3.1.12 Acids & Bases Flashcards

1
Q

what is meant by a ‘Bronsted-Lowry acid’?

A

proton donor

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2
Q

what is meant by a ‘Bronsted-Lowry base’?

A

proton acceptor

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3
Q

what makes an acid strong? what equation can be used to show this?

A
  • strong acid fully dissociates/ionises in solution to release a high [H+]
  • HA + H2O -> H3O+ + A-
    (H3O+ = H+; can be used interchangeably)
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4
Q

what makes an acid weak and what equation can be used to show this?

A
  • weak acids partially dissociate in solution to release low [H+]
  • HA + H2O ⇌ H3O+ + A-
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5
Q

how to name conjugate acids and bases?

A

HA + H2O -> H3O- + A-
- HA = acid
- H2O = base
- H3O- = conjugate acid
- A- = conjugate base

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6
Q

how to calculate pH or [H+]?

A

pH = -log[H+]
[H+] = 10 to the power of -pH

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7
Q

when calculating the pH of strong acids, what do you need to remember?

A

[H+] = HA because there is full dissociation

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8
Q

when calculating the pH of weak acids, what assumptions are made?

A
  1. initial conc. of acid = equilibrium conc. of acid
  2. @ equilibrium, [H+] = [A-]
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9
Q

how do you calculate ka (acid dissociation constant)?

A

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

  • when it is a weak acid [H+] = [A-]
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10
Q

what is ‘pka’ and how do you calculate it?

A

pka determines the strength of an acid
- pka = -log(ka)
- ka = 10 to the power of -pka

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11
Q

how do you use ka and pka to find strengths of weak acids?

A
  • the higher the ka value, the stronger the weak acid because the equilibrium lies more to the right so there is a higher [H+]
  • pka is the inverse of ka, so the higher the pka value the weaker the acid
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12
Q

what is the formula for kw (ionic product of water)?

A

kw = [H+][OH-]

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13
Q

what is the kw of water at standard temp. & pressure?

A

1x10 to the power of -14
units: mol²dm^-6

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14
Q

why does the pH of water change with temperature?

A

H2O ⇌ H+ + OH-
- when temperature increases, equilibrium position shifts to the endothermic side (forward reaction is favoured)
- [H+] increases, so pH decreases

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15
Q

why is water neutral at all pHs?

A

because [H+] = [OH-]

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16
Q

how is an acid buffer made?

A
  • react excess acid with strong alkali e.g. NaOH
  • all the NaOH is neutralised so the remaining acid and salt is left in the buffer
16
Q

what is a buffer solution?

A

a solution that resists changes in pH when small amounts of acid and alkali are added

17
Q

what does a buffer consist of?

A

a mixture of a weak acid and the salt of its conjugate base

18
Q

how do buffer solutions work? use the symbol equation as an example:
CH3COOH ⇌ CH3COO- + H+

A
  • **added OH- **-> pH will increase
  • the added OH- will react with the H+ (H+ and OH- -> H2O)
  • the equilibrium position will shift to the right so [H+] increases so pH decreases back down.
  • added H+ -> pH will decrease
  • equilibrium position will shift to the left so [H+] decreases, so pH increases back up.
19
Q

how do you obtain a pH curve? (explain the calibration process)

A
  • rinse pH probe w/ distilled water
  • dip pH probe into a buffer solution
  • wash pH probe in distilled water AGAIN
  • dip pH probe in buffer solution w/ a different pH
  • calibrate a pH probe and use to measure the initial pH of the alkali in the conical flask
19
Q

how do you obtain a pH curve? (explain the steps post-calibration)

A
  • add acid from the burette in 2cm³ increments and swirl the mixture
  • record pH after every addition; reduce size of the portions of the increments close to the endpoint because the pH would change suddenly
  • repeat until the acid is in excess
  • plot a graph of pH vs. volume (of acid/alkali)
20
Q

what does a strong acid + strong base pH curve look like? (e.g. HCl and NaOH)

A
  • very large vertical section
  • the pH at the equivalence point is 7 because there is full dissociation for both acid + alkali so [H+] = [OH-]
21
Q

what does a strong acid + weak base pH curve look like? e.g. HCl and NH3

A
  • relatively large vertical section
  • pH at equivalence point is LESS than 7. acid has full dissociation and base has partial dissociation, so [H+] > [OH-]
22
Q

what does a strong base + weak acid pH curve look like? e.g. NaOH + HCOOH

A
  • equivalence point pH >7
  • [OH-] = full dissociation; [H+] = partial dissociation
  • vertical section is relatively small
23
Q

what does a weak acid + weak base pH curve look like? e.g. NH3 and HCOOH

A
  • no vertical section due to little dissociation
24
Q

how does an indicator work in the presence of an acid?
H(In) ⇌ H+ + In-
(colour A) (colour B)

A
  • equilibrium position shifts to the left because added H+ reacts with In-
  • colour A is observed
25
Q

how does an indicator work in the presence of an alkali?
H(In) ⇌ H+ + In-
(colour A) (colour B)

A
  • the OH- reacts w the H+ so equilibrium position shifts to the right
  • colour B is observed
26
Q

when does the endpoint occur?

A
  • indicator contains equal [H(In)] and [In-].
  • the colour between the 2 extreme colours of the indicator are observed
27
Q

what is meant by ‘equivalence point’?

A

volume of one solution that exactly reacts with the volume of another solution

28
Q

when should an indicator change colour?

A

it should change colour at a pH range that lies within the vertical section of the pH curve.
- they change colour within a range of minimum of 2 pH units.

29
Q

why is there no indicator for a weak acid + weak base?

A

there is no vertical section.

30
Q

how do you use pH curves to determine ka/pka of a weak acid?

A
  • at 1/2 equivalence point, [A-] = [HA], meaning ka = [H+]
31
Q

(4) explain how a pH curve can be used to determine the ka of an acid.

A
  • find equivalence point on the pH curve and read off hte volume added.
  • half this value to find 1/2 equivalence point
  • read off the pH at 1/2 equivalence point - this is the pka
  • ka = 10 to the power of -pka
32
Q

what are the uses of buffers?

A
  • cosmetic industries - prevents changes in pH so prevents skin irritation
  • blood buffer (made up of H2CO3 + hydrogencarbonate)
33
Q

explain how the blood maintains a pH of 7.4 after vigorous exercise.
H2CO3 ⇌ H+ + HCO3-

A
  • H+ will increase (build up of lactic acid), resulting in a fall in pH, so equilibrium position will shift to the left to oppose the change
  • [H+] will decrease, which increases pH back up to 7.4
34
Q

explain why the concentration of water is not used to calculate ka.

A

[H2O] is always constant