CC2_SM_Mod_2

Solutions Manual: Module 2

The acidic environment

© Pearson Australia (a division of Pearson Australia Group Pty Ltd) 2010.

This page from Chemistry Contexts 2 Teacher’s Resource second edition may be reproduced for classroom use.

CHAPTER 6: ACIDS, BASES AND INDICATORS

Review exercise 6.1 1 a acetic acid, CH3COOH

citric acid, 2-hydroxypropane-1,2,3-tricarboxylic acid

b sulfuric acid, H2SO4

nitric acid, HNO3 c magnesium hydroxide, Mg(OH)2

2 Own answers 3 Own answers

4 a Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

b CO32−(aq) + 2H+(aq) → CO2(g) + H2O(l)

c 2H+(aq) + K2O(s) → 2K+(aq) + H2O(l)

d H+(aq) + OH– (aq) → H2O(l)

5 a 2Cr(s) + 2OH–(aq) + 6H2O(l) → 2[Cr(OH)4]–(aq) + 3H2(g)

b Cr(OH)3(s) + OH–(aq) → [Cr(OH)4]–(aq) 6 Acid solutions have excess hydrogen ions and base solutions have excess hydroxide

ions. These ions are free to move and carry a current.

Review exercise 6.2

1 a K2O ionic bonding; Ga2O ionic; Br2O7 covalent b K2O basic; Ga2O amphoteric; Br2O7 acidic

2 CO2(g) + H2O(l) → H2CO3(aq) 3 Natural sources of sulfur dioxide:

• bacteria decomposing organic matter to make H2S, which is then oxidised atmospherically.

• volcanic gases

• bushfire smoke.





Sources due to human activities:

• burning of fossil fuels

• smelting of sulfide ores such as CuFeS2 and ZnS. 4 SO2—H2SO3; SO3—H2SO4; NO2—HNO3; CO2—H2CO3

5 Increased incidence of low pH in lakes and rivers and continued damage to metal and sandstone buildings.

6 The sulfur oxides in the atmosphere usually come from burning fossil fuels in power stations or smelting sulfide ores in smelting plants. Nitrogen oxide and carbon emissions usually come from motor vehicles. Therefore acid rain usually occurs in densely populated areas or industrial areas.

Review exercise 6.3 1 a Acidic. The solution has a pH between 4 and 6.

b The pH range is 4 to 8. This solution could be neutral, acidic or basic. c The solution is slightly basic. The pH is between 7.5 and 8.

2 a Bromothymol blue: yellow; phenolphthalein: colourless; methyl orange: red b Bromothymol blue: blue; phenolphthalein: pink; methyl orange: yellow

c Bromothymol blue: blue; phenolphthalein: colourless; methyl orange: yellow

Chapter 6 Application and investigation

1 Not all acids are dangerous. Acids are present in foods we eat, e.g. ethanoic acid in vinegar, lactic acid in milk. Also, acids are found in our bodies, e.g. lactic acid, hydrochloric acid. The concentration of acids should be considered.

2 a 2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)

b KHCO3(s) + HNO3(aq) → KNO3(aq) + CO2(g) + H2O(l)

c Fe2O3(s) + 3H2SO4(aq) → Fe2(SO4)3(aq) + 3H2O(l)

d Ba(OH)2(aq) + 2HF(aq) → BaF2(aq) + 2H2O(l)

3 a 2Al(s) + 2NaOH(aq) + 6H2O(l) → 2Na[Al(OH)4](aq) + 3H2(g)

b KOH(aq) + Fe(OH)3(s) → no reaction

c 2KOH(aq) + Zn(OH)2(s) → K2[Zn(OH)4](aq)

d 2NaOH(aq) + SO3(g) → Na2SO4(aq) + H2O(l)

4 a KHC4H4O6 + NaHCO3 → KNaC4H4O6 + CO2(g) + H2O(l)

b Fe2O3(s) + 6HCl(aq) → 2FeCl3(aq) + 3H2O(l)





c Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(l)

d H2SO3(aq) + CaCO3(s) → CaSO3(aq) + CO2(g) + H2O(l)

5 Al(OH)3 reacts with both acids and bases, and therefore is amphoteric.

Reacting with acid: Al(OH)3(aq) + 3HCl(aq) → AlCl3(aq) + 3H2O(l)

Reacting with base: Al(OH)3(aq) + NaOH(aq) → Na [Al(OH)4](aq)

6 Investigation 7 Investigation

8 A useful indicator is one whose colour change occurs over a selected pH range. Methyl orange changes colour too far outside the narrow 7.2–7.6 range for swimming pools. Litmus changes colour over this range, but the colour difference in the 7.2–7.6 range may be too subtle to detect easily. Universal indicator undergoes many colour changes and the change over the 7.2–7.6 range, from yellow-green to green may also be too subtle to detect easily. Bromothymol blue could be a useful indicator. The change for phenolphthalein is above a pH of 7.6 and so this indicator would not be useful.

