Acids - Topic 3 Flashcards

Question

What is a neutral pH?

Answer 1

below pH 7

Answer 2

above pH 7

Answer 3

- pH 0 - 1 - red

Answer 4

- pH 2, 3, 4, 5 & 6 - pH 2 = orange - pH 3 & 4 = yellow - pH 5 & 6 = lime green

Answer 5

- pH 7 - green

Answer 6

- pH 8, 9, 10, 11 & 12 - pH 8 = teal - pH 9, 10, 11 & 12 = blue

Answer 7

- pH 13 & 14 - purple

Answer 8

a substance that is used to distinguish between acidic and alkaline solutions

Answer 9

used to distinguish between acidic and alkaline solutions

Answer 10

- a mixture of several different indicators which is used to measure the approximate pH value of a solution and indicates the strength of the acid or alkali - the strength of the acid or alkali is measured on the pH scale

Answer 11

can be used either in solution or the solution can be dried onto paper to make test papers

Answer 12

Colourless

Answer 13

Colourless

Answer 14

- the higher the concentration of hydrogen ions in an acidic solution the lower the pH so the more acidic the solution is as it is the hydrogen ion that causes acidity - acids release H+ in a solution - the higher the concentration of hydroxide ions in an alkaline solution the higher the pH so the more alkaline the solution is

Answer 15

- Water is a polar solvent and methylbenzene is a non-polar solvent - A polar solvent will have positive and negative charge (in water, oxygen is slightly more negative and the hydrogen is slightly positive) whereas methylbenzene is netral throughout - HCl will dissociate in water into H+ and Cl- ions because HCl is also a polar molecule (Cl is slightly negative and H is slightly positive) - The Cl will be attracted to the slightly positive H in water and the H will be attracted to the slightly negative O in water. - The HCl molecule will dissociate as a result of these strong attractions - With methylbenzene, no such attraction exists, thus HCl will not dissociate and no ions will be made and HCl will remain as a HCl molecule

Answer 16

1. HCl (g) ——> H+ (aq) + Cl- (aq) 2. HCl (g) + H2O (l) ——> H3O+ (aq) + Cl- (aq)

Answer 17

- HCl gas consists of HCl molecules - hydrochloric acid consists of H+ and Cl- ions in solution

Answer 18

- ‘p’ stands for potenz (German word for power) - H stands for H+

Answer 19

logarithmic scale (like Richter scale and loudness in decibels)

Answer 20

- used to measure pH of solution to determine pH to 1 or 2 decimal places - more accurate

Answer 21

- as hydrogen ion concentration in a solution increases by a factor of 10, the pH of the solution decreases by 1 and vice versa - e.g. solution of pH1 contains 10x as many hydrogen ions than a solution that has pH2 (10x as acidic) & pH 1 is 10ox more acidic than pH 3

Answer 22

- solutions of acids made by dissolving certain covalent molecules in water - the water cause the covalent molecules (as it is a polar substance) to break apart in a way that generates hydrogen ions (H+)

Answer 23

concentrated and dilute refers to amount of acid molecules dissolved in water

Answer 24

larger amount of substance in given volume of solution

Answer 25

lesser amount of substance in given volume of solution

Answer 26

- concentrated solution will have large amount of acid molecules dissolved in a certain volume of water

Answer 27

- dilute acid will have relatively few acid molecules dissolved in the water

Answer 28

- ethanoic acid has lower pH than HCl as more of HCl molecules dissociated and so released more hydrogen ions - need a high conc of ethanoic acid and a dilute hydrochloric acid for ethanoic acid to contain more hydrogen ions

Answer 29

- Strong and weak is NOT the same as concentrated and dilute - concentrated and dilute depends on the amount of acid (substance) - strong or weak depends on how many acid molecules dissociate (refers the H+ ion conc. in aq. solutions)

Answer 30

strong or weak depends on how many acid molecules dissociate into separate ions (refers the H+ ion conc. in aq. solutions)

Answer 31

- strong acid fully dissociates in solution (e.g. water) (all acid molecules broken apart and formed separate ions) - acids such as HCl, H2SO4 and HNO3 are strong acids - HCl (g) ——> H+ (aq) + Cl-(aq)

