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Aqa a level chemistry > Acids + bases (Y13) > Flashcards

Acids + bases (Y13) Flashcards

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1
Q

remember to include the charges of ions which gain or loose a proton when writing acid-base equations

A
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2
Q

conjugate acid vs conjugate base

A

conjugate acid = accepts protons
conjugate base = lost a proton

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3
Q

conjugate acid of H2PO4-

A

H3PO4

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4
Q

which cannot function as a Bronsted-Lowry acid

A

CH3COO-

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5
Q

acid dissociation

A

1) remove the proton and write it as H+
2) Remember the charge on the conjugate base

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6
Q

Give an equation to show each stage in the dissociation of sulfuric acid in aqueous solution

A

H2SO4 (aq) –> H2SO4- (aq) + H+ (aq)

HSO4- (aq) –> SO42- + H+ (aq)

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7
Q

weak acid

A

barely dissociates

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8
Q

strong acid

A

fully dissociates

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9
Q

calculate the pH of the solution formed when 50cm3 of water is added to 100cm3 of 0.270mol dm-3 HCl

A

100 + 50 = 150
0.270 x (100 / 150) = 0.180
-log(0.18) = 0.74

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10
Q

calculate the concentration of a solution of H2SO4 with a pH of 0.84

A

this divide by 2

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11
Q

acid =

A

proton donor

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12
Q

base =

A

proton acceptor

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13
Q

acid-base equlibria

A

Acid-base equilibria involves the transfer of protons (H+ ions)

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14
Q

strong acids and bases

A

-strong acid = determined by the amount of H+ ions e.g H2SO4

-base = determined by amount of hydroxide (OH-) ions

-base + acid = salt + water (neutralisation reaction)

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15
Q

neutralisation ionic equation

A

H+ + OH –> H2O (neutralisation ionic equation)

Strong acids completely dissociate into ions in solution

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16
Q

equations pH

A

pH = –log10[H+] -> always give pH values to 2 decimal points

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17
Q

equation H+ ion

A

[H+] = 10^-pH

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18
Q

higher conc of H+ ions =

A

Higher concentration of H+ ions = lower pH (more acidic)

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19
Q

Bronsted-Lowry theory of acids

A

-Acid = substance that can donate a proton (H+)

-base = substance that can accept a proton

e.g HNO3 + NH3 –> NH4+ + NO3- (HNO3 donates a proton to NH3 so is the acid)

H+ ion dissociates from the acid

A H+ ion can only be accepted if a lone pair is present

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20
Q

There is 25cm3 of 0.1mol dm-3 of NaOH. This is neutralized by 32cm3 of H2SO4. Calculate the pH of the acid:

A

-moles of NaOH

-H+ + OH- –> H2O (1 mol in each so don’t multiply by 2)

-moles of H+ ions

-concentration of H+ ion

-pH = –log10[H+]

When calculating pH of an acid multiply by the no. Of moles e.g in H2SO4 but if in solution or dilute whatever then don’

If gives 2 vol e.g 250 and 50 in dilute solution add them –> 50/350 = volume

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21
Q

monotropic strong acid

A

The concentration of hydrogen ions in a monoprotic strong acid will be the same as the concentration of the acid.

-For HCl and HNO3 the [H+ (aq)] will be the same as the original concentration of the acid.

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22
Q

excess acid

A

Work out new concentration of excess H+ ions [H+ ] = moles excess H+ total volume (dm3 ) pH = – log [H+ ] —> Total volume = vol of acid + vol of base added

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23
Q

35cm3 of 0.5 mol dm-3 H2SO4 is reacted with 30cm3 of 0.55 mol dm-3 NaOH. Calculate the pH of the resulting mixture.

A

Moles H2SO4 = conc x vol = 0.5 x 0.035 = 0.0175mol

Moles H+ = 0.0175 x2 = 0.035

Moles NaOH = mol OH- = conc x vol = 0.55 x 0.030 = 0.0165

H+ + OH- –> H2O Moles of H+ in excess = 0.035 -0.0165 = 0.0185 [H+ ] = moles excess H+ total volume (dm3 ) = 0.0185/ 0.065 = 0.28 mol dm-3

pH = – log [H+ ] = -log 0.28 = 0.55

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24
Q

45cm3 of 1.0 mol dm-3 HCl is reacted with 30cm3 of 0.65 mol dm-3 NaOH. Calculate the pH of the resulting mixture.

