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A-Level OCR Chemistry > Module 5: 20 & 21 > Flashcards

Module 5: 20 & 21 Flashcards

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1
Q

What is a Bronsted-Lowry acid?

A

A proton donor

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2
Q

What is a Bronsted-Lowry base?

A

A proton acceptor

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3
Q

What is a conjugate acid-base pair?

A

Contains 2 species that be interconverted by transfer of a proton

HCl + H2O ⇌ H3O+ + Cl-
NH3 + H2O ⇌ NH4+ + OH-

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4
Q

What are some common acids and their conjugate bases?

A 
HCl -> Cl-
HNO3 -> NO3-
H2O -> OH-
NH4+ -> NH3
H3O+ -> H2O
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5
Q

Why is water special?

A

Can act as an acid or a base (accept and donate protons)

- Amphoteric

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6
Q

What is the type of acid linked to?

A

The number of hydrogen ions that can be replaced per molecule in an acid-base reaction

Monobasic acid: HCl & CH3COOH
Dibasic Acid: H2SO4
Tribasic acid: H3PO4

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7
Q

What is the general redox reaction involving acids?

A

Acid + metal -> salt + hydrogen gas (MASH)

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8
Q

What are the 3 neutralisation reactions involving acids?

A

Acid + alkali -> salt + H2O
Acid + carbonate -> salt + H2O + CO2
Acid + metal oxide -> salt + H2O

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9
Q

How can molecules accept protons?

A

Lone pairs of electrons -> dative covalent bonds

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10
Q

What is a strong acid?

A

HA -> H+ + A-

  • strong tendency to donate H+ ions
  • fully dissociate in aqueous solution
  • equilibrium lies to the right
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11
Q

What does the pH scale show?

A

the relationship between pH and hydrogen ion concentration [H+(aq)]

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12
Q

How is pH calculated?

A

pH = -log[H+(aq)]

  • give all answers to 2 decimals places
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13
Q

How can the pH of a strong acid be calculated?

A
  • directly from the conc of the acid because the acid completely dissociates into ions
  • H+ ion conc is equal to the conc of the strong acid in question

HA(aq) -> H+(aq) + A-(aq)
[HA(aq)] = [H+(aq)]

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14
Q

How is the pH changes on dilution calculated (1)?

A

diluted concentration = original moles / volume after dilution

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15
Q

How is the pH changes on dilution calculated (2)?

A

diluted concentration = original conc x (volume before dilution / volume after dilution)

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16
Q

What is a weak acid?

A

HA ⇌ H+ + A-

  • weaker tendency to donate H+
  • partially dissociate in aqueous solution
  • equilibrium lies on the left
  • strength of a weak acid is given by the acid dissociation constant (Ka)
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17
Q

How is Ka calculated?

A

Ka = [H+][A-] / [HA]

units: moldm-3

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18
Q

What is Ka?

A

The acid dissociation constant

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19
Q

What do the Ka values represent?

A
  • The larger the value of Ka, the stronger the acid

- The smaller the value of Ka, the weaker the acid

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20
Q

When does Ka change?

A

Changes with temperature (standard temp is used, 298K)

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21
Q

How is pKa calculated?

A

pKa = -log(Ka)

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22
Q

How is Ka calculated with pKa values?

A

Ka = 10^-pKa

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23
Q

How is [H+(aq)] calculated from Ka?

A

10^-Ka

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24
Q

What do the pKa values indicate about the acid?

A
  • the larger the pKa value, the weaker the acid

- the smaller the pKa value, the stronger the acid

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25
Q

What happens if 2 acids are in the same reaction?

A
  • Weaker acids act as a base and accept the proton

- Stronger acid acts as an acid and donates the proton

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26
Q

What 2 approximations does the weak acid pH calculation require?

A
  1. [H+] = [A-] are equal at equilibrium as the dissociation of water is negligible (produces a very small amount of H+ ions)
  2. [HA]start = [HA]eqm since dissociation eqm lies on the left, little HA dissociates
    - the conc of the undissociated acid is much greater than the [H+] at equilibrium
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27
Q

What happens to the Ka equation when the 2 approximations are applied to it?

A

Ka = [H+]^2 / [HA]

  • concentrations are equal
  • little water dissociates so we can ignore any decrease from dissociation
28
Q

How can Ka of a weak acid be determined experimentally?

A
  • preparing a standard solution of the weak acid of a known concentration [HA]
  • measuring the pH of the standard solution using a pH meter [H+] = 10^-pH
29
Q

When do the approximations break down in validity?