Answer 32

- in weak acid only small proportion of molecules dissociates into ions (means only small proportion of solution mainly contains undissociated molecules and just low number of ions) - acids like ethanoic acid, CH3COOH are weak acids

Answer 33

HCl more dangerous as it is stronger acid than methanoic acid so more of its molecules dissociate into separate ions and release more H+ ions (so has a lower pH) make it more corrosive

Answer 34

- strong acids are more reactive than weak acids - faster reaction - more exothermic - due to difference in H+ ions as strong acids have more H+ ions

Answer 35

- substance that reacts with an acid to form a salt and water only - substances that can accept H+ ions/acids/metal oxides/metal hydroxides - all substances whether soluble or not that neutralise acids

Answer 36

bases normally metal oxides or metal hydroxides

Answer 37

- (sub-group of bases) - soluble bases - also neutralise acids but must be soluble in water

Answer 38

soluble hydroxides - potassium, sodium and ammonium hydroxides Soluble carbonates - potassium, sodium and ammonium carbonates

Answer 39

- all alkalis will either dissolve in water or react with water to produce hydroxide ions (OH-) in a solution - it is the OH- that causes alkalinity

Answer 40

strong alkalis like sodium hydroxide (NaOH) and potassium hydroxide (KOH) will completely dissolve in water and dissociate to give high conc of OH- ions

Answer 41

- weak alkalis like ammonia will only partially dissolve / react with water and dissociate to give a low conc of OH- ions - ammonia undergoes a reversible reaction with water which leads to a low conc of OH- ions and so ammonia is a common alkaline solution of approx. pH = 8

Answer 42

ionic substances where hydrogen is replaced by other positive ions (normally metals)

Answer 43

- substances that dissolve/react with water to release H+ ions - common acids: - H2SO4 - sulphuric acid - HCl - hydrochloric acid - HNO3 - nitric acid - H3PO4 - phosphoric acid - CH3COOH - ethanoic acid

Answer 44

- substances that react with acids to form a salt and water (sometimes CO2 as well) - common bases: - metal oxides e.g. CaO, Al2O3, Na2O, CuO - metal hydroxides e.g. Ca(OH)s, NaOH, KOH - metal carbonates e.g. CaCO3, Na2CO3, CuCO3

Answer 45

- substances that react with water to release OH- ions - common alkalis: - NH3 - ammonia - (water soluble metal hydroxides) e.g. NaOH (sodium hydroxide), KOH (potassium hydroxide), Ca(OH)2 (calcium hydroxide)

Answer 46

- ionic substances formed when acids react with bases - common bases: - sulphates - made from sulphuric acid - chlorides - made from hydrochloric acid - nitrates - made from nitric acid - phosphates - made from phosphuric acid - ethanoates - made from ethanoic acid - citrates - made fro citric acid

Answer 47

metal + acid ——> metal salt + hydrogen

Answer 48

- effervescence - exothermic reaction - metal no longer visible

Answer 49

acid + base (alkali) ——> metal salt + water

Answer 50

- form salt and water - if the base starts off as solid it will dissolve - exothermic —> temp change

Answer 51

metal carbonate + acid ——> metal salt + water + carbon dioxide

Answer 52

- fizz and give off carbon dioxide - effervescence

Answer 53

Ammonia + acid ——> ammonium salts

Answer 54

- when ammonia reacts with HCl it produces white fumes of ammonium chloride

Answer 55

a reaction between an acid and a base

Answer 56

- a reaction of hydrogen ions (H+) from the acid react with hydroxide ions (OH-) from the alkali to form water - H+ + OH- ——> H2O (+ salt)

Answer 57

- when a base is added to an acid solution the pH rises - eventually, when sufficient bas has been added the pH reacher 7 - at this point, neither acid nor bases are in excess - all acid has been neutralised by the base - solution is neutral - as more base is added the pH increases above 7 and so solution becomes alkaline

Answer 58

- neutralisation process can be followed using UI paper or solution and also a pH metre - as the pH of solution increases, colour changes from red for acid to green for neutral and finally to purple for alkali

Answer 59

- (ml = cm3) - 100ml beaker - 25ml measuring cylinder - stirring rod - spatula - pH chart for (UI) and UI paper or solution - pH metre - spotting tile - lime (calcium hydroxide) solid - vinegar (ethanoic acid)

Answer 60

1. Use measuring cylinder to pour 20ml of vinegar into beaker. Add 20ml of water to vinegar. 2. Stir mixture with glass stirring rod. Place small piece of UI paper on spotting tile. Use glass rod to place drop of vinegar solution on indicator paper. Note the colour. 3. Place clean tip of pH metre into solution, allow metre to settle for a value and note this down. 4. Add 0.1g of lime to beaker. Take reading with pH metre to test the solution with fresh piece of UI indicator. 5. Repeat steps 3 & 4. 6. Continue with above steps until you have added 1.4g of lime.

