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 = 1.9 x 10^-11 for Mg(OH)2 and Ksp = 6.0x10^-26 for Sn(OH)2.
What [Sn2+] is required so that the [Mg2+] in solution will be 0.15 M?
I would do this.
(Mg^+2)(OH^-)^2 = 1.9E-11
Plug in 0.15 for Mg and solve for (OH^-)^2.
Then (Sn^+2)(OH^-)^2 = 6.0E-26
Plug in (OH^-)^2 from above and solve for (Sn^+2)
Thanks, I got it right!
To determine the required concentration of Sn2+ ions ([Sn2+]) that will result in a [Mg2+] of 0.15 M, we need to use the solubility product constant (Ksp) values for Mg(OH)2 and Sn(OH)2.
1. Mg(OH)2:
The Ksp expression for Mg(OH)2 is given by:
Ksp = [Mg2+][OH-]^2 = 1.9 x 10^-11
We are given that the [Mg2+] in solution should be 0.15 M. Since the solution contains a large excess of Mg(OH)2, we can assume that [OH-] is significantly smaller compared to [Mg2+]. Thus, we can neglect the contribution of [OH-] and directly equate [Mg2+] to 0.15 M.
[Mg2+] = 0.15 M
2. Sn(OH)2:
The Ksp expression for Sn(OH)2 is given by:
Ksp = [Sn2+][OH-]^2 = 6.0 x 10^-26
Since we want to determine the concentration of [Sn2+] required to achieve a [Mg2+] of 0.15 M, we can assume that [OH-] remains the same and calculate the [Sn2+] accordingly.
Let's define x as the concentration of [Sn2+]. Since each Sn2+ ion reacts with two OH- ions, the concentration of OH- will be 2x.
Substituting the values into the Ksp expression for Sn(OH)2:
6.0 x 10^-26 = (x)(2x)^2 = 4x^3
Solving for x:
4x^3 = 6.0 x 10^-26
x^3 = (6.0 x 10^-26) / 4
x^3 = 1.5 x 10^-26
x = (1.5 x 10^-26)^(1/3)
Using a calculator, we find x ≈ 1.085 x 10^-9
Therefore, to achieve a [Mg2+] of 0.15 M, a concentration of approximately 1.085 x 10^-9 M of Sn2+ ions is required.