A

Moles HCl =mol H+ = conc x vol = 1 x 0.045 = 0.045mol

Moles NaOH =mol OH- = conc x vol = 0.65 x 0.030 = 0.0195

H+ + OH- –> H2O

Moles of H+ in excess = 0.045-0.0195 = 0.0255 [H+ ] = moles excess H+ total volume (dm3 ) = 0.0255/ 0.075 = 0.34 mol dm-3 pH = – log [H+ ] = -log 0.34 = 0.47

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25
Q

conc = mol2dm-6

A

square root concentration

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26
Q

explain why water is neutral at 50 degrees

A

[H+] = [OH-]

dissociation gives 1 H+ and 1 OH-

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27
Q

weak acid

A

doesnt fully dissociate

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28
Q

Explain why an aqueous solution containing propanoic acid and its sodium salt constitutes a buffer system able to minimise the effect of added hydrogen ions.

A

propanoic acid is a weak acid

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29
Q

Calculate the pH at 25 °C of the solution that results from mixing 19.0 cm3 of 2.00 M HCl with 16.0 cm3 of 2.50 M NaOH.

A

1) 0.019 x 2 = 0.038
2) 0.016 x 2.50 = 0.040
3) 19 + 16 = 35cm3 (total volume)
4) 0.040 - 0.038 = 2x10^-3
5) find concentration and put in equation

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30
Q

why is pure water always neutral

A

concentration of H+ ions is equal to the concentration of OH-

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31
Q

another way of displaying Kw

A

Kw = [H+] ^2

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32
Q

Kw at room temperature

A

1 x 10^-14

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33
Q

is the dissociation of water endothermic or exothermic

A

endothermic

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34
Q

what happens if we decrease the temperature of Kw

A

shifts left
favour backwards exothermic reaction
pH increases
Kw decreases

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35
Q

increase temperature of Kw=

A

becomes acidic (decrease pH)

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36
Q

if Ca(OH)2 or anything similar

A

multiply conc by 2 due to 2 moles

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37
Q

strong acids / monotropic basic

A

HCl
HNO3

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38
Q

weak acids / monotropic basic

A

carboxylic acids (COOH)

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39
Q

strong acid / diprotic basic

A

H2SO4

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40
Q

strong bases / monoprotic basic

A

NaOh
KOH

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41
Q

strong bases / diprotic basic

A

Ba (OH)2

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42
Q

weak bases / monoprotic basic

A

NH3

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43
Q

explain why H2O is not shown in the Kw expression

A

H2O is liquid = constant = not included (same for solids)

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44
Q

which statement about pH is correct

A

at temperatures above 298K the pH of pure water is less than 7

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45
Q

what is the concentration of NaOH that has a pH of 14.30

A

1) 10 ^ -14.30 = 5.011 x 10^-15
2) Kw = 1.00 x 10^-14
3) Kw / H+ = 2.00

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46
Q

why is pure water at 10 degrees not alkaline

A

[H+] = [OH-]

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47
Q

true or false -> water can act as an acid or a base

A

true

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48
Q

calculate H+ using Kw and OH-

A

Kw / [OH-]

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49
Q

what is meant by the term strong when describing an acid

A

the ability to fully dissociate and form H+ ions in water

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50
Q

water

A

water is slightly dissociated –> slightly ionised as it breaks apart but then forms again with other molecules. (polar, covalent molecule)

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51
Q

H2O reversible reaction

A

H2O ⇌ H+ + OH- ΔH = endothermic

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52
Q

kc [H2O]

A

Kc [H2O] = [H+] [OH-] –> the concentration of H2O is much greater than the concentration of H+ or OH- as water only slightly dissociates so we say it is approximately constant. (minor ionisation)

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53
Q

Kw calculation

A

Kc [H2O] = constant = Kw

Kw = [H+] [OH-]

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54
Q

temperature on Kw

A

-Only temperature affects Kw value –> forward reaction is endothermic

-Increasing the temperature of Kw will favour the forward endothermic reaction to oppose the increase in temperature. The yield of H+ and OH- will increase so Kw increases.

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55
Q

increasing temperature on Kw

A

Increase in temperature = increase in ions = decrease in pH but the solution is still neutral as [H+] = [OH-] –> increase in temp also increases ionisation energy so there is more energy to break bonds

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56
Q

concentration of 1.0 x10^-14 find pH

A

-square root concentration = 1.0 x 10^-7

=put in log equation

= pH of 7

Rearranged equation to find [H+] = Kw / [OH-]

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57
Q

Find pH of 0.200 moldm-3 of NaOH:

A

-[OH-] = 0.200

-Kw = 1.0 x 10^-14 / 0.200 = 5 x10^-14

–log (5 x 10^-14) = 13.30

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58
Q

Find the concentration of Ba(OH2) with a pH of 13.30:

A

-10^-(13.30) = 5.01 x 10^-4 = H+

-Kw / H+ = 0.200

-0.200 / 2 = 0.100

-you have to divide by 2 because there are 2 OH molecules

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59
Q

Find the pH of 0.15mol dm-3 KOH:

A

-log (0.15) = 0.8239….

