Acids + bases (Y13) Flashcards
(250 cards)
1
Q
remember to include the charges of ions which gain or loose a proton when writing acid-base equations
A
2
Q
conjugate acid vs conjugate base
A
conjugate acid = accepts protons
conjugate base = lost a proton
3
Q
conjugate acid of H2PO4-
A
H3PO4
4
Q
which cannot function as a Bronsted-Lowry acid
A
CH3COO-
5
Q
acid dissociation
A
1) remove the proton and write it as H+
2) Remember the charge on the conjugate base
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)
7
Q
weak acid
A
barely dissociates
8
Q
strong acid
A
fully dissociates
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
10
Q
calculate the concentration of a solution of H2SO4 with a pH of 0.84
A
this divide by 2
11
Q
acid =
A
proton donor
12
Q
base =
A
proton acceptor
13
Q
acid-base equlibria
A
Acid-base equilibria involves the transfer of protons (H+ ions)
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)
15
Q
neutralisation ionic equation
A
H+ + OH –> H2O (neutralisation ionic equation)
Strong acids completely dissociate into ions in solution
16
Q
equations pH
A
pH = –log10[H+] -> always give pH values to 2 decimal points
17
Q
equation H+ ion
A
[H+] = 10^-pH
18
Q
higher conc of H+ ions =
A
Higher concentration of H+ ions = lower pH (more acidic)
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
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
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.
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
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
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
25
conc = mol2dm-6
square root concentration
26
explain why water is neutral at 50 degrees
[H+] = [OH-]
dissociation gives 1 H+ and 1 OH-
27
weak acid
doesnt fully dissociate
28
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.
propanoic acid is a weak acid
29
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.
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
30
why is pure water always neutral
concentration of H+ ions is equal to the concentration of OH-
31
another way of displaying Kw
Kw = [H+] ^2
32
Kw at room temperature
1 x 10^-14
33
is the dissociation of water endothermic or exothermic
endothermic
34
what happens if we decrease the temperature of Kw
shifts left
favour backwards exothermic reaction
pH increases
Kw decreases
35
increase temperature of Kw=
becomes acidic (decrease pH)
36
if Ca(OH)2 or anything similar
multiply conc by 2 due to 2 moles
37
strong acids / monotropic basic
HCl
HNO3
38
weak acids / monotropic basic
carboxylic acids (COOH)
39
strong acid / diprotic basic
H2SO4
40
strong bases / monoprotic basic
NaOh
KOH
41
strong bases / diprotic basic
Ba (OH)2
42
weak bases / monoprotic basic
NH3
43
explain why H2O is not shown in the Kw expression
H2O is liquid = constant = not included (same for solids)
44
which statement about pH is correct
at temperatures above 298K the pH of pure water is less than 7
45
what is the concentration of NaOH that has a pH of 14.30
1) 10 ^ -14.30 = 5.011 x 10^-15
2) Kw = 1.00 x 10^-14
3) Kw / H+ = 2.00
46
why is pure water at 10 degrees not alkaline
[H+] = [OH-]
47
true or false -> water can act as an acid or a base
true
48
calculate H+ using Kw and OH-
Kw / [OH-]
49
what is meant by the term strong when describing an acid
the ability to fully dissociate and form H+ ions in water
50
water
water is slightly dissociated --> slightly ionised as it breaks apart but then forms again with other molecules. (polar, covalent molecule)
51
H2O reversible reaction
H2O ⇌ H+ + OH- ΔH = endothermic
52
kc [H2O]
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)
53
Kw calculation
Kc [H2O] = constant = Kw
Kw = [H+] [OH-]
54
temperature on Kw
-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.
55
increasing temperature on Kw
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
56
concentration of 1.0 x10^-14 find pH
-square root concentration = 1.0 x 10^-7
=put in log equation
= pH of 7
Rearranged equation to find [H+] = Kw / [OH-]
57
Find pH of 0.200 moldm-3 of NaOH:
-[OH-] = 0.200
-Kw = 1.0 x 10^-14 / 0.200 = 5 x10^-14
--log (5 x 10^-14) = 13.30
58
Find the concentration of Ba(OH2) with a pH of 13.30:
-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
59
Find the pH of 0.15mol dm-3 KOH:
-log (0.15) = 0.8239....
