Capítulo 6. Gestión interna
6.4 Gestión de Planificación, Proyectos y Procesos Institucionales
Unit 4 Rates, equilibria and further organic chemistry
Chapter 5 Acid–base equilibria
Below 25°C, Kwis less than 1.0 × 10–14mol2dm–6and so pure water has [H+] < 1.0 × 10–7and its pH is greater than 7.0.
eRemember that the higher the value of [H+], the lower the value of the pH.
6 pKw= 14
Kw= 10–pKw= 1.0 × 10–14mol2dm–6
Kw= [H+] × [OH–], so [OH–] = Kw/[H+] = 1.0 × 10–14/[H+]
a If [H+] = 1.0 × 10–2mol dm–3, [OH–] = 1.0 × 10–14/1.0 × 10–2= 1.0 × 10–12mol dm–3. [OH–] < [H+], so the solution is acidic.
b If [H+] = 2.2 × 10–7mol dm–3, [OH–] = 1.0 × 10–14/2.2 × 10–7= 4.5 × 10–8mol dm–3. [OH–] < [H+], so the solution is acidic.
c If [H+] = 3.3 × 10–10mol dm–3, [OH–] = 1.0 × 10–14/3.3 × 10–10= 3.0 × 10–5moldm–3. [OH–] > [H+], so the solution is alkaline.
eThe pH of the solution can be calculated from [H+]. This can be used to find the pOH and hence [OH–]. For example, for the solution in b:
pH = –log (2.2 × 10–7) = 6.66 pOH = 14 – pH = 14 – 6.66 = 7.34 [OH–] = 10–7.34= 4.6 × 10–8mol dm–3
(If all the numbers are left in full on the calculator, the value obtained is 4.5 × 10–8mol dm–3.) 7 a pH = –log [H+] = –log (4.4 × 10–5) = 4.36
b pH = –log (5.5 × 10–9) = 8.26
c pOH = –log [OH–] = –log (6.6 × 10–2) = 1.18 pH = 14 – pOH = 14 – 1.18 = 12.82
d pOH = –log (7.7 × 10–11) = 10.11 pH = 14 – 10.11 = 3.89
eThe alternative method for calculating the pH in c and d is:
Kw= [H+] × [OH–] = 1.0 × 10–14mol2dm–6, so [H+] = 1.0 × 10–14/[OH–] For the solution in c:
[H+] = 1.0 × 10–14/6.6 × 10–2= 1.52 × 10–13, so pH = –log [H+] = 12.82 8 a [H+] = 10−pH= 10−1.33= 0.0468 mol dm−3
b [H+] = 10−7.00= 1.00 × 10−7mol dm−3 c [H+] = 10−13.67= 2.14 × 10−14mol dm−3 9 a At pH = 7, [H+] = [OH–], so the ratio is 1:1.
b At pH = 10, [H+] = 10–pH= 1.0 × 10–10
[OH–] = 1.0 × 10–14/1.0 × 10–10= 1.0 × 10–4mol dm–3
Ratio [H+]:[OH–] = 1.0 × 10–10/1.0 × 10–4= 1:106or 1 to a million c At pH = 3, [H+] = 10–pH= 1.0 × 10–3
Chapter 5 Acid-base equibria
[OH–] = 1.0 × 10–14/1.0 × 10–3= 1.0 × 10–11mol dm–3
Ratio [H+]:[OH–] = 1.0 × 10–3/1.0 × 10–11= 1.0 × 108:1 or one hundred million to one
eIn these calculations it is assumed that the temperature is 25°C.
An alternative method involves using the expression pH + pOH = 14, in part b:
If pH = 10, [H+] = 10–pH= 1.0 × 10–10
pOH = 14 – 10 = 4, so [OH–] = 1.0 × 10–4mol dm–3 10 a HBr is a strong acid and so is totally ionised:
[H+] = [HBr] = 0.200 mol dm–3 pH = –log 0.200 = 0.70
b LiOH is a strong base and so is totally ionised:
[OH–] = [LiOH] = 0.200 mol dm–3 pOH = –log 0.200 = 0.70
pH = 14 – pOH = 14 – 0.70 = 13.30
c Sr(OH)2is a strong base with two OH–ions per mole:
[OH–] = 2 × [Sr(OH)2] = 2 × 0.0500 = 0.100 mol dm–3 pOH = –log 0.100 = 1.00
pH = 14 – pOH = 14 – 1.00 = 13.00
eSulfuric acid, H2SO4, is a strong acid in its first ionisation only. A 0.0500 mol dm–3solution of the acid will have [H+] slightly greater than 0.0500 but not as high as 0.100 mol dm–3.
