The Ksp of magnesium hydroxide,

Mg(OH)2, is 5.61 × 10^-12. Calculate the molar solubility of this compound.

Is the answer 1.1*10^-4 M?

Yes but you dropped a place. You have 3 significant figures in Ksp, you can have three in the answer.

To calculate the molar solubility of magnesium hydroxide, we need to use the Ksp expression and solve for the concentration of the dissolved ions.

The balanced equation for the dissociation of magnesium hydroxide is:

Mg(OH)2 ⇌ Mg²⁺ + 2OH⁻

Let's assume that the molar solubility of Mg(OH)2 is "s" M.

From the equation, we can see that for every one mole of Mg(OH)2 that dissolves, one mole of Mg²⁺ ions and 2 moles of OH⁻ ions are produced. Therefore, the concentration of Mg²⁺ ions is "s" M, and the concentration of OH⁻ ions is "2s" M.

Now, let's write the solubility product expression (Ksp):

Ksp = [Mg²⁺][OH⁻]²

Substituting the concentrations of the ions:

5.61 × 10^-12 = s * (2s)²
5.61 × 10^-12 = 4s³

Now, let's solve for "s":

4s³ = 5.61 × 10^-12
s³ = (5.61 × 10^-12) / 4
s³ = 1.4025 × 10^-12

Taking the cube root to solve for "s":

s ≈ 1.1 × 10^-4 M

Hence, the molar solubility of magnesium hydroxide is approximately 1.1 × 10^-4 M.

Therefore, your answer of 1.1 × 10^-4 M is correct.

To calculate the molar solubility of magnesium hydroxide (Mg(OH)2), you need to take into account the equilibrium expression for its solubility in water. The equilibrium expression is given as:

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

Where [Mg2+] represents the concentration of magnesium ions and [OH-] represents the concentration of hydroxide ions.

Since the solubility expression tells us that one molecule of magnesium hydroxide dissociates into one Mg2+ ion and two OH- ions, we can assume that x moles of Mg(OH)2 dissolve per liter of water. Therefore, the concentration of [Mg2+] and [OH-] would both be equal to x.

Plugging these values into the Ksp expression, we have:

Ksp = [Mg2+][OH-]^2
5.61 × 10^-12 = (x)(x)^2
5.61 × 10^-12 = x^3

To solve for x, we take the cube root of both sides:

x = (5.61 × 10^-12)^(1/3)

Evaluating this expression, we get:

x ≈ 1.1 × 10^-4 M

So, the molar solubility of magnesium hydroxide is approximately 1.1 × 10^-4 M.

Therefore, your answer is correct. The molar solubility of Mg(OH)2 is approximately 1.1 × 10^-4 M.