Question

The hydroxide ion has the formula OH− . The solubility-product constants for three generic hydroxides are given here. Generic hydroxide Ksp XOH 2.00×10−8 Y(OH)2 2.20×10−10 Z(OH)3 9.20×10−15 Use these values to answer the following questions. Part A Part complete The removal of an ion is sometimes considered to be complete when its concentration drops to 1.00×10−6 M . What concentration of hydroxide would cause Y2+ to "completely" precipitate from a solution? Part B Part complete 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). [OH-]= [

Answer 1

Part A

Given that , Ksp of Y(OH)₂ = 2.20*10^-10

Y²⁺ = 1.00 *10^-6

_{Reaction is :}

Y(OH)₂ Y²⁺ + 2OH^-

Ksp = [ Y²⁺][OH^-]²

2.20*10^-10 = (1.00 *10^-6) [OH^-]²

[OH^-]² =  2.20*10^-4

[OH^-] = 0.0148 M

Part B

Now, pH = 10.5

pOh = 14 - pH = 14 - 10.5 = 3.5

pOH = -log[OH^-]

[OH^-] = antilog(-3.5) = 0.000316 M

For first hydroxide

Ksp = [X⁺][OH^-]

2.00*10^-8 = [X⁺][0.000316)

[X⁺] = 6.33 *10^-5

For second hydroxide

Ksp = [ Y²⁺][OH^-]²

2.20*10^-10 = [ Y²⁺](0.000316)²

[ Y²⁺] = 2.20 *10^-3

For third reaction,

Ksp = [Z³⁺][OH^-]³

9.20*10^-15 = [Z³⁺](0.000316)³

[Z³⁺] = 2.92*10^-4

Thus, decreasing order of cation concentration is:

Y,Z, X or Y> Z > X