-14 – 0.823… = 13.18

-you have to subtract from 14 as total pH is 14

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60
Q

Kw value =

A

1 x 10^-14 mol2dm-6

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61
Q

calculate pH of H2SO4

A

multiply concentration by 2 before putting in log

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62
Q

square root value of Kw

A

-only when the concentration of ions in a neutral solution e,g pure water

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63
Q

pH of a diluted solution
vol = 0.3
HCl = 0.8
pH = 10.0

added 0.3dm3 of water

A

1) Find new conc of H+ ions
2) pH of diluted solution

0.3 x 0.8 = 0.24 H+ ions
New vol = 0.3 + 0.3 = 0.6
new [H+] = 0.24 / 0.6 = 0.4

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64
Q

dilute HCl in water

A

lower conc of H+ = increase in pH

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65
Q

conc of H+ =

A

conc of acid

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66
Q

pH change with solid bases

A

Find pH, concentration of H+, volume, moles of acid and moles of H+ ions for original acid

1) volume remains the same
2) Moles = subtract from reacting base
3) moles H+ = moles of resulting solution
4) moles of H+ x unchanged volume = concentration of H+
5) find pH using log

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67
Q

add solid base to solution of an acid

A

pH will increase
volume of solution will remain the same

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68
Q

add aqueous base to solution of an acid

A

volume increases
H+ ions decreases

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69
Q

Ezra has 0.25dm3 of HCl with concentration of 0.2moldm-3.
He reacts 0.04 moles of NaOH dissolved in 0.12dm3 of water.

How many H+ ions left when the reaction has finished

And what is the volume of the resulting solution

A

Part 1:
0.25 x 0.2 = 0.05
0.05 - 0.04 = 0.01

Part 2:
-0.37

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70
Q

calculating new pH of adding aqueous base

A

1) find change in volume
2) find change in {H+]

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71
Q

find new pH of solution after dilution

30ml HNO3
0.2moldm-3
20ml H2O

A

1) 0.2 x (30/1000) = 0.006
2) add the 2 volumes = 50ml

Conc of H+ ions = moles of H+ ions / volume of solution

0.006 / (50 / 1000) = 0.12

put in log

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72
Q

pH of solution after the reaction

50ml HBr
0.30moldm-3
0.120 moles KOH

A

1) moles = 0.015
2) moles of HBr - moles of KOH = 0.003
3) 0.003 / volume of HBr
4)0.06
5) put in log

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73
Q

doesnt rlly matter if you divide or multiply by 2

A

easier to multiply though

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74
Q

steps to calculating pH of mixed solutions

A

1) mol of H+

2) mol of OH-

3) subtract excess (big number) from limiting to get complete moles

4) excess value / total volume

5) put in log equation to get pH for acid in excess

6) if finding pH of base in excess subtract log answer from 14 or put in Kw equation

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75
Q

Calculate the pH of a solution formed when 50cm3 of 0.100mol dm-3 H2SO4 is added to 25cm3 of 0.150mol dm-3

A

-mol of H+ = 2 x 50/1000 x 0.100 = 0.01

-mol of OH- = 0.025 x 0.150 = 0.00375

-excess = H+ as it is larger so 0.01 - 0.00375 = 0.00625

-0.00625 / (50+25/1000) = 0.0833

-log (0.0833) = 1.079

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76
Q

pH of 25cm3 of 0.25moldm-3 H2SO4. 100cm3 of 0.2moldm-3 NaOH

A

H+ = 0.0125 moles

-OH- = 0.0200 moles

-OH = in excess (0.0200 - 0.0125) = 0.0075

-0.0075 / 125/1000 = 0.06

–log = 1.22

-14 – 1.22 = 12.78

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77
Q

when calculating change in pH

A

-Do normal steps to find pH of excess reagent

-Find original pH of acid and subtract values

-When finding pH change always subtract new value of pH from original value of pH acid

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78
Q

10.35cm3 of 0.100moldm-3 HCl added to 25cm3 of 0.150moldm-3 Ba(OH)2. Calcuate pH of solution at 30 degrees with Kw of 1.47 x 10-14.

A

1) H+ moles = 1.035 x 10^-3

2) OH- moles = 7.5x10^-3

3) 7.5 - 1.035 = 6.465 x 10^-3

4) 6.456 x10^-3 / (10.35 + 25) / 1000 = 0.1828854314…

5) (1.47 x 10^-14) / (0.182….) = 8.038 x 10^-14

6) -log (ANS) = 13.09

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79
Q

Calculate the pH of the solution formed after the addition of 35cm3 of 0.150moldm-3 NaOH to the original 25cm3 of monotropic acid:

