Calculate how many times more soluble Mg (OH) 2 is in pure water Based on the given value of the Ksp, 5.61*10-11, calculate the ratio of solubility of Mg (OH) 2 dissolved in pure H2O to Mg (OH) 2 dissolved in a 0.130 M NaOH solution.

molar solubility in water = 2.41 x 10^-4

molar solubility of NaOH in 0.130M = 3.32 x 10^-9

molar solubility in water = 2.412 * 10^-4 mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg (OH) 2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

The equation for the solubilization reaction of Mg (OH) 2 can be given as:

Mg (OH) 2 (s) → Mg2 + (aq) + 2OH - (aq)

Ksp can then be given as followed:

Ksp = [Mg^2+][OH^-]²

Step 2: Calculate the solubility in water

Mg (OH) 2 (s) → Mg2 + (aq) + 2OH - (aq)

The mole ratio Mg^2 + with OH - is 1:2

So there will react X of Mg^2 + and 2X of OH-

The concentration at equilibrium will be XM Mg^2 + and 2X OH-

Ksp = [Mg^2+][OH^-]²

5.61*10^-11 = X * (2X) ² = X * 4X² = 4X³

X = 2.412 * 10^-4 mol/L = solubility in water

Step 3: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2 + = 0 M

The initial concentration of OH - = 0.130 M

The mole ratio Mg^2 + with OH - is 1:2

So there will react X of Mg^2 + and 2X + 0.130 for OH-

The concentration at equilibrium will be XM Mg^2 + and 0.130 + 2X OH-

The value of "[OH-] + 2X" is, because the very small value of X, equal to the value of [OH-].

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^-]²

5.61*10^-11 = X * (0.130) ²

5.61*10^-11 = X * (0.130) ^2

X = 3.32*10^-9 = solubility in 0.130 M NaOH

Step 4: Calculate how many times Mg (OH) 2 is better soluble in pure water.

(2.412*10^-4) / (3.32*10^-9) = 0.73 * 10^5

Mg (OH) 2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH