5.1.3 Acids, Bases and Buffers

Question 1

Bronsted lowry acid

Answer 1

Substance that can donate a proton

Question 2

Bronsted lowry base

Answer 2

Substance that can accept a proton

Question 3

How to calculate ph

Answer 3

pH = -log [H+]

Question 4

Conjugate base pairing which acts as an acid

Answer 4

Substance with higher Ka

Question 5

Role of H+ in acid metal equations

Answer 5

2H+ + Mg -> Mg2+ +H2

Question 6

Role of H+ in acid alkali equations

Answer 6

H+ + OH- -> H2O

Question 7

Role of H+ in acid carbonate equations

Answer 7

2H+ +CO3 2- -> H2O + CO2

Question 8

Calculating pH of strong acids

Answer 8

Completely dissociate

H+ is equal to conc of acid

Question 9

What do you give pH values to in exams

Answer 9

2 dp

Question 10

how to calculate [H+]

Answer 10

[H+] = 10^-ph

Question 11

Ionic product for water describe

Answer 11

In all aqueous solutions and pure water the neutralisation of water equilibrium occurs
Uses Kw

Question 12

Kw

Answer 12

Kw = [H+] [OH-]

Question 13

Value of Kw at 25 degrees

Answer 13

1 x 10^-14 mol2dm-6

Question 14

What can Kw be used to find

Answer 14

[H+]

[OH-]

Question 15

Finding the pH of pure water

Answer 15

Neutral because H+ = OH-
Kw = [H+]^2
So H+ = square root of Kw

Question 16

Different temps pH of pure water

Answer 16

Changes
Dissociation of water is endothermic as bonds are broken
Increasing temp pushes equilibrium to right
Bigger conc of H+ ions and lower pH

Question 17

Calculating ph of strong bases

Answer 17

Normally get given conc of hydroxide ion
To work out pH work out H+ using Kw
Strong bases completely dissociate into their ions

Question 18

Weak acids

Answer 18

Only slightly dissociate when dissolved in water

Uses Ka

Question 19

Whats the weak acids dissociation expression

Answer 19

Ka= [H+][A-] divided by [HA]
Larger Ka stronger the acid
Products divided by reactants

Question 20

pKa

Answer 20

pKa = -logKa
Ka= 10^-pka

Question 21

Calculating pH of a weak acid

Answer 21

Ka = [H+]^2 divided by [HA]

Question 22

Assumptions made to simplify Ka

Answer 22

1. [H+] = [A-] because they have dissociated according to a 1:1 ratio
2. As the amount of dissociation is small we assume the initial conc of the undissociated acid has remained constant

Question 23

pH of diluted strong acid

Answer 23

H+ = H+ old x old volume divided by new volume

Then pH = -log [H+]

Question 24

pH of diluted base

Answer 24

OH- = OH- old x old volume divided by new volume
H+ = Kw divided by OH-
pH = - logH+

Question 25

What is a buffer solution

Answer 25

One where the pH doesnt change significantly if small amounts of acid or alkali are added to it

Question 26

Basic buffer solution

Answer 26

Made from a weak base and a salt of that weak base (react weak base with a strong acid) Ammonia and ammonium chloride

Question 27

Acidic buffer solution

Answer 27

Made from a weak acid and a salt of that weak acid (reacting weak acid with strong base) Ethanoic acid and sodium ethanoate

Question 28

How could salt content be added

Answer 28

Salt solution added to acid or some solid salt added | Buffer made by partially neutralising a weak acid with alkali and produces a mixture of salt and acid

Question 29

Ethanoic acid buffer

Answer 29

CH3CO2H (reversible) CH3CO2- + H+ | Much higher conc of salt ion than in pure acid

Question 30

If small amounts of acid are added to buffer

Answer 30

Equilibrium shift to left Removing H+ added As theres a large conc of the salt ion in the buffer the conc ratio CH3CO2H/CH3CO2- stays almost constant so pH stays constant

Question 31

If small amounts of alkali are added to buffer

Answer 31

OH- ions will react with H+ ions to form water Equilibrium shifts to right to produce more H+ ions Conc of H+ and pH remains constant but some ethanoic acid molecules are changed to ethanoate ions

