It is possible to manipulate the solubility of some compounds through pH adjustments. What is the appropriate pH that will yield a Molar solubility for Mg(OH)2 of 2.51 x 10-8? The Ksp for Mg(OH)2 is 1.2 x 10-11. (Hint: you may assume the concentration of hydroxide is constant as the solute dissolves, since so little dissolves. Also, it is required to solve for pOH in this problem and then convert to pH.)
.......Mg(OH)2 ==> Mg^2+ + 2OH^-
I......solid........0........0
C......solid........x........2x
E......solid........x........2x
So Ksp = 1.2E-10
You want solubility Mg(OH)2 to be 2.51E-8M = (Mg^2+) = x in the above equation.
Plug those values into Ksp expression and solve for OH^-, then convert to pH.
To find the appropriate pH for a molar solubility of Mg(OH)2, we need to use the solubility product constant (Ksp) and the equilibrium expression for the dissolution of magnesium hydroxide:
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH^-(aq)
The Ksp expression can be written as:
Ksp = [Mg2+][OH^-]^2
Given that the Ksp for Mg(OH)2 is 1.2 x 10^-11, we can assume that the concentration of hydroxide ions, [OH^-], is constant at equilibrium since only a small fraction of the solute dissolves. Therefore, we can treat the concentration of hydroxide ions as "x".
From the balanced chemical equation, we can see that the molar solubility of Mg(OH)2 is equal to twice the concentration of hydroxide ions:
Molar solubility = 2[OH^-] = 2x
Given that the molar solubility of Mg(OH)2 is 2.51 x 10^-8, we can set up the following equation:
2.51 x 10^-8 = 2x
Now, solve this equation for x to find the concentration of hydroxide ions:
x = (2.51 x 10^-8) / 2 = 1.255 x 10^-8
Next, we can calculate the pOH using the equation:
pOH = -log[OH^-]
pOH = -log(1.255 x 10^-8)
Finally, convert the pOH to pH using the equation:
pH = 14 - pOH
pH = 14 - (-log(1.255 x 10^-8))
Calculate the value of pOH using a scientific calculator and subtract it from 14 to find the appropriate pH.