the solubility of Mg(OH)2 at 25C is4.6x10^-3 g/100cm3? what is the solubility product of M g(OH)2?

4.6E-3 g/100 cc = 4.6E-2/1000 cc.

mols = 4.6E-2/molar mass Mg(OH)2.
Then M = mols/L. I will call that S whatever it is.

........Mg(OH)2 ==> Mg^2+ + 2OH^-
I.......solid.......0........0
C.......solid.......s........2s
E.......solid.......s........2s

Substitute the E line into the Ksp expression and calculate the value of Ksp.

1.258

To find the solubility product (Ksp) of Mg(OH)2, we need to use the solubility information provided. The solubility of Mg(OH)2 at 25°C is given as 4.6x10^-3 g/100 cm^3.

The molar mass of Mg(OH)2 can be calculated as follows:
Mg: atomic mass = 24.31 g/mol
O: atomic mass = 16.00 g/mol
H: atomic mass = 1.01 g/mol

Molar mass of Mg(OH)2 = (24.31) + 2(1.01) + 2(16.00) = 58.33 g/mol

Now, let's convert the solubility into moles per liter (mol/L):
Solubility of Mg(OH)2 (g/L) = (4.6x10^-3 g/100 cm^3) * (1 L/1000 cm^3)
= 4.6x10^-5 g/L

Solubility of Mg(OH)2 (mol/L) = (4.6x10^-5 g/L) / (58.33 g/mol)
= 7.88x10^-7 mol/L

The solubility product (Ksp) of Mg(OH)2 can be calculated as the product of the concentrations of Mg2+ and OH- ions in a saturated solution at 25°C, based on the balanced equation for the dissociation of Mg(OH)2:

Mg(OH)2 ⇌ Mg2+ + 2OH-

Since the stoichiometry is 1:1 between Mg2+ and OH-, we can say that:
[Mg2+] = 7.88x10^-7 mol/L
[OH-] = 2 * (7.88x10^-7 mol/L) = 1.576x10^-6 mol/L

Therefore, the solubility product (Ksp) can be calculated as follows:
Ksp = [Mg2+] * [OH-]
= (7.88x10^-7 mol/L) * (1.576x10^-6 mol/L)
= 1.24x10^-12

Hence, the solubility product (Ksp) of Mg(OH)2 at 25°C is 1.24x10^-12.

To find the solubility product (Ksp) of Mg(OH)2, you need to understand the relationship between the solubility (S) and the concentrations of its ions.

The balanced equation for the dissociation of Mg(OH)2 is:
Mg(OH)2 ↔ Mg2+ + 2OH-

The solubility (S) of Mg(OH)2 is given as 4.6x10^-3 g/100 cm3. However, we need to convert this into molarity (M) by using the formula:

Molarity (M) = Mass (g) / Volume (L)

First, convert cm3 to liters by dividing by 1000:
Volume (L) = 100 cm3 / 1000 = 0.1 L

Now, plug in the values into the formula:
S (Mg(OH)2) = 4.6x10^-3 g / 0.1 L = 4.6x10^-2 g/L

Since Mg(OH)2 dissociates into Mg2+ and 2OH-, the concentration of Mg2+ and OH- ions will be twice the solubility of Mg(OH)2:

[Mg2+] = S (Mg(OH)2) = 4.6x10^-2 g/L
[OH-] = 2 × S (Mg(OH)2) = 2 × 4.6x10^-2 g/L = 9.2x10^-2 g/L

Now we can use these concentrations to calculate the solubility product (Ksp) using the formula:

Ksp = [Mg2+] × [OH-]^2

Ksp = (4.6x10^-2 g/L) × (9.2x10^-2 g/L)^2
Ksp = (4.6x10^-2) × (9.2x10^-2)^2

Calculating this value yields the solubility product (Ksp) of Mg(OH)2.