A

1) moles NaOH = 5.25 x 10^-3

2) moles of montropic acid (assuming conc is same as base) = 3.75 x 10^-3

3) difference = 1.5 x 10^-3

4) assume OH- is in excess

5) (3.75 x 10^-3) / (35+25) / 1000 = 0.0625 000 = 0.0625

6) Kw / 0.0625 = 1.6 x 10^-3

7) -log (1.6 x 10^-3) = 12.80

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80
Q

propanoic acid =

A

C3H6O2

1 carbon counts as part of the carboxyl group

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81
Q

weak acids

A

-weak acids and weak bases dissociate slightly in aqueous solutions

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82
Q

Ka

A

Ka = dissociation constant of a weak acid

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83
Q

equations for a weak acid

A

PKa = -log (Ka)

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

Ka = 10^-pKa

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84
Q

calculate Ka

A

10^-pka

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85
Q

calculate pKa

A

-log (Ka)

86
Q

Ka constant

A

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

87
Q

NH3 and COOH

A

NH3 = weak monoprotic base

Carboxylic acid = weak monoprotic acid

88
Q

reversible reaction

A

HA ⇌ H+ + A-

HCOOH –> COOH- + H+

[H+] = [A-]

Ka = [H]^2 / [HA]

89
Q

what is pKA

A

The pKa is the tendency of a chemical to gain or loose protons

90
Q

low pKa and high pKA

A

Lower pKa = stronger acid

Lower pKa = weaker base

High pKA = stronger base

High pKA = weaker acid

91
Q

Calculate the pH of 0.100 mol dm-3 propanoic acid with pKA = 4.87:

A

-0.100 = [HA]

-10^-4.87 = 1.35 x 10^-5 = Ka

-1.35 x 10^-5 = [H+]^2 / 0.100

-[H+]2 = 1.35 x 10^-6

-square root = 1.16 x 10^-3

-pH = 2.93 = weak acid

1) Find Ka

2) input into Ka constant equation

3) Find H+

4) calculate pH

92
Q

calculate HA

A

[HA] = initial moles HA – moles OH- total volume (dm3)

93
Q

[A-]

A

moles OH- added
total volume (dm3
)

94
Q

diluted strong acid

A

[H+] = [H+]
old x old volume
new volume

95
Q

diluted base

A

[OH–]
old x old volume
new volume

96
Q

calculate pH of 0.0131 mol dm-3 of calcium hydroxide at 10 degrees

A

0.0131 x 2 = 0.0262

1 x 10^-14 / 0.0262 = ANS

put in log

97
Q

units for Ka

A

moldm3

98
Q

what is Ka only affected by

A

temperature

99
Q

all weak acids have

A

a low Ka bc it barely dissociates in water

100
Q

the higher the Ka value

A

the stronger the acid

101
Q

state the expression for the acid dissociation constant of carbonic acid

A

Ka = [HCO3+] [H+] / [H2CO3]

102
Q

assumptions made to calculations

A

1) no H+ from ionisation of water so is only from the acid
2) dissociation of weak acid is negligble (little differennce to concentration)

103
Q

dilute an acidic solution but a factor of 10

A

-H+ conc will alter (smaller)
-pH will change (greater)

104
Q

dilution of weak acids

A

-add water (reactant) = shift right to increase conc of products

-pH increase = solution is diluted. But H+ also increases

105
Q

in the case of HCl dilute by a factor of 10 increases the pH by 1 unit. Why dos ethanoic acid react differentyl

A

ethanoic acid = weak acid
when water is added = shift right = increase conc of H+ ions as doesnt increase pH as much

pH decreases slightly

106
Q

calculating mixtures of weak acids and strong bases

A

HA + OH- –> A- + H2O

107
Q

Calculate the pH of a solution formed when 30cm3 of 0.200mol dm3 ethanoic acid (pka 4.76) is added to 100cm3 of 0.100moldm3 NaOH

A

1) Moles of HA

2) Moles of OH

3) Calculate excess moles

4) Calculate conc excess

5) use Kw to find H+

6) calculate pH

–> ignore pKa as base is in excess not the weak acid

108
Q

conc of excess H+

A

[H+ ] = moles excess H+ total volume (dm3

109
Q

conc of excess OH-

A

Work out new concentration of excess OHions [OH-] = moles excess OH- total volume (dm3 ) [H+ ] = Kw /[OH– ] pH = – log [H+ ]

110
Q

conc of excess [HA]

A

Concentration of excess [HA] = (initial moles HA – moles OH-) / total volume (dm3 )

111
Q

conc of A-

A

Work out concentration of salt formed [A-] [A-] = (moles OH- added) / total volume (dm3 )

Rearrange Ka = [H+ ] [A-] / [HA] to get [H+ ]

pH = – log [H+ ]

-H+ = square root Ka x [HA]

112
Q

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 mixture:

