Metal Hydroxides: Evaluating Completeness of Precipitation The hydroxide ion has
ID: 1018217 • Letter: M
Question
Metal Hydroxides: Evaluating Completeness of Precipitation 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 2.40×1010 Z(OH)3 6.00×1015 Use these values to answer the following
Part 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? Express the molar concentration numerically.
Part 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. HelpReset
Explanation / Answer
Ksp XOH 1.60×108 , Ksp Y (OH) 2 2.40×1010 , Ksp Z (OH) 3 6.00×1015
Part A
Ksp for Y(OH)2 is 2.40 x10-10
So first we write out the Ksp equation:
Ksp = [Y2+][OH-]2
For the removal of an ion to be considered complete, its concentration must be 1.0x10-6 M
So [Y2+] = 1.0x10-6
Plugging all this into the Ksp equation above
Ksp = [Y2+][OH-]2
2.40 x10-10 = (1.0x10-6)[OH-]2
[OH-] = 0.0154
B)
pH = 10.5(given)
pOH = 14- pH
= 14-10.5 = 3.5
[OH-] = antilog (-3.5)
= 0.000316 M
We can calculate the amount of X, Y, and Z, using the following equations:
Ksp = [X][OH-]
1.60×108 = [X] [0.000316 M]
[X] = 1.60×108 / [0.000316 M]
[X] = 5.06 x 105
2Ksp = [Y2+][OH-] 2
2.40×1010 = [Y2+] [0.000316 M] 2
[Y2+]= 2.40×1010 / [0.000316 M] 2
[Y2+] = 1.60 x 103
Ksp = [Z3+][OH-]3
6.00 × 1015 = [Z] [0.000316 M] 3
[Z3+] = 6.00 × 1015/ [0.000316 M] 3
[Z3+] = 6.00 x 1015 /3.155 x1011
[Z3+] = 1.90 x 104
Highest to lowest cation concentration
[Y2+],[ Z3+],[ X +]