
Ksp = [Mg2+] * [Cl-]^2
Ksp = (0.23 M) * (0.23 M)^2
Ksp = 0.0121 M^3

Now, let's move on to the second part of the question.

If we want to reduce the concentration of dissolved Mg2+ to 0.00537 M by adding NaCl, we need to determine the concentration of NaCl required.

From the balanced dissociation equation mentioned above, we can see that each mole of MgCl2 that dissolves produces one mole of Mg2+ ions. Therefore, to reduce the concentration of Mg2+ from 0.23 M to 0.00537 M, we need to dilute by a factor of 0.00537/0.23.

Since the chloride ions (Cl-) come from NaCl, the molar ratio of NaCl to MgCl2 is 1:1. This means that the concentration of NaCl required to achieve a final Mg2+ concentration of 0.00537 M will also be 0.00537 M.

To find the value of Ksp for MgCl2, we need to use the molar solubility of magnesium chloride in water. The molar solubility represents the concentration of the dissolved ion in a saturated solution.

Given that the molar solubility of MgCl2 is 0.23 M, we can write the balanced equation for the dissolution of MgCl2 as follows:

MgCl2(s) ⇌ Mg2+(aq) + 2Cl-(aq)

The expression for the equilibrium constant Ksp in terms of the concentrations of the dissolved ions is:

Ksp = [Mg2+][Cl-]^2

Since the stoichiometry of the balanced equation shows that the concentration of Cl- is twice that of Mg2+, we can substitute 2[Cl-] for [Mg2+] in the expression:

Ksp = (2[Cl-])^3

Now, we can substitute the given molar solubility value of 0.23 M into the expression to find the numerical value for Ksp:

Ksp = (2 * 0.23)^3
= 8 * 0.23^3
= 8 * 0.012167
= 0.097336

Therefore, the numerical value for Ksp for MgCl2 is approximately 0.097336.

Now, let's move on to the second part of the question.

To reduce the concentration of dissolved Mg2+ to 0.00537 M by adding NaCl, we need to consider the common-ion effect. When a common ion is added to a solution, it decreases the solubility of the compound containing that ion.

In this case, adding NaCl will provide additional Cl- ions, which will lower the solubility of MgCl2 and subsequently reduce the concentration of dissolved Mg2+.

We can set up an expression using the solubility product constant, Ksp, to find the concentration of Cl- needed to achieve a desired concentration of Mg2+. The balanced equation allows us to determine the stoichiometric relationship between Mg2+ and Cl-:

1 mol MgCl2 produces 1 mol Mg2+ and 2 mol Cl-

So, if we want the concentration of Mg2+ to be 0.00537 M, we'll need twice the concentration of Cl- ions:

[Cl-] = 2 * [Mg2+]
[Cl-] = 2 * 0.00537
[Cl-] = 0.01074 M

Therefore, to achieve a concentration of dissolved Mg2+ of 0.00537 M, you would need a concentration of NaCl of 0.01074 M.