4.2 Double Displacement Reactions

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Video Transcript
In this lesson you will learn:

-Double Displacement Reactions (a.k.a. Exchange, Metathesis, or Double Replacement Reactions)

-How to predict the products and balance double displacement reactions

-How to determine complete ionic and net ionic equations

-The three types of Double Displacement Reactions

1. Precipitation Reactions

2. Gas-forming Reactions

3. Acid-Base Neutralization Reactions


Double Displacement Reactions

Double displacement reactions are also commonly referred to as double replacement reactions, exchange reactions, or metathesis reactions.  Typically two ionic compounds (may also involve acids) are mixed.  The cations and anions trade partners resulting in two new ionic compounds (or possibly an acid).

 

AB + CD → AD + CB

1. Precipitation Reactions

A precipitation reaction is one which produces an insoluble (solid) ionic compound.  The solid ionic compound is referred to as a precipitate.  When a double displacement reaction occurs that results in a precipitate, the solution typically becomes cloudy, but the precipitate often settles out of solution over time.  Consider the following reaction between aqueous lead (II) nitrate and aqueous sodium iodide:

 

Pb(NO3)2(aq) + NaI(aq) → ?

 

The lead (II) ions will combine with the iodide ions and the sodium ions will combine with the iodide ions.  As the lead (II) ions have a +2 charge and the iodide ions are -1, the proper formula for lead (II) iodide is PbI2.  As sodium ions have a +1 charge and nitrate ions a -1 charge, the proper formula for sodium nitrate is NaNO3.

 

Pb(NO3)2(aq) + NaI(aq) → PbI2(s) + NaNO3(aq)

 

A quick look at the solubility rules shows that all nitrate salts are soluble and so the NaNO3 is aqueous.  And while the solubility rules show that most iodides are soluble, Pb2+ is an exception, and therefore PbI2 is insoluble and is a solid.  The formulas for the products are now correct and the solubility rules have been used to determine their phases, but the reaction has yet to be balanced.

How to Write the Net Ionic Equation

Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq)     (Molecular Equation)

 

But strong electrolytes exist fully dissociated in solution and we often re-write the chemical equation to reflect this in what is called the Ionic Equation.  Recall that strong electrolytes include soluble ionic compounds and strong acids/bases.  Pb(NO3)2, NaI, and NaNO3 are all soluble ionic compounds and therefore strong electrolytes and will be split into cations and anions when writing the Ionic Equation.  But PbI2 is an insoluble ionic compound and not a strong electrolyte and will not be split up in the Ionic Equation.

 

Pb2+(aq) + 2NO3-(aq)+ 2Na+(aq) + 2I-(aq) → PbI2(s) + 2Na+(aq) + 2NO3-(aq)    (Ionic Equation)

 

From the Ionic Equation it can be seen that Na+ and NO3- are free ions on both the reactants' and products' side of the reaction.  This means they didn't participate in the reaction at all and are referred to as 'spectator ions.'  These are eliminated to re-write the chemical equation one last time in what is referred to as the Net Ionic Equation:

 

Pb2+(aq) + 2I-(aq) → PbI2(s)     (Net Ionic Equation)

2. Gas-Forming Reactions

The are a couple of examples of double displacement reactions that form gases as products, namely CO2 and SO2.  The are a lot of other reactions that form gases, these are just the two double displacement reactions that form gases.  It turns out that carbonic acid (H2CO3) and sulfurous acid (H2SO3) are somewhat unstable and will spontaneously decompose to form CO2 / water and SO2 / water respectively.

 

H2CO3(aq) → CO2(g) + H2O(l)

 

H2SO3(aq) → SO2(g) + H2O(l)

 

But you are much more likely to encounter carbonic acid and CO2(g) formation and I'll cover an example of it here.  Ultimately, when writing the products of a double displacement reaction, if a product appears to be H2CO3(aq), you should re-write the products to include CO2(g)+ H2O(l) instead.  Take for example the reaction between aqueous hydrochloric acid and aqueous sodium carbonate:

 

HCl(aq) + Na2CO3(aq) → ?

 

The H+ ions will combine with the carbonate ions and the sodium ions will combine with the chloride ions.  As the H+ ions have a +1 charge and the carbonate ions are -2, the proper formula for hydrogen carbonate (a.k.a. carbonic acid is H2CO3.  As sodium ions have a +1 charge and chloride ions a -1 charge, the proper formula for sodium chloride is NaCl.

 

HCl(aq) + Na2CO3(aq) → H2CO3(aq) + NaCl(aq)

 

A quick look at the solubility rules shows that most chloride salts are soluble and sodium is no exception so the NaCl is aqueous. But the carbonic acid should be re-written as CO2(g) + H2O(l), but I'd recommend balancing the equation before substituting this in.

How to Write the Net Ionic Equation

2HCl(aq) + Na2CO3(aq) → H2CO3(aq) + 2NaCl(aq)

 

And now that the reaction is balanced we will replace the H2CO3(aq) with CO2(g) + H2O(l).