A
  1. breaks down for very weak acids or very dilute solutions (pH>6) as the dissociation of water starts to become significant
  2. breaks down for stronger acids with Ka>10^-2
30
Q

Equation for the ionisation of water?

A

H2O ⇌ H+ + OH-

31
Q

Why is the water dissociation special?

A

Water dissociates so very little that we can say it’s mainly undissociated at equilibrium and is essentially a constant

32
Q

Kw = …?

A 
Kw = [H+][OH-]
Kw = 1 x 10^-14
33
Q

Prove that the pH of pure water is 7.00?

A

[H+] = [OH-] (this is only applies to pure water)
Kw = [H+]^2
square root of 1x10^-14 = 7

34
Q

When is an aqueous solution acidic?

A

[H+] > [OH-]

35
Q

When is an aqueous solution neutral?

A

[H+] = [OH-]

36
Q

When is an aqueous solution alkaline?

A

[H+] < [OH-]

37
Q

How do you calculate the pH of strong bases?

A

A strong base completely dissociates in aqueous solution
NaOH -> Na+ + OH-
[NaOH] = [OH-]
- the conc of OH- is equal to that of NaOH
- NaOH is a monobasic base: each mole of NaOH releases one mole of OH- ions

38
Q

What is a buffer solution?

A

A buffer solution is a system that minimises a change in pH when small amounts of an acid or base are added

39
Q

What is the first way a buffer solution can be made?

A

Made from a weak acid and one of its salts

  • a weak acid and its conjugate base
  • e.g. ethanoic acid (CH3COOH) and sodium ethanoate (CH3COONa) as ethanoic ions are the conjugate base of ethanoic acid
40
Q

What is the second way a buffer solution can be made?

A

Made from a weak base and one of its salts

  • a weak base and its conjugate acid
  • e.g. ammonia solution (NH3) and ammonium chloride (NH4Cl) as NH4+ is a conjugate acid of ammonia)
41
Q

What is the difference between a weak acid and a buffer solution?

A
  • A weak acid partially dissociates into its ions (equilibrium lies of the left), a large reservoir of HA and a small reservoir of A-
  • A buffer solution has a larger reservoir of A- than available from the partial dissociation of HA
  • The large A- reservoir is provided by addition of a aqueous solution of one of its salts or via partial neutralisation of the acid
42
Q

How is a buffer created from a weak acid and its salts?

A

By mixing a solution of ethanoic acid and a solution of its salt sodium ethanoate
CH3COOH (component 1 the weak acid) ⇌ H+ + CH3COO-
equilibrium lies on the left, weak acid
CH3COONa -> Na+ + CH3COO- (component 2 the conjugate base)
complete dissociation of the salt when dissolved in water, becomes source of the conjugate base

43
Q

How is a buffer created from partial neutralisation of the weak acid?

A

Reacting excess ethanoic acid solution with sodium hydroxide (adding an aqueous solution of an alkali to an excess of weak acid)
CH3COOH (in excess) + NaOH (an alkali) -> CH3COONa + H2O
- CH3COONa provides CH3COO-, component 2 for conjugate base
- since ethanoic acid is in excess there will be some left unreacted after the neutralisation reaction
CH3COOH (component 1) ⇌ H+ + CH3COO-

44
Q

How do buffer solutions work?

A
  • The weak acid partially dissociates into ions, large reservoir of HA
  • The salt MA dissociates completely into ions generating a large reservoir the conjugate base A-
  • The mixture of the 2 gives a large reservoir of HA from the acid and a large reservoir of A- ions from the salt
45
Q

How do buffer solutions work in addition of acidic substances?

A
  • [H+(aq)] increases
  • H+ ions react with the conjugate base A-
  • The equilibrium position shifts to the left to reduce the concentration of H+ ions
  • Fall in pH is prevented as [H+(aq)] changes very little
46
Q

How do buffer solutions work in addition of alkaline substances?

A
  • [OH-(aq)] increases
  • The small concentration of H+(aq) ions from the weak acid react with OH- ions forming water
  • This decreases the H+ concentration
  • The equilibrium position moves to the right to restore the concentration of H+ ions and pH
  • HA molecules dissociate to restore [H+] and rise in pH is prevented
47
Q

When are buffer solutions most effective?

A

If there are equal concentrations of the weak acid and conjugate base [HA] = [A-], therefore…
- the pH of the buffer solution is the same as the pKa value of HA

48
Q

What does the pH of a buffer depend on?

