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Chemistry > 21. Acids, bases and buffers > Flashcards

21. Acids, bases and buffers Flashcards

1
Q

What is the Bronsted-Lowry definition of an acid

A

An acid is a substance that can donate a proton (H+ ion)

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2
Q

What is the Bronsted-Lowry definition of a base

A

A base is a substance that can accept a proton

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3
Q

What is H3O+ called

A

Hydronium ion

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4
Q

Name a reaction where water acts as a base

A

HCl + H2O —> H3O + +Cl-

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5
Q

Name a reaction where water acts as an acid

A

H2O + NH3 —> OH- + NH4+

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6
Q

What is the equation for the equilibrium established in water

A

H2O (l) H+ (aq) + OH- (aq)

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7
Q

Write the equation for the Kc of the equilibrium

H2O (l) H+ (aq) + OH- (aq)

A

Kc=[H+ (aq)][OH- (aq)] / [H2O (l)]

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8
Q

What is the equation for Kw

A

Kw= [H+ (aq)] [OH- (aq)]

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9
Q

What is Kw (the ionic product of water) at 298K

A

1.0 x 10^-14 mol^2dm^-6

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10
Q

What is [H+ (aq] at 289K)

A
  • H2O dissosciates to one [H+] and one [OH-]
  • so [H+ (aq)] = [OH- (aq)]
  • so 1.0 x 10^-14 =[H+ (aq)]^2
  • so [H+ (aq)] = 1.0 x 10^-7
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11
Q

pH=

A

pH= -log10[H+ (aq)]

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12
Q

The smaller the pH…

A

The greater the concentration of H+ (aq)

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13
Q

Why is the pH of water 7

A
  • [H+ (aq)] = [OH- (aq)]
  • so 1.0 x 10^-14 = [H+ (aq)]^2
  • so [H+ (aq)] = 1.0 x 10^-7
  • pH= -log10[1.0 x 10^-7] = 7.00
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14
Q

To what decimal place should you quote pH to

A

2 decimal places

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15
Q

Write an equation for how a weak acid disassociates

A

HA (aq) = H+ (aq) + A-

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16
Q

Write the equation for Ka

A

[H+ (aq)]eqm [A-(aq)]eqm / [HA (aq)]eqm

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17
Q

The larger the value of Ka…

A

The more dissociated and stronger it is

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18
Q

What is pKa

A

pKa=-log10 Ka

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19
Q

How should you find the pH of a strong acid

A
  • find [H+]

- use pH= -log10[H+ (aq)]

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20
Q

How should you find the pH of an alkaline solution

A
  • find [OH-]
  • use -[H+ (aq)] = 1.0 x 10^-14 / [OH- (aq)]
  • to find find [H+]
  • use pH= -log10[H+ (aq)]
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21
Q

How should you find the pH of a diluted acid

A
  • [H+] = [H+]old x (old volume/new volume)

- use pH= -log10[H+ (aq)]

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22
Q

How should you find the pH of a diluted base

A
  • [OH-] = [OH-]old x (old volume/new volume)
  • use -[H+ (aq)] = [OH- (aq)] = 1.0 x 10^-14
  • to find find [H+]
  • use pH= -log10[H+ (aq)]
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23
Q

How should you find the pH of a weak acid

A

-[H+ (aq)]eqm [A-(aq)]eqm / [HA (aq)]eqm
-[H+] = s.r(Ka x[HA]
use pH= -log10[H+ (aq)]

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24
Q

How should you find the pH of a half neutralised weak acid

A
  • Ka= [H+]

- as [HA]= [A-]

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25
Q

How should you find the pH of strong acid + base

A
  • find moles of [H+]
  • find moles of [OH-]
  • find XS [H+] or [OH-]
  • find pH of the XS
26
Q

How should you find the pH of weak acid + base

A
  • find moles of [HA]
  • find moles of [OH-]
  • find XS [HA] or [OH-]
  • find pH of the XS
27
Q

Describe the titration curve for a strong acid and a strong base

A

-S shaped curve

28
Q

Describe the titration curve for a strong acid and a weak base

A
  • S shaped curve

- however moved down y axis

29
Q

Describe the titration curve for a weak acid and a weak base

A
  • Shaped curve
  • however more condensed towards middle
  • steep change at the start of the reaction
30
Q

Describe the titration curve for a weak acid and a strong base

A
  • S shaped curve
  • however moved up y axis
  • steep change at the start of the reaction
31
Q

What is the equivalence point

A

The point in a titration at which the reaction is just complete

32
Q

What makes an indicator suitable for a titration

A
  • colour change must be sharp to give sharp end point
  • end point given by indicator must be the same as the equivalence point
  • distinct colour change
33
Q

What is a suitable indicator for strong acid and strong base

A

Phenolphthalein or methyl orange, as it changes within the vertical portion of the pH curve