-14 – 0.823... = 13.18
-you have to subtract from 14 as total pH is 14
60
Kw value =
1 x 10^-14 mol2dm-6
61
calculate pH of H2SO4
multiply concentration by 2 before putting in log
62
square root value of Kw
-only when the concentration of ions in a neutral solution e,g pure water
63
pH of a diluted solution
vol = 0.3
HCl = 0.8
pH = 10.0
added 0.3dm3 of water
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
64
dilute HCl in water
lower conc of H+ = increase in pH
65
conc of H+ =
conc of acid
66
pH change with solid bases
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
67
add solid base to solution of an acid
pH will increase
volume of solution will remain the same
68
add aqueous base to solution of an acid
volume increases
H+ ions decreases
69
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
Part 1:
0.25 x 0.2 = 0.05
0.05 - 0.04 = 0.01
Part 2:
-0.37
70
calculating new pH of adding aqueous base
1) find change in volume
2) find change in {H+]
71
find new pH of solution after dilution
30ml HNO3
0.2moldm-3
20ml H2O
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
72
pH of solution after the reaction
50ml HBr
0.30moldm-3
0.120 moles KOH
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
73
doesnt rlly matter if you divide or multiply by 2
easier to multiply though
74
steps to calculating pH of mixed solutions
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
75
Calculate the pH of a solution formed when 50cm3 of 0.100mol dm-3 H2SO4 is added to 25cm3 of 0.150mol dm-3
-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
76
pH of 25cm3 of 0.25moldm-3 H2SO4. 100cm3 of 0.2moldm-3 NaOH
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
77
when calculating change in pH
-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
78
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.
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
79
Calculate the pH of the solution formed after the addition of 35cm3 of 0.150moldm-3 NaOH to the original 25cm3 of monotropic acid:
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
80
propanoic acid =
C3H6O2
1 carbon counts as part of the carboxyl group
81
weak acids
-weak acids and weak bases dissociate slightly in aqueous solutions
82
Ka
Ka = dissociation constant of a weak acid
83
equations for a weak acid
PKa = -log (Ka)
Ka = [H+] [A-] / [HA]
Ka = 10^-pKa
84
calculate Ka
10^-pka
85
calculate pKa
-log (Ka)
86
Ka constant
Ka = [H+] [A-] / [HA]
87
NH3 and COOH
NH3 = weak monoprotic base
Carboxylic acid = weak monoprotic acid
88
reversible reaction
HA ⇌ H+ + A-
HCOOH --> COOH- + H+
[H+] = [A-]
Ka = [H]^2 / [HA]
89
what is pKA
The pKa is the tendency of a chemical to gain or loose protons
90
low pKa and high pKA
Lower pKa = stronger acid
Lower pKa = weaker base
High pKA = stronger base
High pKA = weaker acid
91
Calculate the pH of 0.100 mol dm-3 propanoic acid with pKA = 4.87:
-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
calculate HA
[HA] = initial moles HA – moles OH- total volume (dm3)
93
[A-]
moles OH- added
total volume (dm3
)
94
diluted strong acid
[H+] = [H+]
old x old volume
new volume
95
diluted base
[OH–]
old x old volume
new volume
96
calculate pH of 0.0131 mol dm-3 of calcium hydroxide at 10 degrees
0.0131 x 2 = 0.0262
1 x 10^-14 / 0.0262 = ANS
put in log
97
units for Ka
moldm3
98
what is Ka only affected by
temperature
99
all weak acids have
a low Ka bc it barely dissociates in water
100
the higher the Ka value
the stronger the acid
101
state the expression for the acid dissociation constant of carbonic acid
Ka = [HCO3+] [H+] / [H2CO3]
102
assumptions made to calculations
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
dilute an acidic solution but a factor of 10
-H+ conc will alter (smaller)
-pH will change (greater)
104
dilution of weak acids
-add water (reactant) = shift right to increase conc of products
-pH increase = solution is diluted. But H+ also increases
105
in the case of HCl dilute by a factor of 10 increases the pH by 1 unit. Why dos ethanoic acid react differentyl
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
calculating mixtures of weak acids and strong bases
HA + OH- --> A- + H2O
107
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
