The hydroxide ion has the formula OH^-. The solubility-product constants for gen
ID: 1055466 • Letter: T
Question
The hydroxide ion has the formula OH^-. The solubility-product constants for generic hydroxide are given here. Use these values to answer the following questions. The removal of an ion sometimes considered to be complete when its concentration drops 1.00 times 10^-6 M. What concentration of hydroxide would cause Y^2+ to "completely" precipitate from a solution? Express the molar concentration numerically. [OH^-] = At a pH of 10.5, arrange the solution containing the following generic hydroxides in order of decreasing concentration of the cation remaining in the solution (i.e., in order of increasing completeness or precipitation). Rank from highest to lowest cation concentration. To rank items as equivalent, overlap them.Explanation / Answer
Part-A:
The chemical reaction for the dissociation of Y(OH)2 is
Y(OH)2(s) + H2O --------> Y2+(aq) + 2OH-(aq): Ksp = 3.20x10-10
------------------------------------- S, -------- 2S
Where 'S' is solubility
Ksp = 3.20x10-10 = [Y2+(aq)]x[OH-(aq)]2
=> 3.20x10-10 = S x (2S)2 = 4S3
=> S = cuberoot (3.20x10-10 / 4) = 4.31x10-4 M
Hence equilibrium concentration of OH-(aq) is [OH-(aq)] = 2S = 8.62x10-4 M (answer)
Part-B:
Given pH = 10.5
=> pOH = 14 - 10.5 = 3.5
=> [OH-(aq)] = 10-3.5 = 3.16x10-4 M
For XOH, Ksp = [X+(aq)] x [OH-(aq)] = S x 3.16x10-4 M
=> 2.80x10-8 = S x 3.16x10-4 M
=> S = 2.80x10-8 / 3.16x10-4 M = 8.86x10-5 M
=> [X+(aq)] = S = 8.86x10-5 M
For Y(OH)2, Ksp = 3.2x10-10 = [Y2+(aq)] x [OH-(aq)]2 = S x (3.16x10-4 M)2
=> [Y2+(aq)] = 3.2x10-10 / (3.16x10-4 M)2 = 3.2x10-3 M
For Z(OH)3,, Ksp = 8.2x10-15 = [Z3+(aq)] x [OH-(aq)]2 = S x (3.16x10-4 M)2
=> [Z3+(aq)] = 8.2x10-15/ (3.16x10-4 M)2 = 8.21x10-7 M
Hence the order of decreasing concentration of cations remaining is
[Y2+(aq)] > [X+(aq)] > [Z3+(aq)] (answer)
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