Chemistry iGCSE Section D



Opposite of an acid is a base. Soluble bases are alkalis.

                0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

pH scale             ACID  -  ALKALINE

litmus: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

methyl orange: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

phenolphthalein: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

universal indicator: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

UI has two forms: one is a solution in ethanol and the other is paper that has been soaked in indicator solution and dried. To test: dissolve unknown pH in water and put it onto pH paper.

1 of 11

Reactions of acids

Acid: a substance that dissolves in water to produce hydrogen ions (H+)

Only metals above hydrogen in the reactivity series will react with dilute acid.

Reacting dilute acids with metals:
Dilute HCL and H2SO4 react with metals to form a salt and hydrogen.
   E.g. Mg(s)+2HCl(aq) --> MgCl2(aq)+H2(g)

Reacting dilute acids with bases:
All metal oxides and metal hydroxides can act as bases. Bases neutralise acids to form a salt and water.
   E.g. MgO(s)+2HCl(aq) --> MgCl2(aq)+H2O(l)

Reacting dilute acids with metal carbonates:
Dilute acids react with metal carbonates to form salts, carbon dioxide and water.
   E.g. MgCO3(s)+2HCl(aq) --> MgCl2(aq)+CO2(g)+H2O(l)

2 of 11

Solubility of salts

SOLUBLE SALTS:                                                                      
all common sodium, potassium and ammonium salts
all nitrates
all common chlorides (except silver and lead(II) chloride)
all common sulphates (except barium, calcium and lead(II) sulphate)
sodium, potassium and ammonium carbonates (all other carbonates aren't)

Preparation of soluble salts:

  • dilute acid+base (i.e. metal oxide or metal hydroxide)
  • dilute acid+metal carbonate
  • dilute acid+metal

The ONLY common bases that are soluble in water are sodium hydroxide and potassium hydroxide.

3 of 11

Preparation of soluble salts

Dilute acid+base

1. Acid + insoluble base  (same for dilute acid + metal carbonate/ metal)

  • Heat some dilute acid using a bunsen burner. DO NOT BOIL.
  • Add the insoluble base a little at a time. Stir until the base stops dissapearing.
  • Filter into an evaporating basin
  • Leave the filtrate in a warm place to allow crystals to form
  • Remove the crystals and dry them with filter paper.

2. Acid + soluble base (alkali)

  • Put alkaline solution in a conical flask and add an indicator.
  • Add dilute acid from a burette until the indicator changes colour
  • Add powdered charcoal and shake the mixture to remove the indicator
  • Filter to remove the charcoal and obtain the crystals from the filtrate.
4 of 11

Preparation of insoluble salts

Precipitation reactions
Precipitate: an insoluble solid made by a chemical reaction in an aqueous solution

Mix together 2 aq solutions. 1 solution should have + ion, the other - ion.
E.g. to make silver chloride, mix together silver nitrate solution and sodium chloride solution.
The silver nitrate has Ag+ and the chloride, Cl-. The precipitate is removed by filtration and washed with distilled water and left to dry. Ag+(aq)+Cl-(aq) --> AgCl(s). The ionic equation can be used for any solution containing silver and chloride ions.  

Making ammonium salts
Ammonia reacting with an acid. These are neutralisation reactions. Use an indicator to see when the raction is complete.Use a burette to add ammonia solution slowly to the acid until the indicator (litmus) changes colour. Then repeat without the indicator, using the same volumes of solution. Crystallise in the the same way as usual.

5 of 11

Titration - Acd-alkali

Titration: finding out the exact volume of one solution needed to react with a different solution. (Usually to find the exact volume of an acid with a given volume of an alkaili.)


  • using a pipette put 25.0cm3 of the alkali solution into a conical flask.
  • add a few drops of indicator
  • put the acid into a burette and and note the initial reading
  • add the acid until the alkali changes colour
  • note the final reading of acid in the burette
  • subtract the initial reading from the final reading to find the amount of acid added.
6 of 11

Exothermic and endothermic reactions

exothermic: energy is given out  +∆H

endothermic: energy is taken in  -∆H

heat given out = mass of solution X specific heat capacity of solution X temp rise

E.g. heat capacity: 4.2J, Temp rise: 10.0°C,  moles of acid: 0.005, mass of solution:50g
heat given out = 4.2 X 10 X 50 = 2100J = 2.10kJ
0.005 mol of acid produce 2.10kJ so 1 mol of acid produces 2.10/0.005=420kJ
∆H= -420kJ/mol

For reactions at a constant pressure, the heat change is known as enthalpy change
Enthalpy change per mole is called molar enthalpy change,
Assumptions to make: heat capacity at end is same as water, 4.2J/g/°C, the density of the final solution is 1g/dm3 and there is little heat loss due to surroundings etc.

7 of 11

Heat change during combustion reactions

  • put 100cm3 of water into a copper can
  • measure and record the initial temp of the water
  • fill the spirit burner with alcohol and measure and record its mass
  • place the burner under the copper can and light the wick
  • stir the water constantly until temp rises to 20-30°C
  • measure and record highest temp and the final mass of burner and remaining alcohol

Calculate: the rise of temp of water and the mass of alcohol burnt, the molar enthalpy of reaction and the heat given out.

Calculating bond energy (kJ/mol)
Step 1: calculate the sum of energies for the bonds broken
Step 2: calculate the sum of energies for the bonds made
Step 3: calculate ∆H using the formula ∆H = bonds broken - bonds made

C-H = 412  H-H = 436  O=O = 496 H-Cl = 432  Cl-Cl = 242  C=O = 743  O-H = 463

8 of 11

Rates of reaction

rate of reaction = change of concentration of reactant / time

Collision theory
For a reaction to take place, the reactants must collide with one another and they must have enough energy (activation energy).

Factors that affect the rate of a reaction
Concentration - increases the number of particles of reactant in a given volume so there will be more collisions per second.
Temperature - Increases kinetic energy meaning they have more activation energy.
State of division of a solid reactant - The smaller pieces have a larger surface area and more particles are exposed to the other reactant.

Increases rate of a chemical reaction but is chemically unchanged. They provide an alternative pathway with a lower activation energy so more reactions will take place with lower activation energy. They allow lower pressures/temps to be used.

9 of 11


Heating copper(II) sulfate crystals CuSO4.5H2O

Copper(II) sulfate in the crystals is known as hydrated copper(II) sulfate because it contains water of crystallisation. When copper(II) sulfate is heated it turns into anhydrous copper(II) sulfate.

CuSO4.5H2O(s) blue --> CuSO4(s) white + 5H2O(l)
When water is added to anhydrous copper(II) sulfate, it changes from white to blue.

dynamic equilibrium

  • the rate of the forward reaction - the rate of the backwards reaction
  • the amounts of reactants and proucts remain constant
  • colour or pressure remains constant
10 of 11

Controlling the equilibrium

The quoted H in a reversible reaction is always for the forward reaction (exothermic). By changing the pressure and temperature it is possible to shift the position of equilibrium.

Increase in pressure: shifts to the direction that makes the smaller number of gas molecules
Decrease in pressure: shifts in the direction that makes the larger number of gas molecules gas

Increase in temp: moves in endothermic direction
Decrease in temp: moves in exothermic direction

3H2(g)+N2 <--> 2NH3
(4 molecules) (2 molecules) - exothermic forward reaction

It shifts to the right when pressure is increased and temp is decreased
It shifts to the left when pressure is decreased and temp is increased

11 of 11


No comments have yet been made

Similar Chemistry resources:

See all Chemistry resources »