What occurs when (i) sodium hydroxide is added to silver nitrate; and (ii) ammonia solution is added to the resultant precipitate?

Aug 18, 2015

You know that solubility rules follow a hierarchy, and that complex formation can occur with certain ligands. Here, (a) addition of $N {H}_{3}$ will give rise to a colourless solution; and (b) addition of $H C l$ will give rise to a precipitate.

Explanation:

All hydroxides are insoluble, save those of the alkali metals; all nitrates are soluble. The 1st reaction may therefore be represented as:

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

The down arrow signifies that the silver crashes out of solution as an hydroxide. The "silver hydroxide" is very poorly characterized. Addition of (a) $N {H}_{3}$ will form a complex ion with silver, which we may represent as:

$A g \left(O H\right) \left(s\right) + 2 N {H}_{3} \left(a q\right) \rightarrow$ $A g {\left(N {H}_{3}\right)}_{2}^{+}$(aq) + $O {H}^{-}$(aq)

Addition of $H C l \left(a q\right)$ will result in an acid base reaction with $A g \left(O H\right)$. The silver salt will go up momentarily, however, $A g C l$, a very insoluble species will crash out as a (momentarily) white solid, i.e.

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

On standing you will see the white solid darken as silver is reduced to form finely divided silver metal. This is more of an undergraduate than an A level treatment. If you're interested, what I have represented as $A g \left(O H\right)$ is probably a hydrous oxide of silver, i.e. $A {g}_{2} O \cdot {H}_{2} O$. No one really knows, because it is so insoluble. You will notice that the white precipitate darkens on standing as metallic silver precipitates.

How to remember all this? Do the reaction in the laboratory, carefully note what occurs, and confirm with your prof what is expected.