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
conc of excess H+
[H+ ] = moles excess H+ total volume (dm3
109
conc of excess OH-
Work out new concentration of excess OHions [OH-] = moles excess OH- total volume (dm3 ) [H+ ] = Kw /[OH– ] pH = – log [H+ ]
110
conc of excess [HA]
Concentration of excess [HA] = (initial moles HA – moles OH-) / total volume (dm3 )
111
conc of 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
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:
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
weak acid with half alkali
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
weak acid reacts with strong base
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
excess HA / H+ for mixture
= Ka x [HA] / [A-]
116
moles of A-
Moles of [A-] = moles of [OH-]
117
what is Ka affected by
temperature
acid strength
118
explain why calibrating a pH meter just before it is used improves the accuracy of the pH measurement
Over time/after storage the meter does not give accurate readings
119
what are pH curves
pH curves successively measure the pH after the addition of small volumes of solution from a burette
120
method used to calibrate pH probe
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
what is the purpose of a titration
-calculate the concentration of an unknown solution by titrating it with a solution of known concentration
122
pH probe
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
method of producing titration curves
-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
HCl + NaOH --> NaCl + H2O
CH3COOH + NaOH --> CH3COONa + H2O
125
2H+ + CO3^2-
H2O + CO2
126
H+ + NH3 -->
NH4+
127
indicators used for acid vs base
acid = methyl orange (red) = pH range from 3.2-4.4
base = phenolphthalein = pink for base = 8.2-10.0
128
suggest why you wouldnt except a large rise in pH
base is weak
129
why is it difficult to do a titration of a weak acid against a weak base
no distinctive equvilant point so makes it difficult to identify the end
130
strong acid and strong alkali
acid pH = 0-1
base pH = 12-13
pH at eq = 7
pH range = 3-11
131
strong acid and weak alkali
pH acid = 0-1
pH base = 9-10
pH at eq = 5
pH range = 3-7
132
weak acid and strong alkali
-acid pH = 3-4
base pH = 12-13
pH at eq = 9
133
weak acid and weak alkali
pH at eq = 7
134
memorise each pH curve
135
diprotic acid =
dissociates 2 times
136
end point
-end point = final part of a reaction when the colour change can be detected
137
equivalence point
-equivalence point = pH where you had mixed solutions in correct proportions according to the equation
138
difference between end point and equivilance point
-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
0.01
concordant titres
140
buffer solution
Buffer solution = hold another solution at a certain pH = maintain a specific level of pH
141
points when describing pH curve
-describe start point
-end point
-volume
-equivalence point
142
equivilance point for curves
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
pH = pka
-pH = pka when half the acid/base has been neutralised. This means we can calculate Ka of a weak acid using pH
144
calculate Ka from pH curve
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
1:2 ratio in flask
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
Explain why running a weak base into a strong acid produces the graph:
-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
describe pH curve in detail
-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
Calculate the volume of 0.5moldm3 of NaOH which is added at equivalence to 25cm3 of 0.1moldm-3 H2SO4:
2NaOH + H2SO4 --> NaSO4 + 2H2O
0.5 x v / 2 = 25 x 0.1 / 1
= 0.25V = 25
V = 10cm 3
149
volume at equvilance
-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
by considering the species present explain why the pH at equvilance is not 7
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
constructing pH curve
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
indicator
Indicators can be considered as weak acids. The acid
must have a different colour to its conjugate base
153
how indicators work
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
deciding what indiactor to use
