# What it the basic formula for a double displacement equation?

Aug 20, 2017

color(red)("AB")color(white)(l) +color(white)(l) color(blue)("CD")color(white)(l) rarr color(white)(l) color(red)("A")color(blue)("D")color(white)(l) + color(white)(l) color(blue)("C")color(red)("B"

#### Explanation:

In a double replacement (displacement) reaction, two reactants combine to form two new substances, via switching the cations of each substance:

Let's illustrate this with a sample reaction:

Suppose that sodium hydroxide and hydrochloric acid are mixed in aqueous solution (that is to say, in a solution where water is the solvent).

We have our reactants:

• $\textcolor{red}{\text{NaOH}}$

• color(blue)("HCl"

This will be a double replacement reaction, because there are two substances reacting that can both dissociate into their component ions in solution, which allows for the ions to move around and combine with another species.

Thus, we would have the reaction

$\textcolor{red}{\text{NaOH")(aq) + color(blue)("HCl")(aq) rarr color(red)("Na")color(blue)("Cl")(aq) + overbrace(color(red)("H")color(blue)("OH")(l))^("or H"_2"O, water}}$

Notice how the cations swapped compounds to form two new substances.

$- - - - - - - - - - - - - - - - - - - - - - - - - -$

Something to take note of when working with double displacement reactions is to figure out whether the reaction will even occur or not.

To figure this out, we must refer to a solubility chart like the one below:

In the above equation, we notated a substance that is dissolved in (aqueous) solution with an $\left(a q\right)$. We refer to the above solubility guidelines to figure out if a substance is considered soluble (i.e. has an $\left(a q\right)$) or insoluble (i.e. is just a solid, $\left(s\right)$), based on the ions present.

If all the reactants and products are soluble and are present as dissolved ions in aqueous solution, then *no reaction occurs* because all the ions are simply solvated by water molecules and there is no "drive" to carry out a reaction.

A great example of this necessary "drive" is the formation of a precipitate, which is an insoluble solid that forms in a double-replacement reaction.

Here is an example of a precipitation reaction:

color(red)("Pb(NO"_3")"_2) (aq) + 2color(blue)("KI")(aq) rarr 2color(blue)("K")color(red)("NO"_3)(aq) + overbrace(color(red)("Pb")color(blue)("I"_2)(s))^"solid precipitate"

We knew that ${\text{PbI}}_{2}$ was a solid because it is considered insoluble according to the solubility guidelines above.

$\text{ }$

Here is an example where no reaction occurs:

$\textcolor{red}{\text{NaCl")(aq) + color(blue)("KBr")(aq) rarr color(red)("Na")color(blue)("Br")(aq) + color(blue)("K")color(red)("Cl}} \left(a q\right)$

All the species here are soluble (according to the guidelines), and thus have an $\left(a q\right)$ after them. Since the resulting solution is simply dissociated ions in solution, no new products form and thus no reaction occurs.

$\text{ }$

We can even examine the first example reaction we used above. This is a special type of double replacement reaction called a neutralization reaction, in which an acid ($\text{HCl}$) and a base ("NaOH") (both exist as ions in solution) combine to form a salt and water (which is a liquid and thus is not present as dissolved ions).

Since not all the species are present as ions, a reaction indeed occurs. This is in fact true for all neutralization reactions.