3

Question 1

what is pH

Answer 1

pH is a measure of how acidic or alkaline a solution is. The pH scale goes from 0 to 14 (see Figure 1).
* Anything that forms a solution with a pH of less than 7 is an acid. The lower the pH, the more acidic the solution.
* Anything that forms a solution with a pH of greater than 7 is an alkali The higher the pH, the more alkaline the substance is.
* Neutral substances are neither acidic nor alkaline and have a pH of exactly 7. Pure water is an example of a neutral substance.

Question 2

what is a base

Answer 2

A base is a substance that reacts with an acid to produce a salt and water.

Question 3

what is an Alkali

Answer 3

An alkali is a base that is soluble in water.

Question 4

Acids and bases in solution

Answer 4

When a substance is dissolved in water, the pH of the solution depends on the type of ions that are released by the substance.
* Acids release hydrogen ions (H+) when they are in an aqueous solution.
* Alkalis form OH- ions (otherwise known as hydroxide ions) in water.

Question 5

Concentrations of ions

Answer 5

The higher the concentration of hydrogen ions in a solution, the more acidic it is, so the lower its pH will be. So, as the concentration of hydrogen ions increases, the pH decreases. In alkaline solutions, the higher the concentration of OH- ions, the higher the pH.

Question 6

Measuring the pH of a solution

Answer 6

You can measure the pH of a solution using an indicator. An indicator is a dye that changes colour depending on whether it’s above or below a certain ph.
Indicators are simple to use - add a few drops to the solution you’re testing, then compare the colour the solution goes to a pH chart for that indicator. For example, Universal indicator gives the colours shown in Figure 1.
they can be used either in solution or the solution can be dried on paper to make test papers.

Question 7

Phenolphthalein

Answer 7

• -colourless in acids
• -colourless in neutral Solutions
• -pink in alkalis

Question 8

methyl orange

Answer 8

• -pink in acids
• -orange in neutral Solutions
• -yellow in alkalis

Question 9

universal indicator

Answer 9

universal indicator is a mixture of several different indicators. Universal indicator can indicate the strength of the acid or alkali

Question 10

Red litmus paper

Answer 10

Red litmus paper turns blue when alkalise are added. it’s used to test for alkalis.

Question 11

blue litmus paper

Answer 11

blue litmus paper turns red when acids are added. it is used to test for acids

Question 12

What are neutralisation reactions?

Answer 12

An acid will react with a base to form a salt and water - this is called a neutralisation reaction. The general equation for a neutralisation reaction is shown below.
acid + base → salt + water

Question 13

ionic equation for Neutralisation reactions

Answer 13

Neutralisation reactions between acids and bases can also be shown as an ionic equation in terms of H+ and OH- ions. During neutralisation reactions, hydrogen ions (H+) from the acid react with hydroxide ions (OH) from the base to produce water. The equation for this reaction is:
H+(aq) + OH-(aq) → H2O (l)
When an acid neutralises a base (or vice versa), the solution that’s formed is neutral it has a pH of 7. At pH 7, the concentration of hydrogen ions is equal to the concentration of hydroxide ions. An indicator can be used to show that a neutralisation reaction is over.

Question 14

Investigating neutralisation reactions

Answer 14

You’ve got to know how to investigate how the pH of a solution of dilute hydrochloric acid changes on addition of calcium oxide. Calcium oxide is a base. It reacts with hydrochloric acid to give calcium chloride (a salt) and water. The equation for the reaction is:
2HCI + CaO → CaCl2 + H₂O

Question 14

Dissociation of acids

Answer 14

When acids are added to an aqueous solution they ionise (or dissociate) to produce H+ ions and another type of ion (which is negatively charged).
e.g. Hydrogen chloride dissolves in water to form hydrogen ions and chloride ions:
HCl (g) → H+ (aq) + Cl- (aq)

Question 14

Investigating neutralisation reactions - method

Answer 14

1. Measure out 150 cm³ of dilute hydrochloric acid into a conical flask. Use a pipette or a measuring cylinder for this (see p.321-322).
2. Measure out 0.5 g of calcium oxide using a mass balance.
3. Carefully add the calcium oxide to the hydrochloric acid.
4. Wait for the base to completely react, then record the pH of the solution, using either a pH probe (see page 323) or Universal indicator paper. (You can use a glass rod to spot samples of the solution onto the paper).
5. Repeat steps 2 to 4 until all the acid has reacted. You’ll know you’ve reached this point when unreacted calcium oxide starts to collect at the bottom of the flask.
6. You can then plot a graph to see how pH changes with the mass of base added (see Figure 3).

Question 15

Acid strength

Answer 15

The strength of an acid tells you about the proportion of acid particles that will dissociate to produce H+ ions in solution.