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
dissociation equations for H2SO4
H2SO4 --> H+ + HSO4-
HSO4- ⇌ H+ + SO4^2-
156
NH3 + H2O ⇌
NH4+ + OH-
157
[H+] =
square root Ka x [HA]
158
[H+] =
[HA]
159
Ka and pka values
stronger acids have a smaller Ka than weaker acids
stronger acids have larger pKa than weak acids
160
calculate pH of mixtures with weak acids and strong bases = excess acid
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
calculate pH of 0.20moldm-3 Sr(OH)2
kw / (0.20/2)
put in log
162
explain what is meant by the term buffer solution
solution which resists change in pH despite the addition of small amounts of acid/base
163
why is there a lot of unionised ethanoic acid and lots of ethanoate ions in a buffer solution
ethanoic acid = weak acid so partially dissociates
ethanoate ions = due to salt (CH3COONa) fully ionising
164
state what should be added to propanoic acid to form an acidic buffer
sodium propanoate --> with the salts to reform, the acid and resulting pH change
165
state the two equations that demonsrtate the formation of the buffer solution
CH3CH2COOH ⇌ CH3CH2COO- + H+
CH3CH2COONa --> CH3CH2COO- + Na+
166
what happens to equilbirum when H+ ions are added to the buffer
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
addition of an alkali
H+ + OH- --> H2O
CH3COOH ⇌ CH3COO- + H+
168
why is there a large supply of weak base
partially dissociates = equiilbirium will lie to the left
169
equation to show the formation of an alkaline buffer
NH3 + H2O ⇌ NH4+ + OH-
NH4Cl + NH4+ + Cl-
170
which forms a buffer solution with a pH less than 7
ethanoic acid and sodium ethanoate
171
which forms buffer solution with a pH more than 9
ammonia and ammonium chloride
172
when a small amount of acid is added to a buffer of ammonia and ammonium chloride what happens
hydrogen ions combine with NH3 to make NH4_
173
summary of acidic buffers
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
state how a buffer solution can be amde from solutions of KOH and CH3COOH
give the equation
add excess ethanoic acid to KOH
KOH + CH3COOH --> CH3COOK + H2O
CH3COO- from salt reacts with added H+
175
give the buffer solution and reagent which could be added to a solution of ammonia
-solution that resist a change in pH when small amounts of acid or base are added
-reagent = NH4Cl
176
state and explain the effect on the pH of this buffer solution when a small amount of HCL is added
H+ ions react with salt to reform the acid
equilbirium shifts right to oppose change in pH and resist the change
177
write an equation for the reaction of ethanoic acid when small amount of H+ is added
CH3COO- + H+ ⇌ CH3COOH
178
define buffering capacity
the quantiity of H+/OH- ions it can absorb without a significant change in pH
179
[H+] -> buffering equation
Ka x [HA] / [A-]
depends on the Ka of the buffer and the ratio of acid/base to the salt
180
define a buffer solution
Buffer solution = solution that resists changes in pH when small amounts of acid or alkali are added to them
181
river
What was present in the river to make it a buffer solution --> acid (from rain) and limestone (calcium carbonate)
182
blood
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
acidic buffer
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
how an acidic buffer solution works
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
alkaline buffer solution
-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
uses of buffer
-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
salt
Salt = conjugate base in acid buffers and conjugate acid in alkali buffers
188
dissociation of ethanoic acid
CH3COOH ⇌ CH3COO- + H+
->the acid is in large supply because it is a weak acid and therefore the equilbrium shifts to the left
189
formation of acidic buffers
-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
weak acid and strong base neutralisation
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
ethanoic acid buffer
CH3CO2H ---> (aq) CH3CO2- (aq) + H+
(aq)
192
small amount of alkali added to a buffer
CH3CO2H (aq) +OH- --> CH3CO2
-
(aq) + H2O (l)
193
small amount of acid added to a buffer
CH3CO2
-
(aq) + H + --> CH3CO2H (aq)
194
diluting a buffer solution