11 As it is a strong acid, [H+] = 2.0 mol dm−3, so pH = −log 2.0 = −0.30.
I When diluted 10 times, [H+] = 0.20 mol dm−3, so pH = 0.70.
I When diluted 100 times, [H+] = 0.020 mol dm−3, so pH = 1.70.
I When diluted 1 000 000 times, [H+] = 0.0000020 mol dm−3, so pH = 5.70.
eNote that pH values can be negative. This occurs when [H+] > 1 mol dm−3. They can also be greater than 14, if [OH−] >
1 mol dm−3. It is a fallacy to think that the pH scale goes from 0 to 14.
12 a HOCl H++ ClO–
Ka = = 3.02 × 10–11mol dm–3
b It is assumed that [H+] = [ClO–]:
Ka= [H+]2/[HOCl]
[H+] = Ka× [HOCl] = 3.02 × 10–11 × 0.213 = 2.54 × 10–6mol dm–3 pH = –log (2.54 × 10–6) = 5.60
eThe assumption that [H+] = [ClO–] is not valid for calculations involving buffer solutions. However, in weak acid and in buffer solution calculations, it is assumed that a negligible amount of the weak acid ionises. Thus, [HOCl]eq= [HOCl]initial
= 0.213 mol dm–3.
Always check that your answer to the pH of a solution of an acid is less than 7.
[H+][ClO–] [HOCl]
Chapter 5 Acid-base equibria
13 HNO2 H++ NO2–
Ka =
As with all weak acid calculations, it is assumed that [H+] = [NO2–]:
[H+] = 10–pH= 10–2.02= 9.55 × 10–3mol dm–3= [NO2–]
The assumption can also be made that a negligible amount of the HNO2ionises:
[HNO2]eq= [HNO2]initial= 0.200 mol dm–3 Ka = = 4.56 × 10–4mol dm–3
eThe assumption that [HNO2]eq= [HNO2]initialneed not be made here, as [NO2–] = 9.55 × 10–3mol dm–3and so 9.55 × 10–3 mol dm–3of HNO2must have ionised. Thus, [HNO2]eq= 0.200 – 9.55 × 10–3= 0.190 mol dm–3. This gives a more accurate value of Ka= 4.80 × 10–4mol dm–3.
14 C2H5COOH H++ C2H5COO–
Ka = = 1.35 × 10–5mol dm–3
[H+] = [C2H5COO–] = 10–pH= 10–3.09= 8.13 × 10–4mol dm–3 [C2H5COOH] =
= = 0.0489 mol dm– 3
eCheck that your value of the concentration of a weak acid solution is somewhere between 2 and 0.01 mol dm–3. If it is not, then you probably forgot to square the numerator or failed to treat one of the powers of ten as a negative number.
15 CH2(OH)COOH H++ CH2(OH)COO–
Ka = = 10–pKa= 10–3.83= 1.48 × 10–4mol dm–3 [H+] = [CH2(OH)COO–] and [CH2(OH)COOH] = 1.05 mol dm–3 [H+]2= Ka× [CH2(OH)COOH]
[H+] = Ka× [CH2(OH)COOH] = 4.36 × 10−4× 1.05 = 0.0125 mol dm–3 pH = –log 0.0125 = 1.90
16 a Kb =
eThe term [H2O] is left out because it is the solvent.
b [CH3NH3+] = [OH–] [OH–]2= Kb× [CH3NH2]
[OH–] = Kb× [CH3NH2] = 4.36 × 10−4× 0.200
= 9.34 × 10–3mol dm–3 pOH = –log(9.34 × 10–3) = 2.03
pH = 14 – pOH = 14 – 2.03 = 11.97 [CH3NH3+][OH–]
[CH3NH2] [H+][CH2(OH)COO–]
[CH2(OH)COOH]
(8.13 × 10–4)2 1.35 × 10–5 [H+][C2H5COO–]
Ka [H+][C2H5COO–]
[C2H5COOH]
(9.55 × 10–3)2 0.200 [H+][NO2–]
[HNO2]
Chapter 5 Acid-base equibria
eThe calculation is similar to those for weak acids, with similar assumptions. The difference is that [OH–] is calculated first and used to evaluate the pH.