A

1) Moles CH3CO2H = conc x vol =0.5x 0.055 = 0.0275mol

Moles NaOH = conc x vol = 0.35 x 0.025 = 0.00875

2) Moles of CH3CO2H in excess = 0.0275-0.00875 = 0.01875 (as 1:1 ratio)

3) [CH3CO2H ] = moles excess CH3CO2H / total volume (dm3 ) = 0.01875/ 0.08 = 0.234M

4) [CH3CO2 - ] = moles OH- added / total volume (dm3 ) = 0.00875/ 0.08 = 0.109M

5) [H+] = Ka x 0.234 x 0.109 = 3.64 x 10^-5

6) put in log to get pH

113
Q

weak acid with half alkali

A

When a weak acid has been reacted with exactly half the neutralisation volume of alkali, the above calculation can be simplified considerably. At half neutralisation we can make the assumption that [HA] = [A-] Ka = [H+ ] [CH3CO2 - ] [ CH3CO2H ] So [H+ (aq)] = Ka so pH = Ka

114
Q

weak acid reacts with strong base

A

When a weak acid reacts with a strong base, for every mole of OH- is added, one mole of HA is used up and one mole of A- is formed

e.g HA = 3 OH = 1 A- = 1

115
Q

excess HA / H+ for mixture

A

= Ka x [HA] / [A-]

116
Q

moles of A-

A

Moles of [A-] = moles of [OH-]

117
Q

what is Ka affected by

A

temperature
acid strength

118
Q

explain why calibrating a pH meter just before it is used improves the accuracy of the pH measurement

A

Over time/after storage the meter does not give accurate readings

119
Q

what are pH curves

A

pH curves successively measure the pH after the addition of small volumes of solution from a burette

120
Q

method used to calibrate pH probe

A
  1. Use buffer solutions of known conc
  2. Rinse probe with distilled water between readings
  3. Measure pH of more than one buffer solution (of known conc)
  4. Draw a calibration curve (of the pH of the buffer solution against the pH read on the meter_
121
Q

what is the purpose of a titration

A

-calculate the concentration of an unknown solution by titrating it with a solution of known concentration

122
Q

pH probe

A

identify when the acid and alkali have been mixed in the correct proportions to neutralise = more precise and more significant figures compared to a universal indicator

123
Q

method of producing titration curves

A

-note the intial pH of solution A

-add solution B at 1cm3 intervals

-swirl and note new pH

-add smaller volumes near endpoint

-plot graph of pH over volume of B

124
Q

HCl + NaOH –> NaCl + H2O

A

CH3COOH + NaOH –> CH3COONa + H2O

125
Q

2H+ + CO3^2-

A

H2O + CO2

126
Q

H+ + NH3 –>

A

NH4+

127
Q

indicators used for acid vs base

A

acid = methyl orange (red) = pH range from 3.2-4.4

base = phenolphthalein = pink for base = 8.2-10.0

128
Q

suggest why you wouldnt except a large rise in pH

A

base is weak

129
Q

why is it difficult to do a titration of a weak acid against a weak base

A

no distinctive equvilant point so makes it difficult to identify the end

130
Q

strong acid and strong alkali

A

acid pH = 0-1
base pH = 12-13
pH at eq = 7
pH range = 3-11

131
Q

strong acid and weak alkali

A

pH acid = 0-1
pH base = 9-10
pH at eq = 5
pH range = 3-7

132
Q

weak acid and strong alkali

A

-acid pH = 3-4
base pH = 12-13
pH at eq = 9

133
Q

weak acid and weak alkali

A

pH at eq = 7

134
Q

memorise each pH curve

A
135
Q

diprotic acid =

A

dissociates 2 times

136
Q

end point

A

-end point = final part of a reaction when the colour change can be detected

137
Q

equivalence point

A

-equivalence point = pH where you had mixed solutions in correct proportions according to the equation

138
Q

difference between end point and equivilance point

A

-Difference between end point and equivalence point = end point is the limit where the colour modification happens whereas the equivalence point is the precise limit where the chemical reaction comes to an end in the titration combination

139
Q

0.01

A

concordant titres

140
Q

buffer solution

A

Buffer solution = hold another solution at a certain pH = maintain a specific level of pH

141
Q

points when describing pH curve

A

-describe start point

-end point

-volume

-equivalence point

142
Q

equivilance point for curves

A

Equivalence point is NOT pH 7 for all curves –> it is when a sufficient base has been added to neutralise the acid (half way between straight part of the curve)
-pH falls only small amount till it reaches the equivalence point. PH then drastically/steeply falls from 11.3 to 2.7 which occurs when 25cm3 volume is added

-in this image the base is in the conical flask as the curve starts from alkali pH and the acid is in the burette

143
Q

pH = pka

A

-pH = pka when half the acid/base has been neutralised. This means we can calculate Ka of a weak acid using pH