34
Q

What is a suitable indicator for weak acid and strong base

A

Phenolphthalein, as it changes within the vertical portion of the pH curve

35
Q

What is a suitable indicator for weak acid and weak base

A

No indicator, as it changes within the vertical portion of the pH curve

36
Q

What is a suitable indicator for strong acid and weak base

A

Methyl orange, as it changes within the vertical portion of the pH curve

37
Q

Where is the half neutralisation point on a graph

A

Point half way between 0 and the equivalence point (pH)

38
Q

Explain why at the half neutralisation point pKa=pH

A
  • [HA]=[A-]
  • Ka= [H+ (aq)]eqm [A-(aq)]eqm / [HA (aq)]eqm
  • Ka= [H+]
  • pKa=pH
39
Q

What are buffers

A

Solutions that resist the changes of pH when small amounts of acid or alkali are added to them

40
Q

What are acid buffers

A
  • made from weak acids and a soluble salt of that acid
  • resist change but maintain pH below 7
  • work because the disassociation of a weak acid is an equilibrium reaction
41
Q

What happens when you add an alkali to an acid buffer

A

HA (aq) + OH- (aq) —> H2O (aq) + A- (aq)

Removes OH- to produce water and salt

42
Q

What is a Bronsted-Lowry acid-base reaction

A

A reaction involving the transfer of a proton

43
Q

What is a monoprotic acid

A

An acid that releases one H+ ion per molecule

44
Q

What is a diprotic acid

A

An acid that releases two H+ ions per molecule

45
Q

What is Kw

A

The ionic product of water

46
Q

Why does Kw = 1 x 10^14 or [H+][OH-]

A
  • H20 —> H+ + OH-
  • so Kc = [H+][OH-]/[H2O]
  • so Kc[H2O]= [H+][OH-]
  • [H2O] is a far greater number than [H+] and [OH-] and therefore can be treated as a constant
  • so Kc [H2O] = constant Kw
  • Kw= [H+][OH-] = 1x10^14
47
Q

What is the active ingredient in household bleach

A

chloric acid (I) (HClO)

48
Q

What is the active ingredient in acid-based cleaners

A

hydrochloric acid (HCl)

49
Q

Why can you not mix bleach and acid based cleaners?

A
  • chlorine gas is produced

- HClO (aq) + HCl (aq) —-> Cl2 (g) + H2O (aq)

50
Q

Is the equilibrium reaction H2O (l) H+ (aq) + OH- (aq) endothermic or exothermic

A

Endothermic

51
Q

How is Kw affected by an increase in temperature

A
  • it increases
  • H20 H+ + OH- is endothermic
  • equilibrium shifts right
  • Kw=[H+][OH-]
  • therefore increase in Kw
52
Q

How is the pH of water effected by an increase in temperature

A
  • it decreases
  • H20 H+ + OH- is endothermic
  • equilibrium shifts right
  • increase in [H+]
  • therefore decrease in pH
53
Q

What happens when you add an acid to an acidic buffer

A
  • H+ is added so equilibrium shifts to the left meaning H+ ions are removed, pH change is therefore resisted
  • A- + H+ —> HA
54
Q

Why is a salt necessary in acidic buffers

A
  • A- is needed to combine with the H+ to remove it
  • but there is very little A- so it will run out
  • adding a soluble salt of the solution means it wont run out
55
Q

What are basic buffers

A
  • made from weak base and a soluble salt of that base

- resist change but maintain pH above 7

56
Q

What is an example of a system involving a buffer

57
Q

Give an example of a basic buffer

58
Q

What happens when you add H+ to a basic buffer

A
  • ammonia removes added H+

- NH3 + H+ —->NH4+

59
Q

What happens when you add OH- to a basic buffer

A
  • the ammonium ion removes the OH-

- NH4+ + OH- —-> NH3 + H2O

60
Q

What two assumptions do you make during buffer calculations

A

[A-] = [salt] since salt is fully ionised and HA barely disassociates (no extra A-)
[HA] equilibrium = [HA] initial since little of the weak acid is ionised

61
Q

How to find pH change when acid is added to a buffer

A
  • initial pH
  • find initial moles of A- and HA (without acid added)
  • find number of moles of H+ ions added
  • new A- = A- initial - H+ mol (as H+ + A- —> HA)
  • new HA = HA initial + H+mol (as H+ + A —> HA)
  • Ka= [H+ (aq)]eqm [A-(aq)]eqm / [HA (aq)]eqm
  • pH= -log10[H+ (aq)]
62
Q

How to find pH change when base is added to a buffer

A
  • initial pH
  • find initial moles of A- and HA (without acid added)
  • find number of moles of OH- ions added
  • new HA = HA inital - OH- moles (as H+ + OH- H2O and H+ + A- HA)
  • new A- = A- inital + OH- moles (as (H+ + OH- H2O and H+ + A- HA)
  • Ka= [H+ (aq)]eqm [A-(aq)]eqm / [HA (aq)]eqm
  • pH= -log10[H+ (aq)]

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