Question 32

Calculating pH of buffer solutions

Answer 32

Ka= H+ x A- divided by HA Assume that A- conc is due to added salt only Assume intial conc of acid has remained constant because amount that has dissociated or reacted is small

Question 33

If small amount of alkali is added to a buffer what happens (moles)

Answer 33

Moles of buffer acid would reduce by number of moles of alkali added and the moles of salt would increase by same amount CH3CO2H + OH- -> CH3CO2- + H2O

Question 34

If small amount of acid is added to a buffer what happens (moles)

Answer 34

Moles of buffer salt would reduce by the number of moles of acid added and the moles of buffer acid would increase by the same amount CH3CO2- + H+ -> CH3CO2H

Question 35

Buffering action in blood

Answer 35

Carbonic acid-hydrogencarbonate equilibrium acts as buffer in control of blood pH H2CO3/HCO3- buffer present in blood plasma mainting pH 7.35-7.45

Question 36

Equilibrium in blood

Answer 36

H2CO3 (reversible) H+ + HCO3- | Adding alkali reacts with H+ so shifts to right forming new H+ and more HCO3-

Question 37

Constructing a pH curve

Answer 37

1. Acid into conical flask 2. Measure intial pH with pH meter 3. Add alkali in small amounts noting the volume 4. Stir mixture to equalise pH 5. Measure and record pH 6. Repeat but when approaching end point add smaller volumes of alkali 7. Add until alkali in excess Need to calibrate meter first

Question 38

Calibrating meter in pH curve

Answer 38

Measure known pH of buffer solution pH meters can lose accuracy on storage Put probe in a set buffer and press calibrate Maintain constant temp also to improve accuracy

Question 39

4 main types of titration curves

Answer 39

1. Strong acid and strong base 2. Weak acid and strong base 3. Strong acid and weak base 4. Weak acid and weak base

Question 40

Strong acid strong base titration curve

Answer 40

HCl and NaOH pH at equivalence point 7 Long steep part 3 to 9 work out neutralisation volume from titration data given in question, standard titration calculations

Question 41

Key points to sketching titration curve

Answer 41

Intial and final pH Volume at neutralisation General shape (pH at neutralisation)

Question 42

Where does the equivalence point lie

Answer 42

Lies at mid point of extrapolated vertical point of curve

Question 43

Weak acid strong base

Answer 43

``` CH3CO2H and NaOH at start the pH rises quickly and then levels off, flattened part is buffer region and is formed because buffer solution is made Starts near 3 Steep part >7 (7 to 9) Equivalence point >7 ```

Question 44

Half neutralisation volume

Answer 44

Use Ka at 1/2 HA= A Ka = H+ and pKa = pH If you know Ka can work out pH at half pH of weak acid at half neutralisation will equal pKa

Question 45

Strong acid weak base

Answer 45

HCl and NH3 Equivalence <7 Steep <7 around 4 to 7

Question 46

Weak acid weak base

Answer 46

CH3CO2H and NH3 | No vertical part

Question 47

What can indicators be considered as

Answer 47

Weak acids | Acid must have different colour to conjugate base

Question 48

End point of a titration

Answer 48

[Hln]=[ln-] | Choose indicator whose end-point coincides with equivalence point for titration

Question 49

Applying Le Chateliers to give colour

Answer 49

Hln (reversible) ln- + H+ Colour A. Colour B In acid solution H+ ions push equilibrium towards reactants, colour A is acidic colour In alkaline solution the OH- ions will react and remove H+ ions causing equilibrium to shift to products, colour B is alkaline colour

Question 50

When will an indicator work

Answer 50

If pH range of indicator lies on stteo part of titration curve Indicator will change colour rapidly and the colour change will correspond to neutralisation point

Question 51

When do you use phenolphthalein

Answer 51

In titrations with strong bases but not weak bases | Colour change: colourless acid -> pink alkali

Question 52

When do you use methyl orange

Answer 52

Titrations with strong acids but not weak acids | Colour change: red acid -> yellow alkali (orange end point)