 

2HCl(aq) + Na2CO3(aq) → CO2(g) + H2O(l) + 2NaCl(aq)     (Molecular Equation)

 

But again, strong electrolytes exist fully dissociated in solution and we often re-write the chemical equation to reflect this in the Ionic Equation.  Strong electrolytes include soluble ionic compounds and strong acids/bases.  HCl is a strong acid, and Na2CO3 and NaCl are soluble ionic compounds according to the solubility rules.  But CO2(g) and H2O(l) are molecular compounds and are not strong electrolytes and will not be split up in the Ionic Equation.

 

2H+(aq) + 2Cl-(aq)+ 2Na+(aq) + CO32-(aq) → CO2(g) + H2O(l) + 2Na+(aq) + 2Cl-(aq)    (Ionic Equation)

 

From the Ionic Equation it can be seen that Na+ and Cl- are free ions on both the reactants' and products' side of the reaction.  This means they didn't participate in the reaction at all and are referred to as 'spectator ions.'  These are eliminated to re-write the chemical equation one last time in the Net Ionic Equation:

 

2H+(aq) + CO32-(aq) → CO2(g) + H2O(l)     (Net Ionic Equation)

3. Acid-Base Neutralization Reactions

While not all acid-base neutralization reactions can be considered double displacement reactions and won't fit this pattern, but the ones we consider will involve an acid and a base reacting to form water and a salt.  The most common examples is the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH):

 

HCl(aq) + NaOH(aq) → ?

 

The H+ ions will combine with the hydroxide ions to form water and the sodium ions will combine with the chloride ions to form NaCl.

 

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

 

A quick look at the solubility rules shows that most chloride salts are soluble and Na+ is no exception.  The formulas for the products are now correct and the solubility rules have been used to determine their phases and it just so happens that this equation is already balanced.

How to Write the Net Ionic Equation

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)     (Molecular Equation)

 

But again, strong electrolytes exist fully dissociated in solution and we often re-write the chemical equation to reflect this in the Ionic Equation.  Strong electrolytes include soluble ionic compounds and strong acids/bases.  HCl is a strong acid, and Na2CO3 and NaCl are soluble ionic compounds according to the solubility rules.  But CO2(g) and H2O(l) are molecular compounds and are not strong electrolytes and will not be split up in the Ionic Equation.

 

H+(aq) + Cl-(aq)+ Na+(aq) + OH-(aq) → H2O(l) + Na+(aq) + Cl-(aq)    (Ionic Equation)

 

From the Ionic Equation it can be seen that Na+ and Cl- are free ions on both the reactants' and products' side of the reaction.  This means they didn't participate in the reaction at all and are referred to as 'spectator ions.'  These are eliminated to re-write the chemical equation one last time in the Net Ionic Equation:

 

H+(aq) + OH-(aq) → H2O(l)     (Net Ionic Equation)

Acid-Base Neutralization with a Weak Acid

The last acid-base neutralization involved both a strong acid and a strong base which reacted in a 1:1 ratio.  Both of these made dealing with it a little easier.  So I also wanted to include a more challenging example here involving a weak acid.  In this example we'll look at the reaction of phosphoric acid (H3PO4) with sodium hydroxide (NaOH):

 

H3PO4(aq) + NaOH(aq) → ?

 

The H+ ions will combine with the hydroxide ions to form water and the sodium ions which have a +1 charge will combine with the phosphate ions which have a -3 charge to form sodium phosphate (Na3PO4).

 

H3PO4(aq) + NaOH(aq) → HOH(l) + Na3PO4(aq)

 

A quick look at the solubility rules shows that most phosphate salts are insoluble but Na+ is an exception, and therefore Na3PO4 is soluble and in the aqueous phase.  You'll notice that I've written water has hydrogen hydroxide with the hydrogen ion red and the hydroxide ion blue.  This will make the equation easier to balance to keep track of the hydrogen ions that come from the acid and the hydroxide ions that come from the base.

 

H3PO4(aq) + 3NaOH(aq) → 3HOH(l) + Na3PO4(aq)

 

And now that it is balanced we can re-write it with water written as H2O.

 

H3PO4(aq) + 3NaOH(aq) → 3H2O(l) + Na3PO4(aq)

How to Write the Net Ionic Equation

H3PO4(aq) + 3NaOH(aq) → 3H2O(l) + Na3PO4(aq)     (Molecular Equation)

 

But again, strong electrolytes exist fully dissociated in solution and we often re-write the chemical equation to reflect this in the Ionic Equation.  Strong electrolytes include soluble ionic compounds and strong acids/bases.  NaOH is a strong base, and Na3PO4 is a soluble ionic compound according to the solubility rules.  But H3PO4(aq) is a weak acid, and H2O(l) is a molecular compound and neither is a strong electrolyte and therefore will not be split up in the Ionic Equation.

 

H3PO4(aq) + 3Na+(aq) + 3OH-(aq) → 3H2O(l) + 3Na+(aq) + PO43-(aq)    (Ionic Equation)

 

From the Ionic Equation it can be seen that only Na+ exists as a free ion on both the reactants' and products' side of the reaction.  It is therefore the only spectator ion and the only one eliminated in the Net Ionic Equation:

 

H3PO4(aq) + 3OH-(aq) → 3H2O(l) + PO43-(aq)     (Net Ionic Equation)