9 Investigation

10 Acid rain has a pH less than 5; methyl orange when yellow means pH is greater than 4.5, and litmus when red indicates a pH of less than 7. Therefore the pH of the sample of rain water lies between 4.5 and 7, and so is unlikely to be classified as acid rain.

11 a Na2O, Al2O3, SiO2, P4O10, Cl2O7

b K2O, Ga2O3, GeO2, As2O3, Br2O7 12 CO2(aq) + H2O(l) ↔ H2CO3(aq) carbonic acid

4NO2(g) + 2H2O(l) + O2(g) → 4HNO3(aq) nitric acid

SO2(g) + H2O(l) → H2SO3(aq) sulfurous acid

Cl2O7(l) + H2O(l) → 2HClO4 perchloric acid

13 a SO2(g) + H2O(l) → H2SO3(aq)

b H2SO3(aq) + CaCO3(s) → CaSO3(aq) + CO2(g) + H2O(l)

14 Investigation 15 Investigation

16 Investigation





CHAPTER 7: CHEMICAL EQUILIBRIUM

Review exercise 7.1

1 a NH3(g) + HCl(g) ↔ NH4Cl(s)

b 2SO2(g) + O2(g) ↔ 2SO3(g)

c Ag+(aq) + Fe2+(aq) ↔ Ag(s) + Fe3+(aq)

d 2NH3(g) ↔ N2(g) + 3H2(g) 2 a Ammonia molecules are reacting with hydrogen chloride molecules to form

ammonium chloride at the same rate at which the ammonium chloride is decomposing to ammonia and hydrogen chloride.

b Sulfur dioxide reacts with oxygen to form sulfur trioxide at the same rate at which the sulfur trioxide decomposes to produce sulfur dioxide and oxygen.

c Silver ions are reduced to silver at the same rate at which the iron(III) is reduced to iron(II).

d Ammonia decomposes to nitrogen and hydrogen at the same rate at which hydrogen and nitrogen combine to form ammonia.

3 a If 0.21 mol L–1 H2 remains, 0.79 mol L–1 reacted. Molar ratio of H2 : HI is 1 : 2

∴ 0.79 × 2 = 1.58 moles HI formed

∴ equilibrium conc HI = 1.58 mol L–1 b i and ii

c Forward and reverse reactions will occur at equal but opposing rates. The

concentrations of reactants and products remain constant.





4 In an open container, the CO2 would escape and the reverse reaction would not occur. In a sealed container, increasing the temperature would increase the pressure until equilibrium was established.

Review exercise 7.2

1 a i ↑ [NO], ↑ [H2O], ↓ [NH3] ii Will increase rate of forward reaction until equilibrium is re-established.

b i ↓ [NH3], ↓ [O2], ↑ [H2O] ii Will increase rate of forward reaction until equilibrium is re-established.

2 a i Concentration of H2O will decrease, concentration of CO, H2 will increase. No change to [C(s)].

ii Will increase rate of forward reaction. This increases concentration of CO and H2, decreases [H2O].

b i Concentration of H2O will increase, concentration of CO, H2 will decrease. No change to [C(s)].

ii Will increase rate of reverse reaction. [CO] ↓, [H2] ↓ and [H2O] ↑ until equilibrium is re-established.

3 a i ↑ [CO], ↑ [O2], ↓ [CO2]

ii Will increase rate of reverse reaction.

b i ↓ [CO], ↓ [O2], ↑ [CO2] ii Will increase rate of forward reaction.

4 a The system would reach equilibrium more quickly. b Both forward and reverse rates would increase equally, therefore no change.

Review exercise 7.3

1 CO2(g) ↔ CO2(aq)

CO2(aq) + H2O(l) ↔ H2CO3(aq)

H2CO3(aq) ↔ H+(aq) + HCO3–(aq)

HCO3–(aq) ↔ H+(aq) + CO3

2–(aq) 2 a When the bottles are opened, the pressure in the system is decreased, causing a

shift in the equilibrium to the left, where there are more moles of gas and so the change is counteracted. This results in bubbles of CO2(g).

b Cooling the drink shifts the equilibrium to the right (exothermic direction), meaning that more carbon dioxide will be dissolved into the solution. When the





drink is served cold, the rate at which CO2(g) escapes to the atmosphere is decreased.






1 a Changes in concentration

b Rate of reaction

2 No further change in the colour of the mixture would be detected. 3 At equilibrium, the concentration of all species remains constant, therefore there will be

no change in any observable properties. The dynamic process refers to the equal but opposite reactions occurring at the molecular level.