Answer 61

- the pH gradually increases till 0.8g of lime is added then there is a sudden increase in pH when 0.9 of lime is added and then the pH of the solution curves off as solution becomes saturated - there is an increase in pH as lime is added making the solution more alkaline increasing the pH of the solution

Answer 62

- doesn’t smell as all acid has been reacted

Answer 63

when pH of solution reaches pH 7.0

Answer 64

- to ensure all Ca(OH)2 (alkali) has been fully dissolved in solution so pH is consistent throughout the solution - to make sure acid is reacted as much as possible

Answer 65

- measuring cylinder - has a higher resolution as the increase in graduations on the measuring cylinder are in smaller increments (e.g. to nearest 0.5ml) so level of accuracy increases

Answer 66

- loss of mass of solid onto balance - Ca(OH)2 not fully dissolving in acid

Answer 67

use glass pipette to increase accuracy of measuring liquid

Answer 68

1. excess base method 2. titration 3. precipitation

Answer 69

- can only be used to make a salt if base in insoluble and salt made is soluble

Answer 70

- can’t be used to make sodium, potassium and ammonium salts - can’t be used to make soluble salts

Answer 71

1. Choose suitable starting materials - usually metal oxide is used as base and is reacted with appropriate acid (want to make copper sulfate - so use copper oxide and sulphuric acid 2. Put some acid in beaker and add to it excess base (ensures all acid has been reacted). Gently warm the beaker to speed up the reaction. (if carbonate used reaction much faster and mixture doesn’t need to be warmed) - so add excess copper oxide 3. After reaction finished, there will be a solution containing the salt and some unreacted solid base, the solution can be separated from unreacted base by filtration - copper oxide is residue and copper sulfate solution is filtrate 4. The filtrate contains the salt solution. The solution is then heated and some of the water is evaporated away to make a hot saturated solution (this can be tested for by dipping a glass rod into the hot solution and blowing on it - if crystals appear it is saturated) 5. Take care not to evaporate all the water away - this will result in an anhydrous powder of the salt rather than crystals (which usually contain some water of crystallisation) - if copper sulfate solution has all the water evaporated away then white anhydrous copper sulfate will be formed rather than blue hydrated copper sulfate 6. Leave the hot saturated solution to cool and crystallise. Crystals appear as dissolved solids are less soluble in cold water than in hot water. The slower the cooling, the larger the crystals. 7. To separate the crystals from the solution, the mixture can be filtered and then the crystals should be left to dry the warm place

Answer 72

- contain some water molecules ‘locked’ in solid structure - CuO(s) + H2SO4(aq) ——> CuSO4 (aq) + H2O(l)

Answer 73

- anyhdrous copper sulphate of crystallisation - heating the hydrated copper sulphate turns it anyhdrous - CuSO4 + 5H2O (blue, hydrated) ——> (heat) CuO4 + 5H2O (white, anhydrous)

Answer 74

lab technique where 2 solutions mixed together to determine exact volumes of each which react together

Answer 75

- pipette used to measure out one solution into a conical flask - other solution added to flask from a burette a little at a time, with mixing until the reaction is complete - end-point of the titration

Answer 76

- salts formed when acid and base react together - when soluble bases (alkalis) are used all solutions concerned are colourless and it is impossible to tell when 2 bases have exactly neutralised each other - so precise volumes found by titration - method applies when making sodium, potassium and ammonium salts since these hydroxides and carbonates are all soluble in water