17 a A buffer solution is one in which the pH hardly alters when small amounts of acid or base are added.
eDo not say that the pH is constant. The definition of a buffer as ‘a solution that resists change in pH when small amounts of acid or base are added’ is also acceptable.
b Ka= = = 1.8 × 10–5mol dm–3
molar mass of sodium ethanoate (CH3COONa) = 3.0 + (2 × 12.0) + (2 × 16.0) + 23.0 = 82.0 g mol–1 amount of sodium ethanoate = 5.65 g/82.0 g mol–1= 0.0689 mol
[salt] = 0.0689/0.100 = 0.689 mol dm–3
[H+] = = = 2.61 × 10–5mol dm–3
pH = –log [H+] = –log (2.61 × 10–5) = 4.58
eThe assumptions made for buffer solutions are:
G The salt is totally ionised, so [CH3COO–] = [salt].
G Ionisation of the acid is low and is also suppressed by the anions from the salt, so [CH3COOH]eq= [acid]initial. A useful way of checking your answer in buffer calculations is to compare the amounts of acid and salt in the buffer. If there are more moles of acid than salt, the pH of the buffer will be less (more acidic) than the pKavalue of the weak acid. In this question, pKa= –log Ka= 4.74 and [acid] > [salt], so the calculated pH (4.58) should be less than 4.74.
Similarly, if the pH is greater than the pKaof the weak acid, there will be more moles of salt than weak acid in the buffer.
18 NaOH + CH3COOH → CH3COONa + H2O
When the sodium hydroxide is added, the salt sodium ethanoate is formed. There is excess acid, so the mixture is a buffer solution.
amount of salt = amount of NaOH = 2.00 mol dm–3× 0.050 dm3= 0.100 mol initial amount of acid = 1.00 mol dm–3× 0.150 dm3= 0.150 mol
amount of acid reacted = amount of NaOH added = 0.100 mol amount of acid in excess = 0.150 – 0.100 = 0.050 mol
eYou must not make the mistake of thinking of sodium hydroxide as the conjugate base in the buffer mixture. It reacts with some of the acid to form the anion, CH3COO–, which is the conjugate base of ethanoic acid.
The second common error is to fail to divide the moles of each component by the total volume (which must be in dm3).
This type of buffer calculation is the hardest type tested at A-level.
1.8 × 10–5× 0.25
19 C2H5COOH H++ C2H5COO– Ka= 10–pKa= 10–4.87= 1.35 × 10–5=
=
[H+] = 10–pH= 10–4.50= 3.16 × 10–5mol dm–3
moles of salt = volume × concentration = 0.050 × 1.00 = 0.050
moles of acid = = = 0.117
volume of acid = = = 0.117 dm3= 117 cm3
eAs the acid and the salt are in the same solution in the buffer, the ratio [salt]/[acid] is the same as the ratio (moles of salt)/(moles of acid).
moles of calcium ethanoate = 0.225/2 = 0.1125
molar mass of Ca(CH3COO)2= 40.1 + (4 × 12.0) + (4 × 16.0) + 6 = 158.1 g mol–1 mass of calcium ethanoate = moles × molar mass = 0.1125 × 158.1 = 17.8 g
eIn this example, [CH3COO–] = 2 ×[salt].
21 CH3COOH H++ CH3COO–
Ka= 1.80 × 10–5= =
[H+] = 10–pH= 10–4.44= 3.63 × 10–5mol dm–3 As both salt and acid are mixed in the same solution:
=
= = =
This means that one-third of the acid must be neutralised, forming one-third salt and two-thirds unreacted acid.
The initial volume of acid is 100 cm3and both the acid and the sodium hydroxide are of the same concentra-tion:
eThere are two important points in this question. The first is that the sodium hydroxide reacts with some of the ethanoic acid to form the salt, sodium ethanoate. The second point is that there must be twice as much acid left as salt formed.