Question 16

Strong acids

Answer 16

Strong acids, such as sulfuric (H2SO4), hydrochloric (HCI) and nitric acid (HNO3) ionise almost completely in water-most of the acid particles dissociate to release H+ ions. Strong acids tend to have low pHs (pH 0-2).

e.g. Nitric acid ionises completely in water to form hydrogen ions and nitrate ions:
HNO3 (l) → H+ (aq) + NO3 - (aq)
Sulfuric acid also ionises completely but releases two hydrogen ions for every molecule of sulfuric acid:
H₂SO4(l) → 2H+ (aq) + SO4 2- (aq)

Question 17

Weak acids

Answer 17

Weak acids only partially ionise in water - - if you put a sample of a weak acid in water, only some of the acid molecules will ionise and release H ions. Carboxylic acids are weak acids (they don’t ionise completely when dissolved in water) as are citric and carbonic acids. Weak acids tend to have pHs around 2-6.

for weak acids such as ethanoic acid, CH₃COOH we can show that they only partially dissociate by writing an equation - CH₃COOH ⇌ CH₃COO- + H+

Question 18

concentrated and dilute acids

Answer 18

refers to the amount of acid molecules dissolved in a fixed volume of water a concentrated acid will have a high amount of acid molecules dissolved in a fixed volume of water. a delete acid will have a relatively low number of acid molecules dissolved in a fixed volume of water.

The more grams (or moles) of acid per dm³, the more concentrated the acid is. So you can have a dilute strong acid, or a concentrated weak acid.

Question 19

Concentration and pH

Answer 19

The pH of an acid is dependent on the acid’s concentration - increasing the concentration of H+ ions leads to a decrease in the pH. If the concentration of H+ ions increases by a factor of 10, the pH decreases by 1. So if the H+ ion concentration increases by a factor of 100 (= 10 x 10), the pH decreases by 2 (= 1 + 1), and so on.
Decreasing the H+ ion concentration has the opposite effect - - a decrease by a factor of 10 in the H+ concentration means an increase of 1 on the pH scale.
Eg 1 - A solution with a hydrogen ion concentration of 0.001 mol dm³ has a pH of 3. What would happen to the pH if the hydrogen ion concentration was increased to 0.01 mol dm-3?
The H+ concentration has increased by a factor of 10, so the pH would decrease by 1. So the new pH would be 3 - 1 = 2.
Eg 2 - A solution with a hydrogen ion concentration of 0.25 mol dm-³ has a pH of 0.6. What would happen to the pH if the hydrogen ion concentration was decreased to 0.00025 mol dm-3?
The H+ concentration has decreased by a factor of (10 x 10 x 10 =) 1000, so the pH would increase by 3. So the new pH would be 0.6 + 3 = 3.6.

Question 20

Reaction with metals

Answer 20

Acids can react with metals to produce a metal salt and hydrogen gas. The general equation for the reaction of a metal with an acid is:
acid + metal → metal salt + hydrogen

Question 21

Test for hydrogen

Answer 21

If you hold a lit splint at the open end of a test tube containing hydrogen, you’ll get a “squeaky pop”.
(The noise comes from the hydrogen burning quickly with the oxygen in the air to form H2O.)

Question 22

Reaction with metal carbonates

Answer 22

Metal carbonates react with acids to make a salt, carbon dioxide and water. The general equation for the reaction of metal carbonates and acids is:
metal carbonate + acid → metal salt + carbon dioxide + water
The type of salt produced depends on the type of acid used and the metal in the carbonate.

Question 23

Test for carbon dioxide

Answer 23

If you make a solution of calcium hydroxide in water (called limewater) and bubble gas through it, the solution will turn cloudy if there’s carbon dioxide in the gas. The cloudiness is caused by the formation of calcium carbonate.

Question 24

solubility rules

Answer 24

Question 25

Precipitation

Answer 25

To make a pure, dry sample of an insoluble salt, you can use a precipitation reaction. Precipitation occurs when an insoluble solid (known as a precipitate) forms in a solution. You just need to pick the right two soluble salts and react them together to get your insoluble salt.
You can make lead chloride (an insoluble salt) by mixing together lead nitrate and sodium chloride (both soluble salts).
lead nitrate + sodium chloride → lead chloride + sodium nitrate
Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq)

Question 26

Precipitation - Method

Answer 26

1. Add one spatula of lead nitrate to a test tube. Add some deionised water in order to dissolve it. Shake the test tube thoroughly to ensure that all the lead nitrate has dissolved.
2. Repeat step 1 in a separate test tube, except this time use one spatula of sodium chloride.
3. Tip the two solutions into a small beaker, and stir thoroughly to make sure the solutions are fully mixed. The lead chloride should precipitate out
4. Fold a piece of filter paper and place it into a filter funnel. Place the funnel into the top of a conical flask.
5. Pour the contents of the beaker into the middle of the filter paper. Make sure that the solution doesn’t go above the top of the filter paper- otherwise some of the solid lead chloride could end up in the filtrate.
6. Swill out the beaker with more deionised water, and tip this into the filter paper. This will ensure all the precipitate is taken from the beaker.
7. Rinse the contents of the filter paper with deionised water to make sure that all the soluble sodium nitrate has been washed away.
8. Scrape the lead chloride onto fresh filter paper and leave it to dry in an oven or a desiccator.

