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

Sep 8, 2016

$2 A g N {O}_{3} \left(a q\right) + C u C {l}_{2} \left(a q\right) \rightarrow C u {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 A g C l \left(s\right) \downarrow$

#### 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:

$P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) + C u C {l}_{2} \left(a q\right) \rightarrow C u {\left(N {O}_{3}\right)}_{2} \left(a q\right) + P b C {l}_{2} \left(s\right) \downarrow$

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