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
drawing pH graph
1) initial pH of acid
2) Find final pH of base = -log (Kw / H+ )
196
recognising half-equivlance
1) calculate initial moles of acid and base
2) calculate final moles (excess)
3) determine pH of solution formed
197
pH =
pKa
198
HCO3- equilbrium
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
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) 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
calculate pH of a buffer
1) Find moles of acid
2) Find moles of base
3) Ka x (HA / A)
4) put in log to get pH
201
pH of buffer with solid salt
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
how indicators work
1) position of eq
2) oppose the change
3) observation e.g turn blue
inidcators = weak acids
when weak acids are dissolved in water an equilbirum is established
HLit (aq) --> H+ + LiT-
reversible reaction = oppose change
203
where must colour change occur
between vertical region on pH curve
204
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
-so its not contaminated
-more salt and water is produced therefore decreasing conc of NaOH
205
why would the solution not be resisting changes in pH
not enough salt has been formed so the concentrations are not equal or not enough weak acid remaining
206
explaining pH curve
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
describe and explain the behaviour of the solution formed in region of circle on graph
-buffer is formed where at least half of the weak acid is neutralised by its conjugate base
208
2 ways to make a buffer
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
state how buffer solution can be made from solutions of potassium hydroxide and ethanoic acid
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
calculate the pH of the solution formed when 200cm3 of water are added to 50cm3 of 0.50moldm-3 HCl
0.50 x 50/200
-log (ANS)
1) conc x old/new volume
2) put into log to get pH
211
Calculate the pH of the solution formed when 50 cm3 of water is added to 100 cm3 of 0.200 mol dm-3 NaOH.
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
calculate the pH of the buffer solution prepared by adding 1.00g of sodium ethanoate to 250cm3 of 0.100moldm-3 ethanoic acid
1) find moles for both substances
2) ka x mol / mol
213
the final mixture contained Ca(OH)2
solubility = 0.400gdm-3
Kw = 6.80 x 10^-15
find pH
0.400 / mr (74)
ANS x 2
Kw / ANS
-log (ANS)
214
Describe how you would obtain the pH curve for the titration.
Measure pH (of the acid)
* Add alkali in known small portions
Allow 1 – 2cm3
.
* Stir mixture
* Measure pH (after each addition)
* Repeat until alkali in excess
Allow 27 – 50cm3
.
* Add in smaller increments near endpoint
Allow 0.1 – 0.5cm3
.
215
describe briefly how you would ensure a pH meter reading is accurate
calibrate meter with solution of known pH
216
why would this solution be suitable as a buffer
solution contains both HA and A-
217
n aqueous solution, sulfuric acid acts as a strong acid. The H2SO4 dissociates to
form HSO4
–
ions and H+
ions.
The HSO4
–
ions act as a weak acid and dissociate to form SO4
2–
ions and H+
ions.
Give an equation to show each stage in the dissociation of sulfuric acid in aqueous
solution.
(equation 1 = no EQ)
H2SO4 → HSO4– + H+
HSO4– ⇌ SO42– + H+
218
Some sodium sulfate is dissolved in a sample of the solution from part (d).
Explain why this increases the pH of the solution.
equilibrium shifts left to remove increased SO4^2-
so H+ decreases
219
Use this information to describe a simple test, other than smell, to show that
ammonia is evolved.
concentrated HCl = white fumes
220
Write an equation for the reaction of ethylamine (CH3CH2NH2) with water to
form a weakly alkaline solution.
CH3CH2NH2 + H2O --> CH3CH2NH3+ + OH–
221
why is the solution formed weakly alkaline
partially dissociates so only little OH- forms
222
Give the formula of an organic compound that forms an alkaline buffer solution
when added to a solution of ethylamine.
CH3CH2NH3Cl
223
Use the equation in part (a) to explain how a solution containing sodium
hydrogencarbonate and sodium carbonate can act as a buffer when small amounts
of acid or small amounts of alkali are added.
