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A Level Chemistry > Acid Base Equilibria - Topic 12 > Flashcards

Acid Base Equilibria - Topic 12 Flashcards

(44 cards)

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1
Q

what is a bronsted-lowrey acid

A

a substance that can donate a proton

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2
Q

what is a bronsted-lowrey base

A

a substance that can accept a proton

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3
Q

how to calculate pH

A

pH = - log [H+]

always give pH values to 2dp

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4
Q

how to find the hydrogen ion concentration from pH

A

[H+] = 1 x 10-pH

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5
Q

what expression can be used to calculate hydrogen ions and hydroxide ions concentration (expression for STRONG ACIDS AND STRONG BASES) kw expression

A

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

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6
Q

what is the value for kw at 25 degrees celsius

A

1x10-14 mol2dm-6

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7
Q

how to calculate kw using pKw

A

Kw =10^-pKw

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8
Q

why is pure water neutral

A

because the [H+ (aq) ] = [OH-(aq)]

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9
Q

how to calculate the concentration of hydrogen ions from kw OF PURE WATER

A

[H+(aq) ] = √ Kw

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10
Q

how to work out kw from concentration of hydrogen ions OF PURE WATER

A

Kw = [H+ (aq) ]^2

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11
Q

how can Le Chatelier’s principle predict the change of ph in water when there is an increase in temperature

H20 <——> OH- + H+

A
  • The dissociation of water is endothermic
  • so increasing the temperature would push the equilibrium to the right
  • giving a bigger concentration of H+ ions and a lower pH.
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12
Q

weak acid dissociation expression (ka expression)

A

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

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13
Q

what does a large ka value mean

A

the acid is strong

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14
Q

assumption of ka expression

A

1) [H+ (aq)]eqm = [A- (aq)] eqm because they have dissociated according to a 1:1 ratio.
2) As the amount of dissociation is small we assume that the initial concentration of the undissociated acid has remained constant.

So [HA (aq) ] eqm = [HA(aq) ] initial

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15
Q

what is Ka equal to

A

[H+]

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16
Q

what Is pH equal to

A

pKa

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17
Q

Why does pH increase by 1 unit when diluting a strong acid 10 times , but not for weak acids?

A

Strong acid: Fully dissociates (e.g., HCl → H⁺ + Cl⁻). When diluted, [H⁺] decreases by a factor of 10, so pH increases by 1 unit for every 10× dilution.
Weak acid: Only partially dissociates (e.g., CH₃COOH ⇌ H⁺ + CH₃COO⁻). Diluting the acid shifts the equilibrium, but [H⁺] doesn’t decrease as much, so pH increases by less than 1 unit per 10× dilution.

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18
Q

how to construct a pH curve

Study These Flashcards
A

1) transfer 25cm3 of acid to a conical flask with a volumetric pipette
2) Measure initial pH of the acid with a pH meter
3) Add alkali in small amounts (2cm3) noting the volume added
4) Stir mixture to equalise the pH
5) Measure and record the pH to 1 d.p.
6) Repeat steps 3-5 but when approaching endpoint add in smaller volumes of alkali
7) Add until alkali in excess

19
Q

draw the curve for
Strong acid and strong base
Weak acid and strong base
Strong acid and weak base
Weak acid and weak base

Initial and final pH
Volume at neutralisation
General shape (pH at neutralisation)
equivalence point

Study These Flashcards
A

pg 7 and 8 chemrevise

20
Q

what happens at 1/2 the neutralisation point of weak acid and strong base

Study These Flashcards
A

[HA] = [A-]

So Ka =[H+] and pKa =pH

If we know the Ka we can then work out the pH at 1⁄2 V (neutrlisation point) or vice versa

21
Q

when do we use phenolphthalein and colour change

Study These Flashcards
A

titrations with strong bases but not weak bases-
Colour change: colourless acid —> pink alkali

22
Q

when do we use methyl orange and colour change

Study These Flashcards
A

titrations with strong acids but not weak acids
Colour change: red acid —-> orange alkali

23
Q

why do weak acids have a less negative enthalpy change of neutralisation

Study These Flashcards
A

because energy is absorbed to ionise the acid and break the hydrogen bond in the un-dissociated acid.

