# Why must the silver nitrate test for chloride ions be done in an acidified solution?

In order to determine the presence of the halide, competing equilibria must be taken into account. Silver forms an insoluble hydroxide, which is probably a hydrous oxide. When $p H$ is lowered from $7$, this competing equilibrium is removed, any chloride will precipitate as a curdy white solid as $A g C l$, as a lemon-yellow solid as $A g B r$, and as a bright yellow solid as $A g I$.
The acid of choice would be $H N {O}_{3} \left(a q\right)$. Why do we use this acid and not say $H C l \left(a q\right)$ or $H B r \left(a q\right)$?