Synoptic Questions

Question 1

• State why the equilibrium constant (Kp) for this reaction has no units.

Answer 1

•The amount in moles of products is the same as reactants
• So the units of partial pressure cancel out

Question 2

• A student plans to titrate butanoic acid solution with a solution of ethylamine.
• Explain why this titration could not be done using an indicator.

Answer 2

• This is a weak acid and weak base / alkali titration
• pH change is too gradual / not sharp at the equivalence point so colour change of indicator is difficult to judge

Question 3

• When sodium hydroxide solution is added to aqueous propanoic acid, the solution formed acts as a buffer when between 5 cm? and 15 cmg have been added.
• Explain why the pH stays approximately constant during this part of the experiment.

Answer 3

• OH- reacts with propanoic acid / H+
• Equilibrium for dissociation of CH3CH2COOH moves right to maintain [H+] / replace H+

Question 4

• Methyl orange and universal indicator are not suitable indicators for the titration of solutions of propanoic acid with sodium hydroxide.
• State the reason why each indicator is not suitable.

Answer 4

• Methyl orange - would not change colour at the equivalence point
• Universal indicator - there is a range of colours during the titration / no distinct colour change at equivalence / end point

Question 5

Answer 5

• All have a colour change / pH range within the steep / vertical part of the titration curve

Question 6

Answer 6

• Use a larger mass of NH4NO3
• So temperature change / decrease is greater

Question 7

Answer 7

• Heat gain from the surroundings / incomplete dissolving

Question 8

Answer 8

• 22.45
• The [OH-] is the same at both temperatures so the same number of moles

Question 9

Answer 9

• How buffer solution is made: Add excess ethanoic acid to KOH
• Or add enough KOH to the ethanoic acid so that the acid is partially neutralised
• Or add enough KOH so that the acid contains a mixture of ethanoic acid and ethanoate ions
• Equation: KOH + CH3COOH -> CH3COOK + H2O
• How buffer solution resists pH change: CH3COO- from salt reacts with acid / H+

Question 10

Answer 10

• 22.45
• Same [OH-]

Question 11

Answer 11

• Mechanism to form alkenes: elimination
• Mechanism to form alcohols: nucleophilic substitution

Question 12

Answer 12

• A smaller titre will increase the percentage uncertainty

Question 13

Answer 13

• Extra step: weigh the bottle after oil transfer and record the mass
• Justification: Not all of the oil is transferred / so that the mass of oil left in the bottle is accounted for

Question 14

• Define the mass number of an atom

Answer 14

• The mass number of an atom is equal to the number of protons and neutrons in the nucleus combined

Question 15

Answer 15

Question 16

• By reference to the structure of, and the bonding in, silicon dioxide, suggest why it is insoluble in water (3 marks)

Answer 16

• It has a macromolecular structure
• With lots of covalent bonds which are very strong
• Water cannot supply enough energy to break the strong covalent bonds

Question 17

Answer 17

• As there is a large pH change for a small addition of alkali

Question 18

Answer 18

• Mg

Question 19

Answer 19

• Equation: Q + H+ → QH+

Question 20

• Explain the shape of a BrF3 molecule

Answer 20

• Lone pairs / bond pairs / electron pairs repel each other to be as far apart as possible
• Lone pairs / bond pairs / electron pairs are arranged to be as far apart as possible to minimise repulsion

Question 21

Answer 21

• Standard pressure/100 kPa AND 298K

Question 22

• Draw the displayed formula for the anion formed when sulfur trioxide reacts with water

Question 23

Answer 23

Question 24

Answer 24

Question 25

Answer 25

• Reaction 1 could give rise to further substitution • Reaction 1 could give rise to substitution in different positions • Combination of radicals to get longer chain product

Question 26

• Explain why the enthalpy of lattice dissociation of potassium oxide is less endothermic than that of sodium oxide.

Answer 26

• As the K+ ion is larger than that of Na+ the attraction between the oppositely charged ions is weaker and therefore easier to break