An ore is dissolved generating a solution with 0.10 M Cu 2+ and 0.0010 M Al 3+ .
ID: 516571 • Letter: A
Question
An ore is dissolved generating a solution with 0.10 M Cu2+ and 0.0010 M Al3+ . ( For the purposes of this problem, neglect the hydroxide ion from auto - ionization of water. ) Ksp Cu(OH)2 =1.6 x 10-19 Ksp Al(OH)3 = 4.6 x 10-33
a) As the pH o f the solution increases Cu(OH)2 or Al(OH)3 begin to precipitate first ? What concentration of hydroxide ion is required to begin precipitation of that substance? a) ppt 1st = __________________ [OH-] = ___________________
b) At what concentration of hydroxide ion will be needed to initiate precipitation of the other cation? b) [OH-] = ___________________
c) What percentage of the metal cation that precipitates first will be left in solution when the second hydroxide salt begins to precipitate? c) % = _______
Explanation / Answer
Ksp of Al(OH)3 is lower than the Ksp of Cu(OH)2, thus Al(OH)3 would precipitate out first from its solution
(a)
a) Al(OH)3 would first begins to precipiate from the solution
b) concentration of [OH-] to begin Al(OH)3 precipitation = cube rt.(Ksp/[Al3+])
= cube rt.(4.6 x 10^-33/0.001)
= 1.66 x 10^-10 M
b) concentration of [OH-] to begin Cu(OH)2 precipitation = sq.rt.(Ksp/[Al3+])
= sq.rt.(1.6 x 10^-19/0.1)
= 1.26 x 10^-9 M
c) what would be [Al3+] concentration at 1.26 x 10^-9 M [OH-] concentration,
[Al3+] = 4.6 x 10^-33/(1.26 x 10^-9)^3
= 2.30 x 10^-6 M would remain in solution when IInd precipitate begins to form.
% Al3+ remained = 2.30 x 10^-6 x 100/0.001 = 0.23%
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