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Chemistry > acids and bases > Flashcards

acids and bases Flashcards

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1
Q

bronzed Lowry acid

A

donates a proton

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2
Q

bronzed lowry base

A

accepts a proton

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3
Q

how to calculate pH using [H+]

A

pH= -log[H+]

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4
Q

how do strong acids dissociate

A

they dissociate completely

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5
Q

are the concentration of the acid and [H+] ions linked ?

A

yes in monoprotic acids its the same

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6
Q

how many d.p is pH given to

A

always 2 d.p

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7
Q

how to calculate [H+] from pH

A

10 -pH

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8
Q

what the is the pure water equilibrium

A

H2O (l) <——-> H+ (aq) + OH- (aq)

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9
Q

Kw constant

A

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

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10
Q

why is pure water neutral

A

it is neutral because [H+ (aq) ] = [OH-(aq)] so when neutral Kw = [H+ (aq) ]2

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11
Q

is the dissociation of water endothermic or exothermic

A

the dissociation of water is endothermic so increasing temperature will movie equilibrium to the right giving a bigger concentration of H+

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12
Q

example of strong base dissociating NaOH

A

NaOH –> Na+ + OH- (So to calculate you do a similar thing to water)

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13
Q

how do weak acids dissociate

A

they slightly dissociate into an equilibrium mixture
HA (aq) —-> H+ (aq) + A- (aq)

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14
Q

what does larger Ka mean about acid strength

A

it means stronger acid

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15
Q

pKa to Ka

A

10 -pKa

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16
Q

Ka to pKa

A

-log Ka

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17
Q

what assumption are made when calculating pH of weak acid (1)

A

we assume the [H+]=[A-] at equilibrium because they have dissociated at a 1:1 ratio

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18
Q

what assumption are made when calculating pH of weak acid (2)

A

because the dissociation is so small we assume that the undissociated ions is constant so
[HA(aq)] eqm = [HA(aq)]initial

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19
Q

strong acid and strong base neutralisations

A
  1. work out what’s in excess by working out moles of original acid and base
  2. if excess acid moles work out the new concentration using excess (what is the excess)
    3.if excess moles base work out new conc of [OH-] in excess then use that and Kw to work out H+ excess concentration

important!!!! the volume you use must be of Acid and the base added

important!!!! if its diprotic multiple initial moles of H+ or OH- by 2

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20
Q

weak acid strong base neutralisations

A

1.work out moles of original acid and base added then work out what’s in excess
2.If excess alkali use same method as strong acid and strong base
3. if excess acid workout new conc of [HA]. then work out concentration of salt formed [A-]
[A-]= moles of base divided by total volume . using that find [H+]

21
Q

when is half equivalence

A

when weak acid as been reacted with hard the neutralisation volume of alkali

22
Q

what’s the assumption we make at half neutralisation

A

[HA] = [A-]
so
Ka= [H+]
And pH = pKa

23
Q

how do you work out pH of strong diluted acid

A

[H+]new= [H+]old x old volume/new volume(old and new added)

24
Q

pH of diluted base

A

[OH-]new= [OH-] x old volume/new volume (old and new added)

work out [H=] using Kw

25
Q

what is a buffer solution

A

a solution where pH does not change significantly if small acid or alkali is added to it

26
Q

what is an acidic buffer made of

A

weak acid and salt of that weak acid

27
Q

what is a basic buffer made from

A

weak base and salt of that weak base

28
Q

what is in higher concentration in a buffer the salt of pure acid

A

the salt

29
Q

if a small amount of acid is added to the buffer what happens

A

the equlibrium shifts to left to remove the H+ added
CH3CO2H (aq) <——–> CH3CO2- (aq) + H+ (aq)

As there is a larger concentration of the salt ion in the buffer the ratio of [CH3CO2H]/ [CH3CO2-] stays almost constant, so the pH stays fairly constant.

30
Q

if a small amount of alkali is added to the buffer what happens

A

The OH- ions will react with H+ ions to form water.
H+ + OH - —-> H2O

the equilibrium will then shift to the right to produce more H+ ions

CH3CO2H (aq) CH3CO2- (aq) + H+ (aq)
Some ethanoic acid molecules are changed to ethanoate ions but as there is a large concentration of the salt ion in the buffer, the ratio [CH3CO2H]/ [CH3CO2-] stays almost constant, so the pH stays fairly constant.

31
Q

calculating the pH of buffer solutions

A

use Ka = [H+ (aq)][A- (aq)] /[HA (aq)]

assume that [A-] is due to salt only
we assume the initial acid concentration is the same we assume only a little bit has reacted or dissociated

why can we add moles straight into the equation because they both have the same final volumes so it cancels out

32
Q

3 ways a buffer is made

A
  1. making a buffer by adding salt solution
    2.making buffer by adding solid
    3.partially neutralising a weak acid with alkali and therefore producing a salt (different method)
33
Q

partially neutralising a weak acid with alkali and therefore producing a salt (different method)

A

first find moles for both
then find excess HA
calculate conc of HA using moles of excess acid
calculate conc of A- using moles of OH- added

use Ka to find the [H+] then pH

34
Q

calculate pH change of buffer when a small amount of alkali is added

A

the moles of the buffer acid added will decrease by the same amount of moles as alkali added and the moles of salt would increase by the same amount.

CH3CO2H (aq) +OH- —-> CH3CO2- (aq) + H2O (l)

35
Q

calculate pH change of buffer when a small amount of acid is added

A

the buffer salt will reduce by the moles of acid added and the moles of buffer acid will increase

36
Q

why does adding water to a buffer solution not change its pH

A

it doesn’t change its pH because the ratio of HA to A- will be diluted by the same amount

37
Q

constructing a pH curve (practical)

A
  1. transfer 25cm3 of acid into a volumetric flask with a pipette
    2.measure initial pH of acid with pH meter
    3.add alkali in small amounts (2cm3) noting volume added
    4.stir mixture to equalise pH
  2. measure and record pH to one decimal place
  3. repeat steps 3-5 and when approaching end punt add smaller volumes of alkali
  4. add until alkali is in excess
38
Q

calibrate the pH meter

A

measure the known pH of a buffer solution. This is necessary because pH meters can loose accuracy upon storage. (sometimes calibration is repeated with multiple buffers)

39
Q

strong acid , strong base curve

A

long steep part from around 3 to 9
pH equivalence point at = 7

40
Q

weak acid strong base

A

steep part of the curve is above 7 (normally around 7to 9) so equivalence point is above 7

41
Q

equivalence point

A

mid point of extrapolated vertical section

42
Q

half neutralisation volume (graph)

A

half neutralisation the volume of [HA]= [A-]

so Ka= [H+] and pKa = pH

43
Q

strong acid weak base

A

steep part below 7 (around 4 to 7)
equivalence point below 7

44
Q

weak acid weak base

A

no steep part of the curve

45
Q

how do indicators work

A

In an acidic solution the H+ ions will push equilibrium towards the reactants

In alkaline solution the OH- ions with react to remove H+ ions causing the equilibrium to shift to the products.

HIn (aq) <————->In- (aq) + H+ (aq)

colour A colour B

46
Q

when does indicator has to change colour

A

at the end point of a titration. end point us when [HIn] = [In-]. indicator will work if the pH change of the indicator lies on the steep part of the curve because they colour change will then correspond to the neutralisation point

47
Q

phenolphthalein

A

only strong bases
colourless in acid pink in base

48
Q

methyl orange

A

only works for strong acids
red in acid
yellow in base
orange at end point

Chemistry (10 decks)

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