Question 27

Excess Base - Soluble salts from acids and insoluble reagents

Answer 27

Soluble salts (salts that dissolve in water) can be made by reacting an acid with a metal or an insoluble base (such as a metal oxide, metal hydroxide metal carbonate). When making a soluble salt, the first thing you need to do is choose appropriate reagents to produce that particular salt. You should be able to work out which are the right reagents to choose from the name of the salt you want to produce.

Question 28

Excess Base - Soluble salts from acids and insoluble reagents - Making copper sulfate

Answer 28

You can make copper sulfate (a soluble salt) by adding copper oxide (an insoluble reagent) to warm sulfuric acid.
CuO (s) + H2SO4(aq) → CuSO4 (aq) + H2O(l)
The copper oxide is added in excess and separated out at the end of the reaction using filter paper. Many metal oxides and some metal carbonates are insoluble, so this is also the method you’d use if you’re using any of those.

Question 29

Excess Base - Soluble salts from acids and insoluble reagents - Making copper sulfate - method

Answer 29

1. Put the sulfuric acid in a beaker. Gently warm the dilute acid by placing the beaker in a water bath (see p.326 for more on water baths) speeds up the reaction between the acid and the insoluble base. Carry out this step in a fume cupboard to avoid releasing acid fumes into the room. -this
2. Add the copper oxide (the base) to the acid and stir - they will react with each other to form copper sulfate (a soluble salt) and water.
3. Keep adding the copper oxide until it is in excess. When it reaches excess, the oxide will start to sink to the bottom and won’t react - this shows that all the acid has been neutralised and the reaction has finished. It’s important that the base is in excess, as this ensures that you won’t have any leftover acid in your product.
4. Then you need to filter out the excess copper oxide to separate it from the copper sulfate solution. This ensures that the copper sulfate salt isn’t contaminated with copper oxide once it’s dried. This stage is done using filter paper and a filter funnel
5. At the end of the filtration, you’ll be left with a solution of copper sulfate salt dissolved in water. You can convert this into pure, solid crystals of copper sulfate using crystallisation. To do this you first need to heat the salt solution using a Bunsen burner and an evaporating dish (see p.105), in order to evaporate some of the water and make the solution more concentrated,
6. Then, stop heating it and leave the solution to cool. Blue crystals of hydrated copper sulfate should form, which can be filtered out of the solution and then dried..

Question 30

titration - Making a soluble salt using acid-alkali reactions

Answer 30

Soluble salts can be made by reacting an acid with an alkali. It is not always obvious when the reaction has finished, as there’s no signal that all the acid has been neutralised. You also can’t just add an excess of alkali to the acid, because the salt is soluble and would be contaminated with the excess alkali.
In order to overcome these problems you need to work out exactly the right amount of alkali to neutralise the acid. For this, you need to do a titration using an indicator.

Question 31

titration - Making a soluble salt using acid-alkali reactions - method

Answer 31

1. Measure out a set amount of acid into a conical flask using a pipette. Add a few drops of indicator, then place the flask on a white tile.
2. Fill a burette with alkali and record the volume.
3. Slowly add alkali to the acid, gently swirling the flask all the time, until reach the end point- this is when the you acid’s been exactly neutralised and the indicator changes colour
4. Read the final volume of alkali left in the burette. Calculate the amount of alkali that was needed to neutralise the acid by subtracting the final volume of alkali in the burette from the initial volume.
5. Then, carry out the reaction using exactly the same volumes of alkali and acid but with no indicator, so the salt won’t be contaminated with indicator.
6. The solution that remains when the reaction is complete contains only the salt and water. This is because all the alkali has reacted with all the acid.
7. Slowly evaporate off some of the water and then leave the solution to crystallise. Filter off the solid and dry it - you’ll be left with a pure, dry salt.

Question 32

Electrolytes

Answer 32

Electrolysis requires a liquid to conduct the electricity, called the electrolyte. Electrolytes contain free ions - they’re usually a molten or dissolved ionic compound. In either case, it’s the free ions which conduct the electricity and allow the whole thing to work.

