copper(ii) complex

2008-12-21 3:11 am
when an excess of AgNo3 is added to the following solutions
Na2[CuCl4],[CuCl2(NH3)2],[Cu(NH3)4]Cl2, [CuCl(NH3)3]Cl
which one will produce
the largest amount of AgCl solid? Explain briefly.

回答 (1)

2008-12-21 8:33 am
✔ 最佳答案

In aqueous solutions, only free Cl-(aq) can react with Ag+(aq) ions to form AgCl precipitate.
Ag+(aq) + Cl-(aq) → AgCl(s)
On the other hand, when Cl- ions acts as ligands in stable complexes, no AgCl precipitate is formed when treated with AgNO3 solution. This is because the dissociation of stable complexes is negligible and thus the amount of free Cl-(aq) ions is negligible.

Na2[CuCl4]: In aqueous solution, each mol of Na2[CuCl4] gives 2 mol of Na+(aq) ions and 1 mol of [CuCl4]2-(aq) ions. There are no free Cl-(aq) ions.

[CuCl2(NH3)2]: In aqueous solution, the neutral complex gives no free Cl-(aq) ions.

[Cu(NH3)4]Cl2: In aqueous solution, each mol of [Cu(NH3)4]Cl2 gives 1 mol of [Cu(NH3)4]2+(aq) ions and 2 mol of free Cl-(aq) ions.

[CuCl(NH3)3]Cl: In aqueous solution, each mol of [CuCl(NH3)3]Cl gives 1 mol of [CuCl(NH3)3]+ and 1 mol of free Cl-(aq) ions.

Hence, [Cu(NH3)4]Cl2 will gives the largest amount of AgCl solid. This is because each mole of the compound gives the largest amount of free Cl-(aq) ions.
=


收錄日期: 2021-04-26 12:54:21
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081220000051KK01424

檢視 Wayback Machine 備份