How does metathesis occur between silver(I) nitrate, and copper(II) choride in aqueous solution?

Question

How does metathesis occur between silver(I) nitrate, and copper(II) choride in aqueous solution?

1 Answer

Impact of this question

2248 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-08T12:58:59

$2 A g N O_{3} (a q) + C u C l_{2} (a q) \to C u {(N O_{3})}_{2} (a q) + 2 A g C l (s) ⏐ ↓$ $2AgNO_3(aq) +CuCl_2(aq) rarr Cu(NO_3)_2(aq) + 2AgCl(s)darr$

Explanation:

The silver halide precipitates as a curdy white solid. Copper nitrate remains in solution. This is an expensive way to make copper salts.

A better means might be via the lead salt:

$P b {(N O_{3})}_{2} (a q) + C u C l_{2} (a q) \to C u {(N O_{3})}_{2} (a q) + P b C l_{2} (s) ⏐ ↓$ $Pb(NO_3)_2(aq) +CuCl_2(aq) rarr Cu(NO_3)_2(aq) + PbCl_2(s)darr$

At least this way you don't get silver salts, which have a tendency to reduce. $P b C l_{2}$ $PbCl_2$ is fairly insoluble in cold water, and non-redox active.

Science

Math

Humanities

... and beyond

How does metathesis occur between silver(I) nitrate, and copper(II) choride in aqueous solution?

1 Answer

Explanation:

Related questions

Impact of this question