Question 33

electrode

Answer 33

a solid that conducts electricity and is submerged in the electrolyte.

Question 34

How electrolysis works

Answer 34

In electrolysis, the electrodes are placed into the electrolyte and ions move from one electrode to the other - this allows the conduction of electricity through the circuit. The positive ions (cations) in the electrolyte will move towards the negative electrode (the cathode) and gain electrons. The negative ions (anions) in the electrolyte will move towards the positive electrode (the anode) and lose electrons. As ions gain or lose electrons they become atoms or molecules and are discharged (released) from the electrolyte. These atoms or molecules are the products of electrolysis.

Question 35

Electrolysis and redox

Answer 35

Electrolysis always involves an oxidation reaction and a reduction reaction. Reduction is occurring at the negative electrode (the cathode) as the positive ions (cations) are gaining electrons. Oxidation occurs at the positive electrode (the anode) as the negative ions (anions) are losing electrons.

Question 36

Electrolysis of molten lead bromide

Answer 36

Electrolysis of molten lead bromide
Lead bromide (PbBr2) is an ionic compound, so when it is molten, it will conduct electricity. Electrolysis of lead bromide breaks it down into lead (Pb) and bromine (Br2).

The positive ions are attracted towards the negative electrode. Here, each lead ion gains two electrons and becomes a lead atom. Pb2+ + 2e- -> Pb

The negative ions are attracted towards the positive electrode. Here, bromide ions lose one electron each and form bromine molecules (Br₂). 2Br- -> Br2 + 2e-

Question 37

Electrolysis rules for aqueous solutions

Answer 37

If a salt is dissolved in water there will be some H+ and OH- ions as well as the ions from the salt in the solution. In this situation, the products of electrolysis depend on how reactive the elements involved are.
At the negative electrode (the cathode), if the metal is more reactive than hydrogen the metal ions will stay in solution (e.g. sodium). This is because the more reactive an element, the more likely it is to stay as ions. So, hydrogen will be produced unless the metal is less reactive than it. If the metal is less reactive than hydrogen (e.g. copper), then a solid layer of pure metal will be produced instead.
At the positive electrode (the anode), if OH- and halide ions (Cl-, Br-, I-) are present then molecules of chlorine, bromine or iodine will be formed. If no halide is present (e.g. NO3-, SO4 2-), then the OH- ions are discharged and oxygen and water will be formed - 4OH- -> 2H2O + O2 + 4e-

Question 38

Electrolysis for aqueous solution examples - copper chloride

Answer 38

Question 39

Electrolysis for aqueous solution examples - sodium chloride

Answer 39

Question 40

Electrolysis for aqueous solution examples - sodium sulfate

Answer 40

Question 41

Electrolysis of copper sulphate using inert (graphite) electrodes

Answer 41

A solution of copper sulphate (CuSO4) contains four different ions: Cu 2+, SO4 2-, H+ and OH-. When you electrolyse a solution of CuSO4 with inert electrodes:

Copper metal is less reactive than hydrogen. So, at the cathode, copper metal is produced and coats the electrode: Cu2+ + 2e- → Cu

There aren’t any halide ions present so at the anode oxygen and water are produced. The oxygen can be seen as bubbles: 4OH- → О₂ + 2H₂O + 4е-

Question 42

Electrolysis of copper sulfate using non-inert (copper) electrodes

Answer 42

As the reaction continues, the mass of the anode will decrease and the mass of the cathode will increase. This is because copper is transferred from the anode to the cathode. The reaction is quite slow, so the cell should be left for around 30 minutes in order to achieve a measurable change in mass.
You can measure how the mass of the electrodes has changed during the experiment by finding the difference between the masses of the electrodes before and after the experiment.
Anode – Cu → Cu 2+ + 2e-
Cathode – Cu 2+ + 2e- → Cu

Question 43

Purification of copper using electrolysis of copper sulfate using non-inert (copper) electrodes

Answer 43

Copper can be extracted from its ore by reduction with carbon, but copper made in this way is impure. Electrolysis can be used to purify the copper using an electrochemical cell with copper electrodes.
When copper is purified using electrolysis, the anode starts off as a big lump of impure copper and the cathode starts off as a thin piece of pure copper. The electrolyte is copper sulfate solution (which contains Cu²+ ions). Here’s what happens during the process:

1. Copper in the impure copper anode forms copper ions which dissolve into the electrolyte – Cu → Cu 2+ + 2e-
2. The copper ions move to the pure copper cathode, and react to form a layer of pure copper – Cu 2+ + 2e- → Cu
3. Any impurities from the impure copper anode sink to the bottom of the cell, forming a sludge.