Categories of Chemical Reactions
Senior 2 · Chemistry
Categories of Chemical Reactions
Lesson 4 of 4
22 min
Hydrochloric acid neutralised by sodium hydroxide, sulfuric acid neutralised by potassium hydroxide, nitric acid neutralised by calcium hydroxide: three different reactions with three different salts. Yet all three release almost exactly the same energy, about 57 kJ, for every mole of water formed. Why? Because at the level of the particles, the same change is happening every time.
An ionic equation shows only the particles that actually change in a reaction. It strips away the ions that are just watching. By the end of this lesson you will be able to write ionic equations for precipitation, neutralisation and displacement reactions.
When an ionic compound dissolves in water, its ions separate and move about independently. We say it dissociates:
NaCl(aq) → Na+(aq) + Cl−(aq)
So in an equation, split into ions:
Do not split: solids (s), including precipitates; gases (g); water (l); or metals and other elements.
Write the ionic equation for the reaction between silver nitrate and sodium chloride solutions.
Step 1 AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Step 2 Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + NO3−(aq)
Step 3 Na+ and NO3− are the same on both sides: they are spectator ions.
Step 4 Ag+(aq) + Cl−(aq) → AgCl(s). Atoms: 1 Ag and 1 Cl on each side. Charge: (+1) + ( − 1) = 0 on the left; 0 on the right.
This net equation says that any solution containing silver ions will give a white precipitate with any solution containing chloride ions. That is the basis of the test for chloride ions.
Write the ionic equation for sulfuric acid reacting with sodium hydroxide.
Step 1 H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Step 2 2H+(aq) + SO42−(aq) + 2Na+(aq) + 2OH−(aq) → 2Na+(aq) + SO42−(aq) + 2H2O(l)
Step 3 Cancel Na+ and SO42−.
Step 4 2H+(aq) + 2OH−(aq) → 2H2O(l), which simplifies to H+(aq) + OH−(aq) → H2O(l).
Every reaction between an acid and an alkali has this same ionic equation. That is why they all release about the same energy per mole of water formed.
Zinc is added to copper(II) sulfate solution.
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Zn(s) + Cu2+(aq) + SO42−(aq) → Zn2+(aq) + SO42−(aq) + Cu(s)
The sulfate ion is the spectator. Net: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). Charge: +2 on each side.
Marble chips (calcium carbonate) react with dilute hydrochloric acid.
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
Calcium carbonate is a solid, so it is not split. Splitting the rest and cancelling the two Cl− ions leaves:
CaCO3(s) + 2H+(aq) → Ca2+(aq) + H2O(l) + CO2(g)
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) gives the net ionic equation Ba2+(aq) + SO42−(aq) → BaSO4(s). This is the basis of the test for sulfate ions.
Tip
Only (aq) ionic substances split. Solids, liquids (water) and gases stay whole. Before you finish, do the charge check: add up the charges on each side; if the totals differ, you have made a mistake.
Try each item before you read the solution.
Write the net ionic equation for KOH(aq) + HNO3(aq) → KNO3(aq) + H2O(l).
Solution: Split: K+(aq) + OH−(aq) + H+(aq) + NO3−(aq) → K+(aq) + NO3−(aq) + H2O(l). Cancel K+ and NO3−: H+(aq) + OH−(aq) → H2O(l).
Write the net ionic equation for Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g).
Solution: The chloride ions are spectators. Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g). Charge: +2 on each side.
Sodium hydroxide solution is added to copper(II) sulfate solution, and a pale blue precipitate of copper(II) hydroxide forms. Write the full and ionic equations and name the spectator ions.
Solution: CuSO4(aq) + 2NaOH(aq) → Cu(OH)2(s) + Na2SO4(aq). Net: Cu2+(aq) + 2OH−(aq) → Cu(OH)2(s). The spectator ions are Na+ and SO42−.
Write the ionic equation for sodium carbonate solution reacting with dilute hydrochloric acid.
Solution: Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g). Cancel Na+ and Cl−: CO32−(aq) + 2H+(aq) → H2O(l) + CO2(g).
Summary
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