24
Q

are enthalpies of neutralisation end or exothermic

Study These Flashcards
A

always exothermic

25
what is the end point of titration
the point when the colour of the indicator changes colour
26
when is the end point reached in titration
when the concentration of the acid is = to the concentration of its conjugate base HIn (aq) -----> In- (aq) + H+ (aq) [HIn] = [In-]
27
how do we pick the correct indicator
you pick an indicator whose end-point meets with the equivalence point for the titration.
28
how To calculate the concentration or volume of a solution after dilution and find the pH of the diluted solution
C1V1 = C2V2 after calculation the concentration using this equation you would plug in the concentration of the solution after dilution in the pH = -log(H+) to find the pH C1 = Initial concentration (before dilution) V1 = Initial volume (before dilution) C2 = Final concentration (after dilution) V2 = Final volume (after dilution)
29
what is a buffer solution
a solution which resits changes in pH when small amounts of acid or alkali are added
30
how is an acidic buffer made
by reacting a weak acid and a salt of that weak acid (which is made from reacting the weak acid with a strong base)
31
how is a basic buffer solution made
made from a weak base and a salt of that weak base (which is made from reacting the weak base with a strong acid)
32
how does ethanoic acud resist pH changes when small amounts of acid is added CH3CO2H (aq) <----> CH3CO2- (aq) + H+ (aq)
- If small amounts of acid is added to the buffer Then the equilibrium will shift to the left decreasing the conc of H+ ions - this process keeps the ratio [CH3COOH ] / [ {CH3COO-} ] relatively constant, as the large concentration of both species has the ability to neutralize small amounts of acid. This prevents the pH from changing significantly and remain constant
33
how does the ethanoic buffer resist pH changes when a small amount of alkali is added to the buffer solution CH3CO2H (aq) <----> CH3CO2- (aq) + H+ (aq)
- If small amounts of alkali is added to the buffer. 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) - this process keeps the ratio [CH3COOH ] / [ {CH3COO-} ] relatively constant, as the large concentration of both species has the ability to neutralize small amounts of alkali. This prevents the pH from changing significantly and remain constant
34
what is the pH of blood plasma
7.35 and 7.45
35
what happens to the number of moles of the buffer acid and salt if an alkali was added to the acidic buffer
- If a small amount of alkali is added to a buffer then the moles of the buffer acid would reduce by the number of moles of alkali added - and the moles of salt would increase by the same amount
36
what happens to the number of moles of the buffer acid and salt if an acid was added to the acidic buffer
- If a small amount of acid is added to a buffer then the moles of the buffer salt would reduce by the number of moles of acid added - and the moles of buffer acid would increase by the same amount
37
how to calculate the pH of buffer solutions
pH = pKa + log ([A-]/[HA]) (this is the Henderson hasselbalch equation)
38
how to work out ka from pka
10^-pka
39
how to work out pKa from Ka
pKa = -log Ka
40
Calculate the pH of a buffer made from 45cm3 of 0.1 mol dm-3 ethanoic acid and 50cm3 of 0.15 mol dm-3 sodium ethanoate (Ka = 1.7 x 10-5)
pKa = -log(1.7x10-5) = 4.77 Work out the moles of both solutions Moles ethanoic = conc x vol = 0.1 x 0.045 = 0.0045mol Moles sodium ethanoate = conc x vol = 0.15 x 0.050 = 0.0075 pH = pka + log ([A-]/[HA]) pH = 4.77 + log (0.0075/0.0045) pH = 4.99
41
55cm3 of 0.50 mol dm-3 CH3CO2H is reacted with 25cm3 of 0.35 mol dm-3 NaOH. Calculate the pH of the resulting buffer solution. CH3CO2H+ NaOH ---> CH3CO2Na + H2O Ka is 1.7x10-5 mol dm-3 THIS METHOD OF WORKING ONLY APPLIES TO WHEN NaOH IS ADDED TO A BUFFER
pKa = -log(1.7x10-5) = 4.77 Moles CH3CO2H = conc x vol =0.5x 0.055 = 0.0275 mol Moles NaOH = conc x vol = 0.35 x 0.025 = 0.00875 Moles of CH3CO2H in excess = 0.0275-0.00875 = 0.01875 (as 1:1 ratio) total volume = 55/1000 + 25/1000 = 0.08 [CH3CO2H ] = moles excess CH3CO2H total volume (dm3) = 0.01875/ 0.08 = 0.234 mol dm-3 moles OH- added total volume (dm3) = 0.00875/ 0.08 = 0.109 mol dm-3 pH = 4.7 + log(0.109/0.234) pH = 4.44
42
0.005 mol of NaOH is added to 500cm3 of a buffer where the concentration of ethanoic acid is 0.200 mol dm-3 and the concentration of sodium ethanoate is 0.250 mol dm-3. (Ka = 1.7 x 10-5) Calculate the pH of the buffer solution after the NaOH has been added.
- Work out the moles of acid and salt in the initial buffer solution - Moles ethanoic acid= conc x vol = 0.200 x 0.500 = 0.100mol - Moles sodium ethanoate = conc x vol = 0.25 x 0.500 = 0.125mol - Work out the moles of acid and salt in buffer after the addition of 0.005mol NaOH - Moles ethanoic acid = 0.100 - 0.005 = 0.095 mol - Moles sodium ethanoate = 0.125 + 0.005 = 0.130 mol pH = 4.77 + log (0.13/0.095) pH = 4.91
43
what are the assumptions of Henderson hasselbalch equation
that the initial concentration of the acid has remained constant, because amount that has dissociated or reacted is small.
44
learn conjugate acid and base

A Level Chemistry (23 decks)

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