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Chemistry A2 > Acids and Bases > Flashcards

Acids and Bases Flashcards

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1
Q

Brønsted–Lowry Acid

A

proton donor

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2
Q

Brønsted–Lowry Base

A

proton acceptor

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3
Q

Amphoteric

A

Species that can behave as both acid and base

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4
Q

Conjugate Acid

A

Species created when a base accepts a proton

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5
Q

Conjugate Base

A

Ion or molecule remaining after an acid has lost a proton

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6
Q

Strong Acid

A

Fully ionises to form H+ in water

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7
Q

Weak Acid

A

Partially ionises to form H+ in water

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8
Q

How to determine pH of strong acid

A

[H+]= concentration x molar ratio (based whether acid is mono/di/triprotic)

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9
Q

How to determine pH of weak acid

A

Rearrange Acid Dissociation Constant Equation

Ka = [H+][A-]/[HA] => [H+]^2/[HA]

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10
Q

Assumptions for Ka

A
  • safe to neglect H+ ions produced by ionisation of water so [H+] = [A-]
  • [HA]initial = [HA]equilibrium
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11
Q

Justify assumptions for Ka

A
  • dissociation of acid is large enough to ignore dissociation of water
  • dissociation of acid is small enough that change in concentration of acid at equilibrium can be ignored
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12
Q

Suggest when assumptions for Ka become a greater source of error

A
  • neglecting dissociation of water if weaker acid used

- concentration of acid at equilibrium does not change if stronger acid used

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13
Q

Suggest the need for pKa values

A
  • determine the strength of weak acids

- distinguish between strong acid and concentrated weak acid

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14
Q

Suggest how to interpret pKa

A

Same as pH but for weak acids only, i.e. smaller pKa implies stronger acid

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15
Q

State what is true at half neutralisation point

A

Equilibrium [A-] = [HA] so pH = pKa according to Henderson-Hasslebalch equation

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16
Q

Suggest why ionisation of water is actually 2H2O => H3O+ + OH-

A

H+ is rarely found isolated due to its high charge density so associates with a water molecule

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17
Q

Suggest why the pH of water can change but it always remains neutral

A
  • position of equilibrium for ionisation of water can shift so [H+] can change
  • [H+] and [OH-] remains the same so is always neutral
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18
Q

Suggest why pH scale is logarithmic

A

Covers wide range in [H+]

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19
Q

How to determine pH of strong base

A

Rearrange Ionic Product of Water Equation with known [OH-] from concentration of base
Kw = [H+][OH-] or Kw=[H+]^2 for water
1x10^-14

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20
Q

Suggest why [H2O] is excluded from Kw equation

A

Water is in huge excess compared to [H+] and [OH-] so is considered constant

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21
Q

State assumption for Kw

A

Strong base dissociates completely so [OH-] is proportional to concentration of base

22
Q

Indicator

A

A weak acid or base which changes colour depending on pH of solution if is found in

23
Q

Equivalence Point

A

Volume of acid or alkali added at which point [H+]=[OH-]

24
Q

End Point

A

Volume of acid or alkali added when indicator just changes colour

25
Q

Suggest properties of a suitable indicator for a titration

A
  • colour change must be sharp rather than gradual
  • distinct colour change
  • same result for equivalence and end point
  • changes colour within pH range for equivalence point
26
Q

State what is true at half neutralisation point for indicators

A

Kin= [H+] so pKin = pH

27
Q

Buffer

A
  • solution that resists changes in pH
  • on addition of small amounts of acid or alkali
  • or dilution
28
Q

Acidic Buffer

A
  • WEAK acid and salt of acid

- maintains pH below 7

29
Q

Basic Buffer

A
  • WEAK base and salt of base

- maintains pH above 7

30
Q

Describe what happens when alkali is added to a buffer

A
  • OH- reacts with H+ from conjugate acid
  • [H+] decreases
  • equilibrium shifts to right to replace H+
31
Q

Describe what happens when acid is added to a buffer

A
  • [H+] increases
  • H+ reacts with conjugate base
  • equilibrium shifts to left to remove H+
32
Q

Suggest why in equation for Ka of buffer solution [H+] ≠ [A-]

A

A- introduced from salt of acid and not dissociation alone

33
Q

Assumptions made for Ka of buffers

A
  • All A- come from salt

- [HA]initial = [HA]equilibrium

34
Q

Justify assumptions for Ka of buffers

A
  • dissociation of weak acid is small enough to ignore

- addition of salt means equilibrium lies further to left so less acid dissociates

35
Q

Suggest why pH of buffer remains unaffected by dilution

A

Acid and salt are diluted equally so ratio of [A-]/[HA] remains the same

36
Q

Henderson-Hasslebalch Equation

A

pH = pKa + log [A-]/[HA]

37
Q

Suggest why a buffer is not formed if the same number of moles of acid and alkali are combined

A
  • neutralised

- only salt and water formed

38
Q

State what is needed for a buffer to be formed

A
  • acid/alkali in excess

- conjugate base present

39
Q

How to determine if an indicator is suitable for a titration when given its pH range for colour change

A
  • type of acid-base titration
  • pH at equivalence is > 7 using sketch
  • compare to range indicator changes colour
40
Q

How to derive Kw equation from Kc of water

A

Kc = [H+][OH-]/[H2O]
Kc * [H2O] =[H+][OH-]
since H2O is in huge excess it is considered a constant
Kw = [H+][OH-]

41
Q

State an essential feature of an acid-base reaction

A

Transfer of protons

42
Q

Suggest why a strong acid cannot be used to make a buffer solution

A
  • ionise completely

- buffer requires both acid and conjugate base

43
Q

Suggest why a buffer stops working if too much acid or alkali is added

A

Acid/Conjugate base runs out

44
Q

How to work out volume of water added to change pH by known amount in dilution of acid

A

C1V1=C2V2 since moles of H+ does not change

V2 - V1 = volume of water added

45
Q

Suggest why indicators cannot be used in weak acid - weak base titration

A
  • no sharp change in pH during weak acid - weak base titration
  • indicator needs sharp change in pH to change colour quickly
46
Q

Key points for acid-base calculations

A
  • do not use C1V1=C2V2 for neutralisation but instead write out molar equation ALWAYS
  • do not put moles straight into equations but instead convert to concentration
47
Q

Suggest how to prepare a buffer from equal concentrations of monoprotic acid and strong base where pH = pKa

A
  • pH = pKa at half neutralisation point

- add volumes in 2:1 ratio with excess acid

48
Q

When do buffers work best

A
  • pH = pKa
  • since [A-] = [HA]
  • prevents change in pH in either direction
49
Q

Describe how to calculate pH when acid is added to alkali

A
  • write out balanced equation for reaction ALWAYS
  • work out moles of acid and alkali reacting
  • determine whether acid or alkali is in excess
  • when alkali in excess work out pH using Kw with excess [OH-]
  • when acid is excess CHECK IF BUFFER SOLUTION IS MADE - DO NOT work out pH of excess weak acid
  • if buffer solution if formed with excess acid then determine concentration of salt and excess acid and use Henderson-Hasselbalch equation
50
Q

Value for Kw at 298K

A

1x10-14

Chemistry A2 (12 decks)

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