Acids+ Bases Flashcards

(38 cards)

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

Neutralisation of an acid by an alkali

A

H3O+(aq) + OH-(aq) -> 2H2O(l)

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

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4
Q

Monobasic acid

A

1 hydrogen ion can be replaced per molecule e.g. HCl

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5
Q

Dibasic acid

A

2 hydrogen ions can be replaced per molecule e.g. H2CO3

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6
Q

Tribasic acid

A

3 hydrogen ions can be replaced per molecule e.g. H3BO3

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7
Q

Redox reactions between acids and metals

A

Acid + metal -> salt + hydrogen

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8
Q

Neutralisation with carbonates

A

Acid + carbonate -> salt + water + carbon dioxide

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9
Q

Neutralisation with metal oxides

A

Acid + base -> salt +water

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10
Q

Neutralisation with alkalis

A

Acid + alkali -> salt + water

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11
Q

pH < 7

A

Acidic (red)

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12
Q

pH = 7

A

Neutral (green)

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13
Q

pH > 7

A

Alkaline (blue)

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14
Q

Low value of [H+]

A

High pH

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15
Q

High value of [H+]

A

Low pH

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16
Q

pH calculation

A

-log [H+(aq)]

17
Q

Calculation for [H+] (strong acids)

18
Q

Strong acid

A

Completely dissociates in aqueous solution

19
Q

Weak acid

A

Only partially dissociates in aqueous solution

20
Q

What is Ka

A

Acid dissociation constant

21
Q

Ka calculation

A

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

22
Q

When does Ka change

A

With temperature (standard = 25degrees)

23
Q

pKa in terms of Ka

A

pKa = - log Ka

24
Q

Ka in terms of pKa

25
Stronger acid Ka and pKa
Larger Ka and smaller pKa
26
Weak acid Ka and pKa
Smaller Ka and larger pKa
27
Weak acid Ka calculation
[H+(aq)]^2/ [HA(aq)]
28
2 assumptions when calculation pH of weak acids
1) neglect H+ ions from dissociation of water [H+]eqm = [A-]eqm 2) [HA]eqm = [HA]initial
29
pH of weak acid calculation
[H+(aq)] = root (Ka x [HA(aq)] pH= -log[H+]
30
Determining Ka experimentally
- prepare standard solution of weak acid of known concentration - measure the pH of the standard solution using a pH meter
31
Justifying weak acid assumptions
1) at 25degrees [H+] from dissociation of water = 10^-7. pH>6 = significant [H+] 2) not justified for stronger weak acids with Ka>10^-2 moldm^-3
32
Kw
``` Ionic product of water -ions in water multiplied - [H+] [OH-] -1 x 10^-14 = [H+]^2 ```
33
[H+(aq)] > [OH-(aq)]
acidic solution
34
[H+(aq)] = [OH-(aq)]
Neutral solution
35
[H+(aq)] < [OH-(aq)]
Alkaline solution
36
pH calculation for strong bases
[H+] = Kw/[OH-] pH= -log[H+]
37
[H+] for pH x
10^-x moldm^-3
38
[OH-] for pH x
10^-14-x moldm^-3