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Chemistry A2 > Acids and Bases > Flashcards

Acids and Bases Flashcards

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
Suggest properties of a suitable indicator for a titration
- 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
State what is true at half neutralisation point for indicators
Kin= [H+] so pKin = pH
27
Buffer
- solution that resists changes in pH - on addition of small amounts of acid or alkali - or dilution
28
Acidic Buffer
- WEAK acid and salt of acid | - maintains pH below 7
29
Basic Buffer
- WEAK base and salt of base | - maintains pH above 7
30
Describe what happens when alkali is added to a buffer
- OH- reacts with H+ from conjugate acid - [H+] decreases - equilibrium shifts to right to replace H+
31
Describe what happens when acid is added to a buffer
- [H+] increases - H+ reacts with conjugate base - equilibrium shifts to left to remove H+
32
Suggest why in equation for Ka of buffer solution [H+] ≠ [A-]
A- introduced from salt of acid and not dissociation alone
33
Assumptions made for Ka of buffers
- All A- come from salt | - [HA]initial = [HA]equilibrium
34
Justify assumptions for Ka of buffers
- dissociation of weak acid is small enough to ignore | - addition of salt means equilibrium lies further to left so less acid dissociates
35
Suggest why pH of buffer remains unaffected by dilution
Acid and salt are diluted equally so ratio of [A-]/[HA] remains the same
36
Henderson-Hasslebalch Equation
pH = pKa + log [A-]/[HA]
37
Suggest why a buffer is not formed if the same number of moles of acid and alkali are combined
- neutralised | - only salt and water formed
38
State what is needed for a buffer to be formed
- acid/alkali in excess | - conjugate base present
39
How to determine if an indicator is suitable for a titration when given its pH range for colour change
- type of acid-base titration - pH at equivalence is > 7 using sketch - compare to range indicator changes colour
40
How to derive Kw equation from Kc of water
Kc = [H+][OH-]/[H2O] Kc * [H2O] =[H+][OH-] since H2O is in huge excess it is considered a constant Kw = [H+][OH-]
41
State an essential feature of an acid-base reaction
Transfer of protons
42
Suggest why a strong acid cannot be used to make a buffer solution
- ionise completely | - buffer requires both acid and conjugate base
43
Suggest why a buffer stops working if too much acid or alkali is added
Acid/Conjugate base runs out
44
How to work out volume of water added to change pH by known amount in dilution of acid
C1V1=C2V2 since moles of H+ does not change | V2 - V1 = volume of water added
45
Suggest why indicators cannot be used in weak acid - weak base titration
- no sharp change in pH during weak acid - weak base titration - indicator needs sharp change in pH to change colour quickly
46
Key points for acid-base calculations
- 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
Suggest how to prepare a buffer from equal concentrations of monoprotic acid and strong base where pH = pKa
- pH = pKa at half neutralisation point | - add volumes in 2:1 ratio with excess acid
48
When do buffers work best
- pH = pKa - since [A-] = [HA] - prevents change in pH in either direction
49
Describe how to calculate pH when acid is added to alkali
- 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
Value for Kw at 298K
1x10-14

Chemistry A2 (12 decks)

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