This means that one-third of the acid must react.
22 a CH3NH3++ H2O CH3NH2+ H3O+
As H3O+ions are formed, the solution will be acidic (pH < 7).
eThe equation is similar to that of the ammonium ion reacting with water.
b CN–+ H2O HCN + OH–
eAs OH–ions are formed, the solution will be alkaline (pH > 7).
c CO32–+ H2O HCO3–+ OH–, forming an alkaline solution.
eA further reaction can take place:
HCO3–+ H2O H2CO3+ OH–, but this occurs to a very small extent because the OH–ions produced in the first reaction suppress this second reaction.
d I–+ H2O → no reaction, and so the solution will be neutral (pH = 7 at 25°C). HI is a strong acid (stronger than HCl), and so I–is too weak a conjugate base to react with an acid as weak as water.
eI–ions will react with very strong acids, such as concentrated sulfuric acid, to form HI (which is subsequently oxidised to iodine by concentrated sulfuric acid).
23 a
eNote the rule of two to determine the pH from the strong-base value.
G Ammonia is a weak base, so the starting pH ≈ 11 (the strong-base value of 13 – 2).
G The pH of the equivalence point is ≈ 5 (the strong-base value of 7 – 2).
G The pH after an excess of strong acid has been added ≈ 1 (the strong-acid value).
Note also that as the concentration of the acid is twice that of the alkali, the equivalence point is when 20 cm3of HCl has been added to 40 cm3of NH3solution.
The whole pH range of the indicator must lie on the vertical section of the graph. In this example, where a strong acid is being added to a weak base, the pH of the vertical section is from about pH 7 to pH 3. The ranges for methyl orange, bromophenol blue, bromocresol green and methyl red all lie fully within pH 7–3 and so would be good indicators for the titration.
eA common error is to state that the pKindvalue must be somewhere on the vertical section. The whole range (usually pKind± 1) must lie within the vertical section.
12
10 8
6 4
2 0
10 20 30 40 0
Buffer
Volume of HCl/cm3 pH
Chapter 5 Acid-base equibria
b A solution is a buffer when it contains significant amounts of both the acid and its conjugate base.
Here that is when some, but not all, of the ammonia has been neutralised by the hydrochloric acid. The solution then contains both NH3 and NH4+. It is best as a buffer when half the ammonia has been neutralised, that is when 10 cm3 of hydrochloric acid have been added.
24 a
eNote the initial rapid increase in pH followed by the flatter (buffered region), then the rapid rise around the equivalence point, flattening off to the pH value for strong alkali.
b The pKaequals the pH when half the acid has been neutralised. As the acid and the alkali are both 0.100 mol dm–3, half-neutralisation will occur when 10 cm3of NaOH has been added. Thus, pKa= pH at 10 cm3= 4.65.
Ka= 10–pKa= 10–4.65= 2.24 × 10–5mol dm–3
c When the acid is fully neutralised, [salt] = 1–2[acid]initial= 0.050 mol dm–3. Neutralisation is at the point halfway up the vertical section of the graph. The pH at this point is 9.05, so the pH of the salt solution is 9.05.
eThe addition of sodium hydroxide increases the volume of the solution. At the equivalence point, the volume has gone from 20 cm3to 40 cm3, so the concentration of the salt is 0.050 mol dm–3.
d The full colour change (the pH range) of the indicator must be within the vertical part of the titration curve.
Here the vertical part goes from 7.1 to 11.0 and therefore thymol blue or phenolphthalein would be suitable indicators.
eDo learn to spell ‘phenolphthalein’ correctly. A missing ‘h’ or the ending ‘-ene’ is acceptable to examiners, but ‘phenyl-’
or ‘-palin’ is not.
25 HThy H++ Thy–
yellow blue
I When acid is added, [H+] is increased. The equilibrium is shifted to the left and the indicator turns yellow.
I When alkali is added, the OH–ions react with the H+ions from the indicator and reduce the value of [H+].
This drives the equilibrium to the right and the colour changes from yellow to green and finally to blue.