Acid: Increase in concentration of H+ ions, equilibrium moves to the left
Alkali: OH- reacts with H+ ions, equilibrium moves to the right (to replace the H+ions)1Concentration of H+ remains (almost) constant
224
How buffer solution resists pH change
CH3COO– (from salt) reacts with (added) acid/H+
225
buffer solution must contain
high conc of HA and A-
acid which doesnt dissociate
226
why cant strong acids and weak acids be buffers
strong acid = too low HA and fully dissociates
weak acid = A- (conjugate base) is too low concentration
227
weak acid (aq) + salt (conjugate base) = buffer
weak acid = aqueous standard state = provides high conc of undissociates HA
salt = provides high conc of conjgate base
228
also make buffer by adding strong base to weak acid
half neutralise
this forces salt to be made
CH3COOH + NaOH --> CH3COONa + H2O
229
1 mol of weak acid
0.55 mol of base
neutralise half of weak acid
230
What substance could be added to a solution of sodium cyanide,
NaCN (aq), to make a buffer solution?
HCN
CN- = conjugate base
231
buffer solution + equilbrium shifts
232
add acid to buffer
H+ increases
equilbrium shifts left to reduce amount of H+ ions
233
base added to buffer
OH- + H+ --> H2O
reacts with acid = reduces H+ ion conc so equilbrium shifts right to increase H+ ion conc
234
making alkaline buffer solutions
weak base NH3 with strong acid
or
salt of conjugate acid e.g NH3 + NH4Cl
235
Use this equation to explain how a solution containing carbonic acid (H2CO3) and hydrogencarbonate (HCO3-) can act as a buffer when small amounts of acid or alkali are added.
when small amounts of acid are added H+ increases so eq shifts left to counteract this
when small amounts of base are added OH- reacts with H+. H+ decreases so eq shifts right to replenish this
236
calculate buffer = known solid + acid
find moles
moles / volume to get concentration (assume solid salt doesnt change volume)
use regular Ka equation to find [h+]
237
mol HA in buffer =
mol initial HA - mol NaOH
238
equation for acid added to buffer solution
H+ + A- --> HA
239
equation for base added to buffer solution
HA + OH- --> A- + H2O
240
add HCl =
moles of HA increases
moles of A- decreases
241
explain why dilution with a small volume of water does not affect the pH of a buffer solution
[H+] is constant
volume canceled out
values of Ka, HA, A- = unchanged by addition of water
[HA]/[A]
242
a buffer solution has a constant pH even when diluted. Use a mathematical expression to explain this
[HX] / [X-]
243
a student plans to titrate butanoic acid solution with a solution of ethylamine. Exlain why this could not be done using an indicator
weak acid and weak base titration
pH change is too gradual
244
suggest why chloroethanoic acid is a stronger acid than ethanoic acid
chlorine = more electronegative
withdraws electrons making the OH bond more polar
245
rate determine step =
when all the reactants are used up
246
acid vs base buffer
Acidic buffer solutions contain a weak acid and the salt of
that weak acid.
Basic buffer solutions contain a weak base and the salt of
that weak base.
247
equations to show pH = pka from graph
[A-] = [HA]
[HA] = [H+]
pH = log [H+]
248
explain why with the dilution of a small volume of water the pH of the buffer solution doesnt change
Dilution of a buffer solution with a small volume of water generally does not significantly change the pH because the ratio of the weak acid to its conjugate base (or weak base to its conjugate acid) within the buffer remains relatively constant.
While a small amount of water added to a buffer will technically change the concentrations of both the acid and its conjugate base, the change in the ratio [A-]/[HA] is minimal, especially for dilute buffers. This is because the dilution affects both the acid and its conjugate base proportionally, leading to a negligible change in the overall pH
249
why does the conc of NaOH slowly decrease when left open to air
NaOH reacts with CO2 in the air
250
why would an end point be difficlt to judge using an indicator
pH is gradual = no sharp end point
an indicator would change colour over a range of values
