A saturated solution of Mg(OH)2 is prepared having a large excess of Mg(OH)2. Sn(NO3)2 is added to the solution. Ksp = 1.8 10-11 for Mg(OH)2 and Ksp = 5.1 10-26 for Sn(OH)2.
(a) What [Sn2+] is required to start the precipitation of Sn(OH)2?
(b) What [Sn2+] is required so that the [Mg2+] in solution will be 0.14 M?
First determine the OH^- already in the solution from Mg(OH)2.
.......Mg(OH)2 ==> Mg2+ + 2OH^-
I.......solid.......0......0
C.......solid.......x......2x
E.......solid.......x......2x
Ksp = (Mg^2+)(OH^-)^2
Solve for x = Mg and OH = 2x.
Then go through the Sn(OH)2 the same way, subtitute OH from the Mg(OH)2 calculation and solve for (Sn^2+).
b. Go back to Mg(OH)2, substitute 0.14M for (Mg^2+) and solve for OH^-. Substitute that OH^- into the Sn(OH)2 and solve for (Sn^2+)
Thank you! I got this one right.
To determine the [Sn2+] required to start the precipitation of Sn(OH)2, we need to compare the solubility product constant (Ksp) of Sn(OH)2 with the product of the concentrations of Sn2+ and OH- ions.
(a) Starting with the Ksp expression for Sn(OH)2:
Ksp = [Sn2+][OH-]^2
Since we have a large excess of Mg(OH)2, we can assume all the OH- ions come from Mg(OH)2, and therefore the concentration of OH- ions is equal to the concentration of Mg2+ ions, which is twice the concentration of Sn2+ ions. Let's denote the concentration of Sn2+ as x M.
So, [Sn2+] = x M
[OH-] = 2x M
Now we can substitute these values into the Ksp expression:
1.8 × 10^-26 = (x)(2x)^2
1.8 × 10^-26 = 4x^3
Solving for x gives us:
x^3 = (1.8 × 10^-26) / 4
x^3 = 4.5 × 10^-27
x ≈ 7.5 × 10^-9 M
Therefore, the [Sn2+] required to start the precipitation of Sn(OH)2 is approximately 7.5 × 10^-9 M.
(b) To find the [Sn2+] required to have a [Mg2+] of 0.14 M in solution, we need to set up another equation using the Ksp values and concentrations.
Using the Ksp expression for Mg(OH)2:
Ksp = [Mg2+][OH-]^2
Since [OH-] = 2[Mg2+], we have:
1.8 × 10^-11 = (0.14)(2x)^2
1.8 × 10^-11 = 0.56x^2
x^2 ≈ (1.8 × 10^-11) / 0.56
x^2 ≈ 3.214 × 10^-11
x ≈ √(3.214 × 10^-11)
Therefore, the [Sn2+] required for the [Mg2+] in solution to be 0.14 M is approximately √(3.214 × 10^-11) M.