Answer 77

Setting up the burette: - Rinse out the burette with distilled water, then rinse with little sulphuric acid provided - Clamp the burette and make sure tap is closed. Carefully pour more acid from a beaker into the top of the burette using small funnel - Place the beaker under the burette tap and then open the tap to fill the jet. Make sure no air bubbles are trapped in the jet. - Remove the funnel from top, the read initial volume from burette scale (make sure eyes level with bottom of meniscus). Record volume. Using the pipette: - Collect some sodium carbonate solution in beaker. - Fill the pipette then measure out 25.0cm3 of sodium carbonate solution into small conical flask. repeat this 3 times into 3 separate flasks. - Add drops of methyl orange indicator to each flask (should be yellow) The titration: - 1st titration is rough. After you have approximate value do 2 accurate titrations. - Stand the conical flask with the alkali and indicator in it on a white tile underneath the burette. The jet should be just inside mouth of flask. - Use left hand to control burette hand and right hand tot swirl flask gently as you add acid at about 1cm3 at a time. Stop adding acid when indicator changes to permanent red/pink colour. Repeat titration using fresh lot of alkali in a clean flask. - Use results from rough titration to anticipate where end point will be. - As you get nearer to end point, add acid one drop at a time until indicator orange-pink colour. Remember to record all burette readings. - Repeat until you get 2 titrations where volumes of acid used agree to within 0.1cm3 Making sodium sulphate crystals: - Now you know accurate volumes of sodium carbonate and sulphuric acid needed to neutralise each other you can mix these to make a solution of the salt - Pipette another 20.0cm3 of sodium carbonate solution into evaporating basin. Add it to the exact volume of sulphuric acid required from burette. (don’t use indicator) - Heat solution gently with Bunsen burner over tripod and gauze until you get to crystallisation point (test with cold glass rod), then set aside to cool and crystallise slowly at room temperature

Answer 78

1. volumes needed to make neutral solution are known 2. an indicator would make the salts impure

Answer 79

- to evaporate away some of water to make a hot saturated solution

Answer 80

white crystals

Answer 81

1. Collect some hydrochloric acid. Use acid to fill burette carefully, with tape closed. Fill the nozzle of the burette by running the solution through the tap. Record the start volume reading on the burette. 2. Collect some sodium hydroxide in a different labelled 100cm3 glass beaker. Measure exactly 25.0cm3 of acid using pipette filler. Run this acid into clean conical flask. 3. Add 5 drops of phenolphthalein in different labelled 100cm3 beaker. Indicator is pink at start (pH is very low) but when solution is acidic it will change to colourless. 4. Start running the acid into the conical flask from the burette. After adding 23cm3 add acid one drop at a time, swirling contents of flask to mix them. When one drop of acid turns mixture colourless stop adding acid. Record final volume on burette and use this value and the starting value to work out the amount of acid added. 5. Refill the burette and clean out the conical flask with water. Repeat whole experiment using same volumes. 6. When finished pour solution containing salt into large beaker 7. Salt can be obtained from solution by heating solution evaporating away some of water and then leaving it to cool and crystallise.

Answer 82

when soluble salts are prepared from an acid and soluble reactant

Answer 83

- excess base method relies upon the facts the base and salt can be easily separated by filtration - while this is largely true for most bases and their corresponding salt, it is not always the case - sodium, potassium and ammonium compounds are always soluble - means that the sodium, potassium and ammonium bases can’t be separated from corresponding salts by filtration so titration is used instead

Answer 84

e.g. making sodium chloride 1. Choose suitable starting chemicals - one must be an alkali (either sodium hydroxide, potassium hydroxide or ammonia). The other must be the appropriate acid. -e .g. to make sodium chloride sodium hydroxide and hydrochloric acid must be used 2. Use a glass pipette and place exactly 25.0cm3 of the alkali into a conical flask. Add a few drops of indicator. - e.g. 25.0cm3 of sodium hydroxide is measured by pipette into a conical flask and a few drops of phenolphthalein is added, it will turn pink as sodium hydroxide is added 3. Put the acid in the burette. Add the acid to the alkali in the conical flask a bit at a time. Swirl the conical flask to make sure that the acid and alkali mix and react. 4. When the alkali is nearly neutralised add the acid a drop at a time (dropwise) until the indicator changes colour. The alkali is now neutralised. Record the vol of acid added from the burette. - e.g. hydrochloric acid is added to the sodium hydroxide until one drop of the acid turns the phenolphthalein colourless 5. Repeat the whole experiment again using the exact same volume but without the indicator 6. The solution made contains the salt dissolved in water. Heat the mixture, evaporate away some of the water and make a hot saturated solution 7. Leave the solution to cool and condense. Filter the crystals off from the solution and leave to dry in a warm place.

