What volume in liters will be required to dissolve 60.9 grams of Mg(OH)2? Ksp = 1.8×10-11

After doing the math, i got 1.58e-4, but its not right, so i have no idea what to do now.

To determine the volume of solution required to dissolve a given amount of Mg(OH)2, you need to consider the solubility product constant (Ksp) and the molar mass of Mg(OH)2.

In this case, you have already calculated a value, but if it is not correct, we will go through the steps again to ensure accuracy.

To solve this problem:

1. Write the balanced equation for the dissolution of Mg(OH)2:
Mg(OH)2 ⇌ Mg2+ + 2OH-
The solubility product expression is: Ksp = [Mg2+][OH-]²

2. Write the expression for the solubility in terms of moles per liter (M):
[Mg2+] = x M
[OH-] = 2x M

3. Write the Ksp expression in terms of x:
Ksp = (x)(2x)² = 4x³
Ksp = 1.8×10^(-11)

4. Solve for x (concentration in M):
4x³ = 1.8×10^(-11)
x³ = 4.5×10^(-12)
x ≈ 2.9×10^(-4) M

5. Calculate the moles of Mg(OH)2:
moles = mass / molar mass
molar mass of Mg(OH)2 = (24.31 g/mol + 2 × 16.00 g/mol + 2 × 1.01 g/mol) = 58.33 g/mol
moles = 60.9 g / 58.33 g/mol ≈ 1.046 mol

6. Calculate the volume of solution required to dissolve the given amount of Mg(OH)2 in liters:
V = moles / [Mg(OH)2]
V = 1.046 mol / (2.9×10^(-4) M) ≈ 3600 L

Therefore, the volume of solution required to dissolve 60.9 grams of Mg(OH)2 is approximately 3600 liters.

1.58e-4 what? mL? L? Does it sound reasonable that you can dissolve almost 61 g Mg(OH)2 in less than 1 mL (even less than 1 L)?

If you had shown your work I could tell instantly what you did wrong. As it is I sit here and type all this out and let you look and see if you can find the error. Wasted time on my part. I could have used that time to help others while YOU did the typing.
...........Mg(OH)2 ==> Mg^2+ + 2OH^-
I..........solid.......0.........0
C..........solid.......x.........2x
E..........solid.......x.........2x

Ksp = 1.8E-11 = (x)(2x)^3 = 4x^3
x = ? = M = mols/L
Then grams = mols x molar mass = grams/L which is approx 0.01 g but you need to do that more accurately.
That's about 0.01g/L so how many L will it take to dissolve 60.9g? That's 1 L x (60.9/0.01) = ? L. Remember this is approximate.