Question

How do ions behave in solution?

Answer 1

: We know that some ions example halide ion, copper ion, ammonium ion and zinc ion, we can easily test them in solutions how they behave.
Also Halide ions will always react with the silver nitrate $AgN{{O}_{3}}$ to give a colored precipitation of silver halide. For example whenever chloride ion will react with the $AgN{{O}_{3}}$ to give byproduct as silver chloride $AgCl$ which is insoluble in nature with white precipitate.
Whereas copper ions are usually in their $+2$ electronic state and are with blue color solution. In order to test them for a simple basification process we have to add some sodium hydroxide as well as a blue precipitate in form of $copper\left( II \right)\text{ }hydroxide~Cu{{\left( OH \right)}_{2}}$ will generate.

Complete answer:
Firstly whenever ions are present in solution they need to be pulled apart by each other and then by dissolving properties of water we acquire their current state.
When the salt $NaCl$ in solution dissolves in water the ions of sodium and chloride are pulled away from each other instantly.
$NaCl\left( s \right)\text{ }\to \text{ }NaCl\left( aq \right)$
OR
$NaCl\left( s \right)\text{ }\to ~N{{a}^{+}}\left( aq \right)~+~C{{l}^{-}}\left( aq \right)$
Here takes place the double reaction and which can be defined as; a reaction which occurs when the cation and anion of given two ionic compounds are exchanged. The diagram below illustrates how this swap takes place.
Whenever the chemical formula of each ionic compound is written correctly we can balance the equations like any other chemical equation by default making sure the number of atoms for each of the elements is the same from left and right.
Then after we acquire the simplified way of balancing a double replacement reaction you can balance A, B, C and D which in meaning is handling the cation and anion as a group rather than that as an individual element.

Note: Note that the one reason for which this concept is important that is for many reaction which take place are been in between dissolve ion, for example the double replacement reaction which has production of precipitates which has the solubility and are quantitative that can be determined by experiment.
For the quantitative treatment of solubility we need to consider the reaction.