26 a HBr and HCl are strong acids, which means that they are totally ionised. The neutralisation reaction for both is H++ OH–→ H2O, so the enthalpy of neutralisation is the same.
b The enthalpy change for neutralisation of a weak acid can be considered as the sum of two reactions:
HA H++ A– ΔH = x kJ mol–1
As alkali is added, the OH–ions react with the H+ions and so the first equilibrium is driven to the right.
The overall ΔHneut= x + (–57) kJ mol–1
If the first reaction is endothermic and the value of x is small, the enthalpy of neutralisation of the weak acid is only a little less exothermic than that of a strong acid. This is true regardless of the extent of ionisation of the weak acid. What controls the difference between the two enthalpies of neutralisation is the value of x.
c Hydrocyanic acid, HCN, is a weak acid and the CN–ions are only weakly hydrated in solution. This means that the value of x is much more endothermic than the value for ethanoic acid and so the enthalpy of neutralisation is significantly less exothermic than –57 kJ mol–1.
eThe value of ΔHneutfor a weak acid depends mostly on how endothermic its enthalpy of ionisation is, rather than on the extent of ionisation.
27 a Chloric(V) acid, HClO3:
b Hydrocyanic acid, HCN:
c Sulfurous acid, H2SO3:
28 The acid is slightly ionised:
HOCl H++ ClO–
As the degree of ionisation is very low, [ClO–] is also low.
When OH–ions are added, they react with undissociated HOCl molecules:
HOCl + OH–→ ClO–+ H2O
However, as [ClO–] was initially very small, even small amounts of OH–will cause its value to increase signif-icantly.
Since [H+] =
the value of [H+] and, hence, the pH will alter significantly.
If H+ions are added, the equilibrium is driven to the left, but as [ClO–] is small initially, its value decreases significantly and so the ratio and hence the value of [H+] will change markedly.
eFor a solution to be a buffer, the concentrations of both the weak acid and its conjugate base must be large relative to any amounts of added H+or OH–.
29 e Try Googling ‘blood buffer solutions’.
[HOCl]
[ClO–] Ka× [HOCl]
[ClO–]
O S
O
H O H
C N
H
O Cl
O
H O
Chapter 5 Acid-base equibria
Summary worksheet (www.hodderplus.co.uk/philipallan)
1 D A higher pH means that the solution is less acidic or more alkaline. The correct answer, therefore, is the solution with the lowest [H+]. The numbers in options A, B and C are all larger than 1.0 × 10−7, so they are all more acidic and have a lower pH than the solution in option D.
2 A A buffer solution must consist of an acid/base conjugate pair, i.e. a weak acid and its salt or a weak base and its salt. If the methanoic acid in option B is in excess, the solution will contain methanoic acid and sodium methanoate, as a result of some of the methanoic acid having been neutralised. So this is a buffer solution. Ammonia is a weak base and the ammonium ions in ammonium sulfate are its conjugate acid, so option C is also a buffer. Carbon dioxide reacts with water to form carbonic acid which is a weak acid and the hydrogencarbonate ion is its conjugate base, so option D is also a buffer solution. The answer to this negative question is option A, as this mixture is not a buffer. The solution will consist of either a strong acid and its salt or a strong base and its salt, depending on whether the hydrochloric acid or the sodium hydroxide is in excess.
3 B This is another negative question. It may be helpful to write ‘true’ or ‘false’ beside each option. Option A is a true statement — it is the definition of an acidic solution. Statement B is false. The pH of a neutral solution is only 7 at 25°C. It is always worth checking the other options. Option C is a true statement. A weaker acid will have a smaller value of Ka, say 10−6 compared with, say, 10−5for the less weak acid. Thus pKaof the weaker acid is –log Ka= +6; –log Kaof the less weak acid is +5, which is smaller than the pKaof the weaker acid. Statement D is true, as all aqueous solutions contain both ions and, at 25°C, [H+] × [OH−] = 1.0 × 10−14.
4 C The pH of a buffer solution of a specified acid/base conjugate pair depends on the ratio of [acid] to [conjugate base]. This does not change on dilution. Option A is incorrect because the pH of an acid rises (because less acidic) on dilution. Option B is incorrect because diluting an alkali decreases its pH, as the solution becomes less alkaline. On dilution, the pH of a weak acid changes less than that of a strong acid, so option D is incorrect.
