Chemical Changes

Question 1

What ions do acids and alkalis produce in aqueous solutions?

Answer 1

Acids produce H+ ions
Alkalis produce OH- ions

Question 2

How do acids and alkalis affect indicators like litmus methyl orange and phenolphthalein?

Answer 2

Litmus
Litmus solution:
Alkaline = blue
Acidic = red
Litmus paper:
Blue litmus paper goes red in acidic & stays blue in alkaline.
Red litmus paper goes blue in alkaline & stays red in acidic.
Methyl orange
Alkaline = yellow
Acidic = red
Phenolphthalein
Alkaline = pink
Acidic = colourless

Question 3

CORE PRACTICAL - Investigate the change in pH on adding powdered calcium hydroxide or calcium oxide to a fixed volume of dilute hydrochloric acid

Answer 3

Method:
Add dilute HCl to the beaker and measure pH. Add weighed mass of calcium hydroxide and stir then record pH. Keep adding weighed masses of calcium hydroxide until there is no more change to the pH.
Analysis:
Draw a line graph with mass added on the horizontal axis and with pH on the vertical axis. Draw a line of best fit (remember to ignore any anomalies).

Question 4

Explain the term dilute

Answer 4

Lesser amount of substance in a given volume of a solution.

Question 5

Explain the term concentrated

Answer 5

Larger amount of substance in a given volume of a solution.

Question 6

What is the difference between weak and strong acids and name some examples?

Answer 6

Strong acids fully dissociate in aqueous solutions, releasing more H+ ions while weak acids only partially dissociate.
Strong acids = hydrochloric, nitric and sulfuric acids
Weak acids = ethanoic, citric and carbonic acids

Question 7

What is the ionic equation for a neutralization reaction between an acids and an alkali?

Answer 7

H + (aq) + OH- (aq) -> H2O(l)

Question 8

What are the 5 neutralisation equations?

Answer 8

acid + metal → salt + hydrogen gas
acid + metal oxide → salt + water
acid + metal hydroxide → salt + water
acid + metal carbonate → salt + water + carbon dioxide
ammonia + salt → ammonium salt

Question 9

Describe the chemical test for hydrogen

Answer 9

Use a burning splint held at the open end of a test tube of the gas. Creates a ‘squeaky pop’ sound.

Question 10

Describe the chemical test for carbon dioxide

Answer 10

Bubble the gas through the limewater (calcium hydroxide solution) and it will turn milky (cloudy).

Question 11

Explain what happens if soluble salts are prepared from an acid and an insoluble reactant?

Answer 11

Excess of the reactant is added: This is to ensure your volume of acid reacts completely.

Excess reactant is removed: This is done by filtration of the insoluble reactant and is done so that you are left with just a salt and water.

The remaining solution is only salt and water: This is because all your acid has fully reacted and you have filtered off your other reactant, and that the only products of your reaction are a salt and water. If you have used a carbonate, you would still only have a salt and water remaining as carbon dioxide gas would have been given off into the atmosphere.

Question 12

Explain what if soluble salts are prepared from an acid and a soluble reactant?

Answer 12

Titration must be used: Both reactants are liquids/soluble, so if you have an excess of one you would not be able to easily remove it from your mixture of products, this means you need to measure the exact amount of volumes that react, which is easily done using a titration. You can then mix the exact proportions of the two reactants.
The exact amount of acid has thus been added to the soluble reactant, meaning that the leftover solution is only salt and water, no acid or alkali, because they have been completely neutralised.

Question 13

CORE PRACTICAL - Investigate the preparation of pure, dry hydrated copper sulfate crystals

Answer 13

Add an excess of copper oxide (insoluble) to your acid (sulfuric acid H2SO4 - as you are making copper SULFATE). Use a filter and filter paper to filter off any copper oxide that hasn’t reacted (your solution should be blue as copper sulfate solution has been formed). Evaporate off the water by placing your final solution in a water bath.

