The hydroxide ion has the formula OH. The solubility-product constants for three

Question

The hydroxide ion has the formula OH. The solubility-product constants for three generic hydroxides are given here. Generic hydroxide Ksp XOH 1.60×108 Y(OH)2 4.10×1010 Z(OH)3 8.20×1015 Use these values to answer the following questions. A.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? B.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. To rank items as equivalent, overlap them. Y(OH)2 Z(OH)3 XOH

Explanation / Answer

A. [OH-] concentration to precipitate Y2+

Ksp = [Y2+][OH-]^2

4.10 x 10^-10 = (1 x 10^-6)[OH-]^2

[OH-] required = 0.020 M

B. at pH 10.5

pOH = 14 - pH = 14 - 10.5 = 3.5

pOH = -log[OH-]

[OH-] = 3.16 x 10^-4 M

Let us calculate cation precipitated at this [OH-] concentration,

For XOH

[X+] = Ksp/[OH-] = 1.60 x 10^-8/3.16 x 10^-4 = 5.06 x 10^-5 M

For Y(OH)2

[Y2+] = Ksp/[OH-]^2 = 4.1 x 10^-10/(3.16 x 10^-4)^2 = 4.10 x 10^-3 M

For Z(OH)3

[Z+] = Ksp/[OH-]^3 = 8.2 x 10^-15/(3.16 x 10^-4)^3 = 2.60 x 10^-4 M

So the order for decreasing cation concentration remaining in solution : XOH > Z(OH)3 > Y(OH)2

Ranking on basis of cation precipitated from solution (decreasing order) : Y(OH)2 > Z(OH)3 > XOH

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