Calculate the solubility of magnesium hydroxide (Mg(OH)2), ksp=1.1x10^-11?
i have done this problem so many times and keep getting it wrong, i know the answer is 1.4x10^-4 but how do i get there? i did an ice chart and tried doing the cube root but its wrong. help!
Mg(OH)_2 ---------> Mg^2+ + 2OH^-
I. X 0 0
C. -X X 2X
K_sp <=========> [Mg^2+] [OH^-]^2
1.1*10^-11 = X (2X)^2
X = 3_/ 2.75*10^-12
X= 1.4*10^-4
hope i helped you :)
Cheers!
I have calculated by different cheap method and found answer to be correctly as 1.44x10^-4
To calculate the solubility of magnesium hydroxide (Mg(OH)2), we need to start by writing the balanced chemical equation for its dissociation:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
The solubility product constant (Ksp) is defined as the product of the concentrations of the dissociated ions raised to the power of their stoichiometric coefficients. In the case of Mg(OH)2, the Ksp expression can be written as follows:
Ksp = [Mg2+][OH-]^2
To calculate the solubility, we assume x moles of Mg(OH)2 dissolve, which means we will have x moles of Mg2+ and 2x moles of OH-.
Substituting these values into the Ksp expression, we get:
Ksp = [x][2x]^2
1.1x10^-11 = 4x^3
Rearrange the equation and solve for x:
x^3 = 1.1x10^-11 / 4
x^3 = 2.75x10^-12
Now take the cube root of both sides:
x = (2.75x10^-12)^(1/3)
x ≈ 1.39x10^-4
Therefore, the solubility of magnesium hydroxide (Mg(OH)2) is approximately 1.39x10^-4.
To calculate the solubility of magnesium hydroxide (Mg(OH)2), you need to use the solubility product constant (Ksp) and set up an equilibrium expression. Let's go through the steps together:
Step 1: Write the balanced equation for the dissociation of magnesium hydroxide (Mg(OH)2).
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)
Step 2: Define the solubility of Mg(OH)2 as "s". This represents the amount of Mg2+ ions and OH- ions that dissolve.
Step 3: Write the equilibrium expression for the dissociation reaction.
Ksp = [Mg2+][OH-]^2
Step 4: Substitute the solubility "s" into the equilibrium expression.
Ksp = (s)(2s)^2 = 4s^3
Step 5: Substitute the given Ksp value.
1.1x10^-11 = 4s^3
Step 6: Solve for "s"
To find the solubility (s), you need to take the cube root of (Ksp/4).
s = (∛(Ksp/4)) = (∛(1.1x10^-11/4))
Calculating this value will give you the solubility of magnesium hydroxide (Mg(OH)2).
s = 1.4x10^-4 (approximately)
Therefore, the solubility of magnesium hydroxide is approximately 1.4x10^-4.