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
How should you find the pH of strong acid + base
- find moles of [H+] - find moles of [OH-] - find XS [H+] or [OH-] - find pH of the XS
26
How should you find the pH of weak acid + base
- find moles of [HA] - find moles of [OH-] - find XS [HA] or [OH-] - find pH of the XS
27
Describe the titration curve for a strong acid and a strong base
-S shaped curve
28
Describe the titration curve for a strong acid and a weak base
- S shaped curve | - however moved down y axis
29
Describe the titration curve for a weak acid and a weak base
- Shaped curve - however more condensed towards middle - steep change at the start of the reaction
30
Describe the titration curve for a weak acid and a strong base
- S shaped curve - however moved up y axis - steep change at the start of the reaction
31
What is the equivalence point
The point in a titration at which the reaction is just complete
32
What makes an indicator suitable for a titration
- 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
What is a suitable indicator for strong acid and strong base
Phenolphthalein or methyl orange, as it changes within the vertical portion of the pH curve
34
What is a suitable indicator for weak acid and strong base
Phenolphthalein, as it changes within the vertical portion of the pH curve
35
What is a suitable indicator for weak acid and weak base
No indicator, as it changes within the vertical portion of the pH curve
36
What is a suitable indicator for strong acid and weak base
Methyl orange, as it changes within the vertical portion of the pH curve
37
Where is the half neutralisation point on a graph
Point half way between 0 and the equivalence point (pH)
38
Explain why at the half neutralisation point pKa=pH
- [HA]=[A-] - Ka= [H+ (aq)]eqm [A-(aq)]eqm / [HA (aq)]eqm - Ka= [H+] - pKa=pH
39
What are buffers
Solutions that resist the changes of pH when small amounts of acid or alkali are added to them
40
What are acid buffers
- 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
What happens when you add an alkali to an acid buffer
HA (aq) + OH- (aq) ---> H2O (aq) + A- (aq) | Removes OH- to produce water and salt
42
What is a Bronsted-Lowry acid-base reaction
A reaction involving the transfer of a proton
43
What is a monoprotic acid
An acid that releases one H+ ion per molecule
44
What is a diprotic acid
An acid that releases two H+ ions per molecule
45
What is Kw
The ionic product of water
46
Why does Kw = 1 x 10^14 or [H+][OH-]
- 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
What is the active ingredient in household bleach
chloric acid (I) (HClO)
48
What is the active ingredient in acid-based cleaners
hydrochloric acid (HCl)
49
Why can you not mix bleach and acid based cleaners?
- chlorine gas is produced | - HClO (aq) + HCl (aq) ----> Cl2 (g) + H2O (aq)
50
Is the equilibrium reaction H2O (l) H+ (aq) + OH- (aq) endothermic or exothermic
Endothermic
51
How is Kw affected by an increase in temperature
- it increases - H20 H+ + OH- is endothermic - equilibrium shifts right - Kw=[H+][OH-] - therefore increase in Kw
52
How is the pH of water effected by an increase in temperature
- it decreases - H20 H+ + OH- is endothermic - equilibrium shifts right - increase in [H+] - therefore decrease in pH
53
What happens when you add an acid to an acidic buffer
- H+ is added so equilibrium shifts to the left meaning H+ ions are removed, pH change is therefore resisted - A- + H+ ---> HA
54
Why is a salt necessary in acidic buffers
- 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
What are basic buffers
- made from weak base and a soluble salt of that base | - resist change but maintain pH above 7
56
What is an example of a system involving a buffer
the blood
57
Give an example of a basic buffer
NH4+Cl-
58
What happens when you add H+ to a basic buffer
- ammonia removes added H+ | - NH3 + H+ ---->NH4+
59
What happens when you add OH- to a basic buffer
- the ammonium ion removes the OH- | - NH4+ + OH- ----> NH3 + H2O
60
What two assumptions do you make during buffer calculations
[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
How to find pH change when acid is added to a buffer
- 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
How to find pH change when base is added to a buffer
- 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)]