The concentration of Mg2+ ion in a solution is 0.010M. At what pH will Mg(OH)2 first precipitate? (Given Ksp Mg(OH)2 equals 1.2*10-11)
........Mg(OH)2 ==> Mg^2+ + 2OH^-
Ksp = (Mg^2+)(OH^-)^2
Plug in 0.01 M for (Mg^2+) and solve for OH-, then convert OH^- to pH.
thanks I understand now
To determine at what pH Mg(OH)2 will first precipitate, we need to calculate the concentration of hydroxide ions (OH-) in the solution. Please note that we are assuming the concentration of Mg2+ ion remains constant.
The balanced equation for the dissociation of Mg(OH)2 is:
Mg(OH)2 ⇌ Mg2+ + 2OH-
The equilibrium expression for the solubility product constant (Ksp) of Mg(OH)2 is:
Ksp = [Mg2+][OH-]^2
Given Ksp = 1.2 x 10^-11 and the concentration of Mg2+ = 0.010 M, we need to find the OH- concentration.
Let's assume "x" is the concentration of the OH- ions.
Since one mole of Mg(OH)2 dissociates to give two moles of OH-, the concentration of OH- is 2x.
Using the equilibrium expression:
1.2 x 10^-11 = (0.010)(2x)^2
1.2 x 10^-11 = 0.020x^2
Rearranging the equation:
x^2 = (1.2 x 10^-11) / 0.020
x^2 = 6 x 10^-10
Taking the square root of both sides:
x ≈ √(6 x 10^-10)
x ≈ 7.75 x 10^-6 M
Now, we have the concentration of hydroxide ions (OH-) in the solution. To find the pH at which Mg(OH)2 will first precipitate, we can use the equation:
pOH = -log[OH-]
pOH = -log(7.75 x 10^-6)
pOH ≈ 5.11
Since pH + pOH = 14, we can calculate the pH:
pH = 14 - pOH
pH ≈ 14 - 5.11
pH ≈ 8.89
Therefore, Mg(OH)2 will begin to precipitate when the pH of the solution reaches approximately 8.89.