Copper and Bismuth
What would you observe if S1 was treated with 6 M NaOH rather than
concentrated NH$_3$? The relevant portion of the separation scheme is below, for
your reference.
(S1) Cu$^{2+}$, Bi$^{3+}$
(mostly colorless, may be light blue.)
conc. NH$_3$
(P4) Bi(OH)$_3$
(white solid)
[Cu(NH$_3$)$_4$]$^{2+}$
(deep blue solution)
8
1 point
If S1 was treated with 6 M NaOH instead of NH$_3$, the supernatant would:
Turn deep blue
Stay mostly colorless
9
1 point
If S1 was treated with 6 M NaOH instead of NH$_3$, the precipitate would:
would form (there would be a precipitate)
not form (there would be no precipitate)
10
2 points
If S1 was treated with 5 drops of 6 M NaOH instead of NH$_3$, what compounds
would form?
Multiple answer, select all that apply. There is a penalty for selecting incorrect
answers.
[Cu(OH)$_4$]$^{2+}$(aq)
Bi(OH)$_3$(s)
[Cu(NH$_3$)$_4$]$^{2+}$(aq)
Cu(OH)$_2$(s)
