What [OH-] should be maintained in a solution if, after precipitation of Mg2+ as solid magnesium hydroxide, the remaining [Mg2+] is to be at a level of 1ug-L-1?
Mg(OH)2(s) ==> Mg^2+ + 2OH^-
Ksp = (Mg^2+)(OH^-)^2
You want [Mg^2+] = 1 ug/L. Convert to M = moles/L
moles Mg^2+ = 1E-6 g/L x 1 mol/24.3 g = ? which you substitute in the equation above. You know Ksp, you plug in this amount for Mg in mols/L and solve for OH^-
Post your work if you get stuck.
The Ans is 1.6 E -2 but using that approach doesn't work
You just assumed that approach was wrong. You COULD HAVE posted your work and let me look at it.
What did you use for Ksp? Using a value of 5.61E-12 I found on the web my answer is 1.12E-2 but I found other values for Ksp also. Did you use 1.05E-11 for Ksp or something close to that? Let me know if you want to wrap this up tonight.
They used 1.8E-11
The correct answer, using the numbers in the problem, is as follows:
Ksp = 1.8E-11 = [Mg^2+][OH^-]^2
If [Mg^2+] = 1 ug/L that is 1E-6/24.3 = 4.12E-8 mols Mg^2+/L and
[OH^-] = sqrt (Ksp/[Mg^2+]) = sqrt (1.8E-11/4.12E-8) = 0.021 M.
I suspect there is a typo in the problem. Let me know how things turn out please.
To determine the [OH-] concentration needed to achieve a certain [Mg2+] concentration after precipitation of magnesium hydroxide, we need to consider the solubility equilibrium of magnesium hydroxide.
The solubility equilibrium for magnesium hydroxide (Mg(OH)2) is represented by the following chemical equation:
Mg(OH)2 ⇌ Mg2+ + 2OH-
The equilibrium constant expression for this reaction, Ksp, is given by:
Ksp = [Mg2+][OH-]^2
Given that the remaining [Mg2+] concentration should be 1 μg/L, which is equivalent to 1x10^-9 mol/L, we can substitute the values into the Ksp expression and solve for [OH-].
Ksp = (1x10^-9)([OH-])^2
Now, we need to determine the value of Ksp for magnesium hydroxide. The Ksp value for Mg(OH)2 is approximately 1.8x10^-11 at 25°C.
1.8x10^-11 = (1x10^-9)([OH-])^2
Rearranging the equation and solving for [OH-]:
([OH-])^2 = (1.8x10^-11) / (1x10^-9)
([OH-])^2 = 1.8x10^-2
Taking the square root of both sides:
[OH-] ≈ √(1.8x10^-2)
[OH-] ≈ 0.134 M
Therefore, to maintain a magnesium ion concentration of 1 μg/L after precipitation of magnesium hydroxide, the [OH-] concentration in the solution should be approximately 0.134 M.