144
Q

calculate Ka from pH curve

A

1) Find half of the equvilance point and draw across to get pH

2) pH = Pka

3) pH = 5

4) ka = 10^-5

5) This would be the opposite for the other half of the graph

145
Q

1:2 ratio in flask

A

If the Flask has 25cm3 of 0.10moldm-3 HNO3 and the burette has 50cm3 of 0.20moldm-3 NaOH –> it is a 1:2 ratio so the volume should be divided by 2 meaning the line from acid is shorter

146
Q

Explain why running a weak base into a strong acid produces the graph:

A

-at the beginning of the titration you have an excess HCl

-the shape of the curve will be the same as the strong base and strong acid

-after the equivalence point that things become different

-a buffer solution is formed containing excess NH3 and NH3Cl resulting in a larger increase of pH

147
Q

describe pH curve in detail

A

-at the beginning of the curve, pH falls quickly as acid is added

-curve gets less steep (buffer solution is being set up as ammonia and ammonium chloride)

-equivalence point is acidic (pH 5)

148
Q

Calculate the volume of 0.5moldm3 of NaOH which is added at equivalence to 25cm3 of 0.1moldm-3 H2SO4:

A

2NaOH + H2SO4 –> NaSO4 + 2H2O

0.5 x v / 2 = 25 x 0.1 / 1

= 0.25V = 25

V = 10cm 3

149
Q

volume at equvilance

A

-Ca x Va / na (A = acid)

1) write the balanced equation

2) write out formula and subsitute values

3) rearrange to calculate missing values

150
Q

by considering the species present explain why the pH at equvilance is not 7

A

since the acid is weak when reacting the salt and water at equivaltn with form OH- and RCOOH ions which will raise the pH

151
Q

constructing pH curve

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
152
Q

indicator

A

Indicators can be considered as weak acids. The acid
must have a different colour to its conjugate base

153
Q

how indicators work

A

HIn (aq)—> In- (aq) + H+ (aq)
colour A colour B
We can apply Le Chatelier to give us the
colour.
In an acid solution the H+
ions present will
push this equilibrium towards the reactants.
Therefore colour A is the acidic colour.
In an alkaline solution the OHions will
react and remove H+
ions causing the
equilibrium to shift to the products. Colour B
is the alkaline colour.

154
Q

deciding what indiactor to use

A

An indicator will work if the pH range of the indicator lies on the steep part of the titration curve. In this case
the indicator will change colour rapidly and the colour change will correspond to the neutralisation point.

Only use phenolphthalein in titrations with strong
bases but not weak bases- Colour change: colourless acid  pink alkali

Use methyl orange with titrations with
strong acids but not weak acids
Colour change: red acid  yellow alkali
(orange end point)

155
Q

dissociation equations for H2SO4

A

H2SO4 –> H+ + HSO4-

HSO4- ⇌ H+ + SO4^2-

156
Q

NH3 + H2O ⇌

A

NH4+ + OH-

157
Q

[H+] =

A

square root Ka x [HA]

158
Q

[H+] =

A

[HA]

159
Q

Ka and pka values

A

stronger acids have a smaller Ka than weaker acids

stronger acids have larger pKa than weak acids

160
Q

calculate pH of mixtures with weak acids and strong bases = excess acid

A

1) [H+]
2) [OH-]
3) find excess
4) excess acid / volume = [HA]
5) [OH-] / volume = [A-]
6) [H+] = Ka x [HA] / [A-]
7) put in log to get pH

161
Q

calculate pH of 0.20moldm-3 Sr(OH)2

A

kw / (0.20/2)
put in log

162
Q

explain what is meant by the term buffer solution

A

solution which resists change in pH despite the addition of small amounts of acid/base

163
Q

why is there a lot of unionised ethanoic acid and lots of ethanoate ions in a buffer solution

A

ethanoic acid = weak acid so partially dissociates

ethanoate ions = due to salt (CH3COONa) fully ionising

164
Q

state what should be added to propanoic acid to form an acidic buffer

A

sodium propanoate –> with the salts to reform, the acid and resulting pH change

165
Q

state the two equations that demonsrtate the formation of the buffer solution

A

CH3CH2COOH ⇌ CH3CH2COO- + H+

CH3CH2COONa –> CH3CH2COO- + Na+

166
Q

what happens to equilbirum when H+ ions are added to the buffer

A

when H+ ions are added they react with the salt to reform the acid. This means equilbirum will shift to the left to oppose the change in pH

167
Q

addition of an alkali

A

H+ + OH- –> H2O
CH3COOH ⇌ CH3COO- + H+

168
Q

why is there a large supply of weak base

A

partially dissociates = equiilbirium will lie to the left

169
Q

equation to show the formation of an alkaline buffer

A

NH3 + H2O ⇌ NH4+ + OH-
NH4Cl + NH4+ + Cl-

170
Q

which forms a buffer solution with a pH less than 7

A

ethanoic acid and sodium ethanoate

171
Q

which forms buffer solution with a pH more than 9

A

ammonia and ammonium chloride

172
Q

when a small amount of acid is added to a buffer of ammonia and ammonium chloride what happens

