The Cu2+ ion can be separated from Ag+,
Ca2+, and K+ in aqueous solution by
1. precipitation of Cu2+ as Cu(OH)2(s) with
6 M NaOH(aq).
2. precipitation of Ag+, Ca2+, and K+ as
the carbonates.
3. None of these procedures will separate
Cu2+ from the other ions.
4. precipitation of Cu2+ as CuCl2(s) with 6
M HCl(aq).
5. precipitation of Cu2+ as CuS(s) at pH 1.
The Cu2+ ion can be separated from Ag+,
Ca2+, and K+ in aqueous solution by
1. precipitation of Cu2+ as Cu(OH)2(s) with 6 M NaOH(aq).
Cu(OH)2 will ppt but so will Ag(OH)2/Ag2O.
2. precipitation of Ag+, Ca2+, and K+ as
the carbonates.
CuCO3 will ppt with Ag2CO3 and K^+ will not\
3. None of these procedures will separateCu2+ from the other ions.
I think this is the best answer.
4. precipitation of Cu2+ as CuCl2(s) with 6 M HCl(aq).
Absolutely not. CuCl2 is soluble.
5. precipitation of Cu2+ as CuS(s) at pH 1.
The solubility of Ag2S and CuS are more or less the same at pH=1 so that will not separate the two.
6. By telling them a really bad joke and hoping they leave the solution out of sheer disappointment.
The option that can separate the Cu2+ ion from Ag+, Ca2+, and K+ in aqueous solution is:
1. Precipitation of Cu2+ as Cu(OH)2(s) with 6 M NaOH(aq).
This method takes advantage of the low solubility of Cu(OH)2 compared to the other metal hydroxides. When 6 M NaOH(aq) is added to the solution, Cu2+ ions will react with hydroxide ions to form Cu(OH)2 precipitation, while Ag+, Ca2+, and K+ ions will remain in solution as their corresponding hydroxides are more soluble. This precipitation allows for the separation of Cu2+ ion from the other ions.
To determine which procedure can separate the Cu2+ ion from Ag+, Ca2+, and K+ in aqueous solution, we need to consider the solubility of the compounds formed during the reactions. Let's break down each option:
1. Precipitation of Cu2+ as Cu(OH)2(s) with 6 M NaOH(aq):
To see if Cu2+ can be separated from the other ions, we need to check the solubilities of the resulting compounds. Cu(OH)2 is sparingly soluble, while AgOH, Ca(OH)2, and KOH are more soluble. Therefore, this procedure will not separate Cu2+ from the other ions.
2. Precipitation of Ag+, Ca2+, and K+ as the carbonates:
Similarly, we need to compare the solubilities of the carbonates formed. Ag2CO3 is slightly soluble, CaCO3 is moderately soluble, and K2CO3 is highly soluble. Once again, this procedure will not separate Cu2+ from the other ions.
3. None of these procedures will separate Cu2+ from the other ions:
This option suggests that none of the given procedures can separate Cu2+ from Ag+, Ca2+, and K+. However, we need to verify this statement based on the solubilities of the compounds formed during each reaction.
4. Precipitation of Cu2+ as CuCl2(s) with 6 M HCl(aq):
To determine the solubility of CuCl2, we need to consult a solubility table. According to the table, CuCl2 is soluble in water. This procedure will not separate Cu2+ from the other ions either.
5. Precipitation of Cu2+ as CuS(s) at pH 1:
Again, let's refer to a solubility table to check CuS's solubility. CuS is insoluble in water, so this procedure is capable of separating Cu2+ from Ag+, Ca2+, and K+ in aqueous solution.
Therefore, option 5, precipitation of Cu2+ as CuS(s) at pH 1, is the correct procedure to separate Cu2+ from Ag+, Ca2+, and K+ ions in the given solution.