The hydroxide ion has the formula OH. The solubility-product constants for three
ID: 880243 • Letter: T
Question
The hydroxide ion has the formula OH. The solubility-product constants for three generic hydroxides are given here.
The removal of an ion is sometimes considered to be complete when its concentration drops to 1.00×106 M. What concentration of hydroxide would cause Y2+ to "completely" precipitate from a solution?
Express your answer with the appropriate units
At a pH of 10.5, arrange the solutions containing the following generic hydroxides in order of decreasing concentration of the cation remaining in the solution (i.e., in order of increasing completeness of precipitation).
Rank from highest to lowest cation concentration.
Y(OH)2 Z(OH)3 XOH
Generic hydroxide Ksp XOH 2.20×108 Y(OH)2 2.80×1010 Z(OH)3 8.70×1015Explanation / Answer
A) In the question, it tells you that the Ksp for Y(OH)2 is 2.10 x10^-10, so first write out the Ksp equation:
Ksp = [Y2+][OH-]^2
Also, it tells you that in order for the removal of an ion to be considered complete, its concentration must be 1.0x10^-6. So you know:
[Y2+] = 1.0x10^-6
Plugging all this into the Ksp equation above, you get:
Ksp = [Y2+][OH-]^2
2.80 x10^-10 = (1.0x10^-6)[OH-]^2
[OH-] = 0.0167 M concentration of hydroxide would cause Y2+ to "completely" precipitate from a solution
B) For this part, it tells you that pH= 10.5, so you can calculate [OH-] in the following way:
pOH = 14- pH
= 14-10.5
= 3.5
[OH-] = antilog(-3.5)
= 0.000316 M
From here, you can calculate the amount of X, Y, and Z, using the following equations:
(1) Ksp = [X+][OH-]
2.20 x 10^-8 = p[X+][0.000316]
[X+] = 6.96 x 10^-5
(2) Ksp = [Y2+][OH-]^2
2.8 x 10^-10 = [Y2+][0.000316]^2
[Y2+] = 2.80 x 10^-3
(3) Ksp = [Z3+][OH-]^3
8.70 x 10^-15 = [Z3+][OH-]^3
[Z3+] = 2.76 x 10^-4
the values for X, Y, and Z tells the concentration of the cations.
So the order of generic hydroxides in order of decreasing concentration of the cation remaining in the solution Highest -----------> Lowest : Y, Z, X
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