5 C The pH of the salt of a weak acid and a strong base will be around 9, and the vertical part of the pH curve will be about ±2 pH units from this value, i.e. from 7 to 11. This means that an indicator must change colour completely within the range 7 to 11, so its pKind value should be between 8 and 10. The only indicator in the list that satisfies this condition is thymol blue, option C.
6 B The conjugate acid of a base is obtained by adding an H+ion. H3O+is the conjugate acid of H2O, not of the OH−ion. O2−is the conjugate base of OH−, not its conjugate acid. NaOH contains OH−ions and is the base itself.
7 C Option A is incorrect as the pH does alter (but only slightly) when even small amounts of acid or base are added. Option B is incorrect because the pH will change significantly if large amounts of acid or base are added. At first sight, option D looks true. However, it is wrong, as a buffer could also be a mixture of a weak base and its salt.
8 B All aqueous solutions contain some OH−ions, so option A is incorrect. The calculation of the concentration of OH−ions from pH can be carried out in two ways:
pOH = 14 − pH = 14 − 2 = 12
[OH−] = 10−pOH= 10−12= 1.0 × 10−12mol dm−3 or
[H+] = 10−pH= 10−2= 0.010 mol dm−3
[OH−] = Kw/[H+] = 1.0 × 10−14/0.010 = 1.0 × 10−12mol dm−3
Chapter 5 Acid-base equibria
9 A In this reaction, the sulfuric acid has protonated the HNO3molecule. This means that sulfuric acid is acting as an acid and nitric acid as a base. This can only happen if sulfuric acid is a stronger acid than nitric acid.
Therefore, option B is wrong. Option C is a true statement about the two substances but the reaction does not show this, so it is not the answer to this question. Option D is incorrect because sulfuric acid is reacting as an acid.
10 D Ammonia is a weak base. Hydroiodic acid, HI, is an even stronger acid than hydrochloric acid, HCl. The pH of the salt of a weak base and a strong acid is less than 7. Option A is the pH of a salt of a weak acid and a strong base. Option B is incorrect as many salts have pH values greater than or less than 7. Option C is incorrect as hydroiodic acid is a strong acid.
11 A The amount of acid = mass/molar mass = 0.20/158.1 = 0.00127 mol. [acid] = moles/volume = 0.00127/0.100 = 0.0127 mol dm−3. As it is a strong acid, the pH = −log [acid]= −log 0.0127 = 1.90. In option B, the mass has been divided by 100. Here, two errors have been made — not converting mass to moles and using the volume in cm3 not dm3. In option C the number of moles has been calculated correctly but this has been used as the concentration. In option D, the correct number of moles has been divided by 100 cm3, rather than by 0.100 dm3.
Chapter 5 Acid-base equibria
1 a The four structural isomers are:
1,1-dichloropropane 2,2-dichloropropane
1,2-dichloropropane 1,3-dichloropropane
b 1,2-dichloropropane contains an asymmetric carbon atom: four different groups are attached to the middle carbon (a –CH2Cl group, a Cl, an H and a –CH3group). The two optical isomers are:
eMake sure that you draw the two optical isomers as object and mirror image. Each structure should have one bond shown as a wedge and one as a dashed or dotted bond. The other two bonds must not be at 180° to each other.
2 The conditions for geometric isomerism are:
I a double bond (this restricts rotation)
I two different groups on each of the carbon atoms of the double bond
1-chloropropene fulfils both of these conditions as one of the carbon atoms of the double bond has a –CH3
group and an H atom and the other has a Cl atom and an H atom. 3-chloropropene has two hydrogen atoms on one of the doubly bonded carbon atoms, so geometric isomerism cannot occur.
eA ring structure can also cause geometric isomerism because rotation is impossible without totally twisting the ring.
The two chlorine atoms in 1,2-dichlorocyclohexane can be located above the ring (cis) or one above and one below (trans).
A double bond consists of a σ-bond, with a head-on overlap of two atomic orbitals, and a π-bond, which is formed by a side-by-side overlap of two p-orbitals. Rotation about the double bond would involve breaking the π-bond. This process
A double bond consists of a σ-bond, with a head-on overlap of two atomic orbitals, and a π-bond, which is formed by a side-by-side overlap of two p-orbitals. Rotation about the double bond would involve breaking the π-bond. This process