Question 14

How to carry out an acid-alkali titration to prepare a pure, dry salt? (9)

Answer 14

1. Wash burette using the acid and then water.
2. Fill burette to 100cm3 with acid with the meniscus’ base on the 100cm3 line.
3. Use 25cm3 pipette to add 25cm3 of alkali into a conical flask, drawing alkali into the pipette using a pipette filler.
4. Add a few drops of a suitable indicator to the conical flask (e.g.: phenolphthalein which is pink when alkaline and colourless when acidic).
5. Add acid from burette to alkali until endpoint is reached (as shown by indicator.
6. The titre (volume of alkali needed to exactly neutralise the acid) is the difference between the first (100cm3) and second readings on the burette).
7. Repeat the experiment to gain more precise results 8. To prepare a pure, dry salt – you warm the salt solution to evaporate the water.
8. Crystals form.

Question 15

What is the solubility of common types of substances in water?

Answer 15

Sodium - all soluble
Potassium - all soluble
Ammonium - all soluble
Nitrates - all soluble
Chlorides - all soluble except… Silver, Lead
Sulfates – all soluble except… Lead, Barium, Calcium
Carbonates - only Sodium, Potassium, Ammonium soluble
Hydroxides – only Sodium, Potassium, Ammonium soluble

Question 16

Predict, using solubility rules, whether or not a precipitate will be formed when named solutions are mixed together

Answer 16

First, work out what the products of your reaction will be. Then, determine if any salts formed are soluble/insoluble. Any INSOLUBLE salts will form as a precipitate (as any soluble salts will remain in solution).

Question 17

What is the method used to prepare a pure, dry sample of an insoluble salt? (4)

Answer 17

1. Mix the two solutions needed to form the salt
2. Filter the mixture using filter paper, which the insoluble salt will be left on
3. Wash the salt using distilled water
4. Leave the salt to dry on filter paper (water will evaporate, speed this process up by drying it in an oven).

Question 18

Describe the term electrolysis

Answer 18

The breaking down of an ionic compound using electrical energy.

Question 19

Describe the term electrolyte

Answer 19

An ionic substance is melted or dissolved, the ions are free to move about within the liquid or solution.

Question 20

What are the movements of ions during electrolysis?

Answer 20

During electrolysis, positively charged ions (cations) move to the negative electrode (cathode), and negatively charged ions (anions) move to the positive electrode (anode).Ions are discharged at the electrodes producing elements.

Question 21

Explain the formation of the products in the electrolysis

Answer 21

When you have a ionic solution (NOT a molten ionic compound), your solution will contain: the ions that make up the ionic compound, and the ions in water (OH- and H + ).
At the cathode (-):
Hydrogen is produced UNLESS the + ions in the ionic compound are from a metal less reactive than hydrogen. If the metal is less reactive, it will be produced instead.
At the anode (+):
Oxygen will be produced UNLESS the ionic compound contains halide ions (Cl- , Br- , I - ). If there are halide ions, the halogen will be produced instead (e.g. Cl2 ).

Question 22

What is oxidation and reduction?

Answer 22

Oxidation - loss of electron, gain of oxygen
Reduction - gain of electrons, loss of oxygen

Question 23

How can electrolysis be used to purify Copper from Copper sulfate?

Answer 23

Set up: Anode is made of impure copper (that you are purifying). Cathode is made of pure copper. The solution is copper sulfate.
What happens: Cu2+ ions from the anode move to the cathode, where they gain electrons and are discharged as pure copper. Impurities form as sludge below the anode.
The cathode will increase in mass as it gains pure copper, whilst the anode will lose mass as copper ions are lost (they replace the ones from the CuSO4 solution that go to the cathode) and so are impurities.

Question 24

CORE PRACTICAL - Investigate the electrolysis of copper sulfate solution with inert electrodes and copper electrodes

Answer 24

With inert electrodes:
At the cathode Cu (s) is produced (Cu is less reactive than hydrogen). At the anode O2 is produced (SO4 2- ions are not halide ions). This leaves H + and SO4 2- ions in the solution, which will react to form H2SO4 - sulfuric acid.
With Copper electrodes:
The Cu2+ ions deposited as Cu at the cathode from the solution are replaced by Cu2+ ions from the anode, meaning the concentration of Cu2+ ions in the solution remains constant.