Answer 85

- both reactants are liquids/soluble so if you have an excess of one you would not be able to easily remove it from the mixture of products - titration allows you to measure the exact amount of volumes that react - you can then mix the exact proportions of the two reactants that react - the exact amount of acid therefore, has 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

Answer 86

- excess base and filtration - acid + base ——> salt + water

Answer 87

- H2SO4, particularly when hot, is corrosive - compounds of copper are toxic

Answer 88

1. Measure correct volumes of of dilute H2SO4 into a small beaker, then add correct volume of water 2. Add excess of copper oxide (insoluble solid) to H2SO4 acid which react to form copper sulfate 3. Warm the mixture gently without boiling and stir for 2-3mins, until the copper oxide stops dissolving in and reacting with the acid - the acid has now been neutralised 4. Use filter and filter paper to filter and recover the unreacted copper oxide. Collect the filtrate (copper sulfate solution) in an evaporating dish 5. Place the filtrate in a hot water bath to evaporate the water until the solution is saturated 6. Use the ‘glass rod’ test to decide when saturation has occurred 7. Leave the solution for 3-4 days so that it can cool and crystallise 8. Remove the crystals with a glass rod and dry them using filter paper

Answer 89

- plastic has a low melting point so it would melt - it’s insulating

Answer 90

To speed up process of copper oxide dissolving and reaction in acid

Answer 91

- to ensure all the acid had been reacted so that the solution is pure - to ensure that the neutralisation reaction had occurred

Answer 92

blue - contains copper which is transition metal

Answer 93

- copper oxide which was not dissolved - ripped filter paper by mistake

Answer 94

- it tells us when the solution is saturated - dip glass rod into mixture, take out and blow on it —> if crystals appear it’s saturated

Answer 95

solid blue CuSO crystals appeared

Answer 96

contains water of crystallisation - makes the crystals blues

Answer 97

1. Wash burette using acid then water 2. Fill burette to 100cm3 with acid with the meniscus’ base on the 100cm3 line 3. Use 25cm3 glass pipette to add 25cm3 of alkali into conical flask, drawing alkali into pipette using pipette filler 4. Add few drops of suitable indicator into conical flask 5. Add acid from burette to alkali until end-point is reached 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 experiment to gain more precise results 8. To prepare pure, dry salt - warm solution to evaporate water 9. Crystals form

Answer 98

- Precipitation reactions are ‘swapping partner’ reactions. - In these reactions an insoluble solid is formed which is called a precipitate. e.g. mixing solutions of lead (II) nitrate and potassium iodide produces lead (II) iodide and potassium nitrate - Lead (II) iodide is insoluble and so it precipitates out of the mixture as a yellow solid - To work out the identity of a precipitate you need to know the solubility rules

Answer 99

- State symbols are usually used in precipitation reactions - E.g,. sodium chloride(aq) + silver nitrate(aq) —> sodium nitrate(aq) + silver chloride(s)

Answer 100

- wouldn't stay dissolved - crystals formed - 62g would come out of solution

Answer 101

- as the solution cools down the solvent is less able to hold the solute in solution

Answer 102

- as the temp falls the solute becomes less soluble - therefore it cannot remain all dissolved so some comes out of the solution

Answer 103

- used to make insoluble salt from 2 solutions (soluble → insoluble) - excess base method won’t work as both base and salt cannot be easily separated

Answer 104

1. Mix the 2 solutions (both must be soluble and one solution must contain the cation and the other must contain the anion), they will react and form an insoluble salt which is the precipitate. 2. Filter the mixture using filter paper. The residue is the precipitate (insoluble salt) 3. The insoluble salt will contain impurities from the solution. To remove these impurities, wash the solid with distilled water. 4. Leave the damp solid to dry in a warm place e.g. a drying oven to evaporate the water

Answer 105

to ensure your volume of acid reacts completely and solution is pure

Answer 106

this is done by filtration of the insoluble reactant and is done so that you are left with just a salt and water

Answer 107

- this is because all of 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

Acids - Topic 3 Flashcards

(158 cards)