A
  1. The value of Ka of the weak acid, main effect = ‘coarse tuning’ of the buffer, determines the pH region
  2. The ratio of acid to salt or acid to conjugate base (conc), small effect = ‘fine tuning’ adjusts pH to actual value required
49
Q

How do you calculate the pH of a buffer solution?

A 
[H+] = Ka x ( [HA] / [A-] )
pH = -log[H+]
50
Q

What happens to the pH value calculation when conc of acid = conc of salt in a buffer solution?

A

[H+] = Ka

so pH = pKa

51
Q

How do you calculate the pH of buffer solution made by partial neutralisation?

A
  1. Write a balanced equation for neutralisation reaction (+ mole ratios)
  2. Neutralisation reactions…
    a) calculate the number of moles of reactants
    b) determine what chemical is in excess (usually an acid)
    c) determine the number of moles of the salt (a product)
    d) determine number of moles of acid left present in buffer solution (HA start - HA reacted = HA present in buffer)
    e) calculate con of HA and A- in buffer solution (c = n/v)
  3. Calculate the pH of buffer solution
52
Q

What is the main use for buffers in the body?

A

pH of blood plasma being maintained between 7.35 and 7.45 by a natural buffer system

53
Q

What is the buffer system that is involved in blood plasma?

A

Carbonic acid-hydrogen carbonate buffer system
H2CO3(aq) ⇌ HCO3^-(aq) + H+(aq)
carbonic acid ⇌ hydrogen carbonate + proton

54
Q

What happens if the pH of the blood plasma decreases below 7.35?

A
  • if the level drops below pH 7.35 we develop a condition called acidosis.
  • having to breathe more rapidly to expel the acidic gas CO2
55
Q

What happens if the pH of the blood plasma rises above 7.45?

A

a rise above pH 7.45 leads to a condition called alkalosis
which can cause muscle spasms, light-headedness and
nausea.

56
Q

What happens in the blood plasma on addition of an acid/H+?

A
  • [H+] increases
  • H+ ions react with the conjugate base HCO3- forming H2CO3
  • equilibrium position shifts to the left, removing extra H+ ions and restoring [H+] and pH
57
Q

What happens in the blood plasma on addition of an alkali/OH-?

A
  • OH- ion conc increases
  • H+ ions react with OH- ions to form water
  • H2CO3 dissociates, equilibrium shifts to the right to restore H+ ion concentration and pH
58
Q

How is a build up of H2CO3 prevented?

A
  • Most materials released into the blood are acidic and the hydrogen carbonate ions effectively remove these and resist a change in blood plasma pH.
  • A build-up of the weak acid carbonic acid, H2CO3, is prevented as it is converted to carbon dioxide and exhaled by the lungs: H2CO3 ⇌ CO2 + H2O
59
Q

What are indicators?

A

Indicators are weak acids, acid and the conjugate base have different colours
HA(colour 1) ⇌ H+ + A-(colour 2)

60
Q

Explain how the Methyl Orange indicator works?

A

At the end point of the titration, the indicator contains equal concentrations of both HA and A- and appears orange
HA (red) ⇌(eqm is orange) H+ + A-(yellow base colour)

61
Q

What happens when an indicator is added to an acid?

A
  • [H+ (aq)] increases

* Equilibrium position moves to the left, colour 1 is seen

62
Q

What happens when an indicator is added to an alkali?

A
  • OH- react with H+(aq) in equilibrium (to form H2O)
  • [H+(aq)] decreases
  • equilibrium position moves to the right, colour 2 is seen
63
Q

What happens at an end point?

A

At the end point: indicator contains equal concentrations of HA and A-

64
Q

Calculating the pH of indicators

A

since at the END POINT ONLY [HA] = [A-]
Ka = [H+]
-logKa = -log[H+]
pKa = pH

65
Q

What is the equivalence point?

A

The volume of one solution that exactly reacts with a measured volume of the second solution. The amounts used match the stoichiometry of the reaction.

66
Q

Describe the pH titration curve: strong acid + strong base

A
  • pH rises slowly, then rises steeply just before equivalence point
  • Volume at equivalence point = Volume of alkali required to react exactly with volume of acid
  • equivalence of = centre of vertical section, pH = 7
    [H+(aq)] = [OH-(aq)]
67
Q

Describe the pH titration curve: strong acid + weak base

A
  • pH rises slowly, then rises steeply just before equivalence point

A-Level OCR Chemistry (21 decks)

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