A

hydrogen ions combine with NH3 to make NH4_

173
Q

summary of acidic buffers

A

HA ⇌ A- + H+ and NaA –> NA+ + A-

H+ + A- –> HA

addition of an alkali reacts with HA (HA + OH- –> A + H2O)

ratio of HA to A- remains unchanged since H+ = Ka

174
Q

state how a buffer solution can be amde from solutions of KOH and CH3COOH

give the equation

A

add excess ethanoic acid to KOH

KOH + CH3COOH –> CH3COOK + H2O

CH3COO- from salt reacts with added H+

175
Q

give the buffer solution and reagent which could be added to a solution of ammonia

A

-solution that resist a change in pH when small amounts of acid or base are added

-reagent = NH4Cl

176
Q

state and explain the effect on the pH of this buffer solution when a small amount of HCL is added

A

H+ ions react with salt to reform the acid
equilbirium shifts right to oppose change in pH and resist the change

177
Q

write an equation for the reaction of ethanoic acid when small amount of H+ is added

A

CH3COO- + H+ ⇌ CH3COOH

178
Q

define buffering capacity

A

the quantiity of H+/OH- ions it can absorb without a significant change in pH

179
Q

[H+] -> buffering equation

A

Ka x [HA] / [A-]

depends on the Ka of the buffer and the ratio of acid/base to the salt

180
Q

define a buffer solution

A

Buffer solution = solution that resists changes in pH when small amounts of acid or alkali are added to them

181
Q

river

A

What was present in the river to make it a buffer solution –> acid (from rain) and limestone (calcium carbonate)

182
Q

blood

A

Our blood is a buffer solution –> buffers act as shock absorbers against sudden changes of pH by converting injurious strong acids and bases into harmless weak acids salts

(H2CO3 (aq) –> H+ (aq) + HCO3- (aq) )

Acedosis = too much aciditiy in blood

Alkalosis = too alkali in blood

183
Q

acidic buffer

A

A buffer solution is prepared by mixing a solution of ethanoic acid with a solution of sodium ethanoate

Acidic buffers:

-pH less than 7

-made from a weak acid and one of its salts/conjugate base or a weak acid and a strong base

e.g ethanoic acid and sodium hydroxide

(HA and A-)

-reversible reaction

184
Q

how an acidic buffer solution works

A

CH3COOH –> CH3COO- + H+ (ethanoic acid –> sodium ethanoate + H+) -> equilbrium

-contains lots of unionised ethanoic acid

-lots of ethanoate ions from the sodium ethanonate

-some hydrogen ions making the solution acidic

-adding sodium ethanonate to this will increase the concentration of the conjugate base (CH3COO)

-adding a strong acid to the solution –> increase H+ ions so equilbirium moves to the left to reduce H+ ions resisting changes in pH

185
Q

alkaline buffer solution

A

-pH more than 7

-made from a weak base and one of its salts

-reversible reaction

e.g NH3 (aq) + H2O (l) –> NH4+ (aq) + OH- (aq)

-if an acid is added to this solution then equilbirium moves to replace the removed OH- ions. H+ ions combined with OH- to form H2O

Buffer solution = ratio [HX] / [X-]

186
Q

uses of buffer

A

-in swimming pools to prevent chemicals from irritating our skin

-soda to prevent acidic flavouring from damaging our teeth

-in rivers to protect against effects of acid rain

-in our blood to keep internal pH constant

187
Q

salt

A

Salt = conjugate base in acid buffers and conjugate acid in alkali buffers

188
Q

dissociation of ethanoic acid

A

CH3COOH ⇌ CH3COO- + H+

->the acid is in large supply because it is a weak acid and therefore the equilbrium shifts to the left

189
Q

formation of acidic buffers

A

-add sodium ethanoate to ethanoic acid = cause position of equilibrium to shift left therefore we have large amount of un-ionised ethanoic acid

-if the salt is added (CH3COONa) it will also dissociate completely

190
Q

weak acid and strong base neutralisation

A

Work out new concentration of excess HA
[HA] = initial moles HA – moles OH- total volume (dm3
)
Work out concentration of salt formed [A-]
[A-] = moles OH- added
total volume (dm3
)
Rearrange Ka = [H+
] [A-] to get [H+
]
[HA]
pH = – log [H+
]

191
Q

ethanoic acid buffer

A

CH3CO2H —> (aq) CH3CO2- (aq) + H+
(aq)

192
Q

small amount of alkali added to a buffer

A

(aq) + H2O (l)

193
Q

small amount of acid added to a buffer

A

(aq) + H + –> CH3CO2H (aq)

194
Q

diluting a buffer solution

A

Diluting a buffer solution with water will not change its pH

This is because in buffer equation below the ratio of [HA]/[A-] will stay constant as both
concentrations of salt and acid would be diluted by the same proportion.

195
Q

drawing pH graph

A

1) initial pH of acid
2) Find final pH of base = -log (Kw / H+ )

196
Q

recognising half-equivlance

A

1) calculate initial moles of acid and base
2) calculate final moles (excess)
3) determine pH of solution formed

197
Q

pH =

A

pKa

198
Q

HCO3- equilbrium

A

HCO3- ⇌ H+ + HCO3^2- = equation for addition of acid to HCO3-

-Adding an acid to a buffer solution will increase the concentration of H+ ions so equilbirum shifts to the left

-Adding an alkali to a buffer solution will cause OH- to react with H+ ions so equilbirum moves to the right to replace H+ions

-concentration of H+ ions remain constant

199
Q

Calculate the pH of 0.500L buffer solution composed of 0.700M formic acid HCOOH, Ka = 1.77 x 10^-4 and 0.500M sodium formaite (HCOONa)

Calculate pH after adding 50.0ml of 1.00M NaOH solution. (M = molarity = concentration)

A

A) pH = pka + log (base / acid) –> henderson hasslbach equation

Pka = 3.752 + log (0.5 / 0.7) = 3.606 or (1.77 x 10^-4) x (0.700) / (0.500)

B) moles HCOOH = 0.7 x 0.5 = 0.350

Moles HCOONa = 0.5 x 0.5 = 0.250

Moles NaOH = 0.05 x 1 = 0.05

1:1 ratio

0.350 - 0.05 = 0.300

0.250 - 0.05 = 0.300

PH = 3.752 + log (0.3 / 0.3) = 3.754

200
Q

calculate pH of a buffer

A

1) Find moles of acid

2) Find moles of base

3) Ka x (HA / A)

4) put in log to get pH

201
Q

pH of buffer with solid salt

A

1) Find moles of acid

2) Find moles of solid base (mass / mr)

3) put in Ka equation

4) put in log to get pH

202
Q

how indicators work

1) position of eq
2) oppose the change
3) observation e.g turn blue

A

inidcators = weak acids
when weak acids are dissolved in water an equilbirum is established
HLit (aq) –> H+ + LiT-

reversible reaction = oppose change

203
Q

where must colour change occur

A

between vertical region on pH curve

204
Q

suggest why the pH probe is washed with distilled water between each of the calirbation measruement

explain why the volume of NaOH added is smaller at end point

A

-so its not contaminated

-more salt and water is produced therefore decreasing conc of NaOH

205
Q

why would the solution not be resisting changes in pH

A

not enough salt has been formed so the concentrations are not equal or not enough weak acid remaining

206
Q

explaining pH curve

A

1) initially….low conc of H+ ions
2) As these react with alkali pH changes rapidly
3) At same time sodium ethanonate is formed
4) Once moderate amount is present the solution acts as a buffer
5) until such time as its capacity is exhuated which is near to end point of reaction

207
Q

describe and explain the behaviour of the solution formed in region of circle on graph

A

-buffer is formed where at least half of the weak acid is neutralised by its conjugate base

208
Q

2 ways to make a buffer

A

1) to a solution of weak acid add its salt

2) to excess weak acid add a strong base until the acid is partially neutralised

209
Q

state how buffer solution can be made from solutions of potassium hydroxide and ethanoic acid

A

1) enough potassium hydroxide was added so roughly half of the ethanoic acid was neutralised

2) CH3COOH + KOH –> CH3COOK + H2O

3) H+ ions react with CH£COO- so equilbirum shifts left

210
Q

calculate the pH of the solution formed when 200cm3 of water are added to 50cm3 of 0.50moldm-3 HCl

A

0.50 x 50/200
-log (ANS)

1) conc x old/new volume
2) put into log to get pH

211
Q

Calculate the pH of the solution formed when 50 cm3 of water is added to 100 cm3 of 0.200 mol dm-3 NaOH.

A

OH-
] in original NaOH solution = 0.200
[OH-
] in diluted solution = 0.200 x old volume = 0.200 x 100 = 0.1333
new volume 150
[H+] = Kw = 10-14 = 7.50 x 10-14
[OH-
] 0.1333
pH = -log (7.50 x 10-14) = 13.12

212
Q

calculate the pH of the buffer solution prepared by adding 1.00g of sodium ethanoate to 250cm3 of 0.100moldm-3 ethanoic acid

A

1) find moles for both substances
2) ka x mol / mol

Aqa a level chemistry (18 decks)

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