4 concentration, pressure, temperature

5 a ↓ [NO]; ↑ [O2]; ↑ [NO2] b Concentration of all species will double.

c ↑ [O3]; ↓ [O2]; ↓ [NO2]





6 a ↓ [HBr]; ↓ [O2]; ↑ [H2O]; ↑ [Br2]

b ↑ [HBr]; ↑ [O2]; ↓ [H2O]; ↓ [Br2]

7 a i Shift equilibrium to the right. ii Shift equilibrium to the left.

b i Shift equilibrium to the left. ii Shift equilibrium to the right.

8 a ↑Ca(HCO3)2(s); ↓CaO(s); ↓ [H2O]; ↓ [CO2] after initial increase of CO2

b ↓Ca(HCO3)2(s); ↑CaO(s); ↑ [CO2]; ↑ [H2O] after initial decrease of H2O

c ↓Ca(HCO3)2(s); ↑CaO(s); ↑ [CO2]; ↑ [H2O] d no change

e ↑Ca(HCO3)2(s); ↓CaO(s); ↓ [CO2]; ↓ [H2O]

9 a i Equilibrium will shift right to counteract the change: ↓ [NO2]; ↑ [SO3]; ↑ [NO].

ii The forward reaction rate will increase. b i There are equal numbers of gaseous molecules on each side, so pressure will

increase the concentration of all species, but the system will remain at equilibrium.

ii Forward and reverse reactions increase equally. c i Concentrations of all species will decrease, but the system will remain at

equilibrium. ii Forward and reverse reaction rates decrease equally.

d i The equilibrium will shift left in an endothermic direction: ↑ [SO2], ↑ [NO2], ↓ [SO3], ↓ [NO].

ii Reverse reaction is favoured.

e i No change as a catalyst does not affect the position of equilibrium. ii Both forward and reverse reaction rates will increase.

10 a Pressure is decreased, so equilibrium will shift to counteract the change. Therefore CO2(g) is produced and bubbles form.

b It is an exothermic reaction, so warm temperatures cause the equilibrium to shift to the left and form CO2(g).

c The H+ in the orange juice causes increased production of carbonic acid.

HCO3–(aq) + H+(aq) ↔ H2CO3(aq)





The ↑ [H2CO3] causes the production of more CO2(aq):

H2CO3(aq) ↔ CO2(aq) + H2O(l)

Therefore the ↑ [CO2(aq)] results in the production of more CO2(g) causing drinks to go flat, as follows:

CO2(aq) + 19kJ ↔ CO2(g)

11 Investigation

CHAPTER 8: THE HISTORICAL DEVELOPMENT OF IDEAS OF ACIDS AND BASES

Review exercise 8.1

1 a SO2(g) + H2O(l) → H2SO3(aq)

b Na2O(s) + H2O(l) → 2NaOH(aq) 2 Hydrochloric acid, HCl; hydrobromic acid, HBr; cyanic acid, HCN

3 Arrhenius theory states that bases are substances which deliver hydroxide ions in water. NaCN is a substance that acts as a base but does not contain a hydroxide group.

4 a H2SO4(aq) → 2H+(aq) + SO42–(aq)

or H2SO4(aq) → HSO4–(aq) + H+(aq), then HSO4

–(aq) ↔ SO42– + H+(aq)

b HNO3(aq) → H+(aq) + NO3–(aq)

5 a 1

b 2 c 3

d 1 e 1

f 1 6 a 1

b 3 c 2

7 HOOCCOOH(aq) ↔ HOOCCOO–(aq) + H+(aq)

HOOCCOO–(aq) ↔ –OOCCOO–(aq) + H+(aq)

Review exercise 8.2

1 a A B–L acid is a proton donor (or has a proton ‘pulled’ from it).





b A B–L base is a proton acceptor (or ‘takes’ a proton).

2 a HCl, HCO3–, NH4

+, HClO4, HSO4–

b F–, SO42–, NH3, PO4

3–, H2O

3 a An amphiprotic substance can either accept or donate a proton, and therefore it can act as both an acid and a base.

b HPO42–(aq) + H2O(l) ↔ H2PO4

–(aq) + OH–(aq)

HPO42–(aq) + H2O(l) ↔ PO4

3–(aq) + H3O+(aq)

4 Acid HSO4–(aq) + OH–(aq) ↔ SO4

2–(aq) + H2O(l)

Base HSO4–(aq) + H3O+(aq) ↔ H2SO4(aq) + H2O(l)

5 a A conjugate base is the species that results when a B–L acid has lost its proton

b A conjugate acid is the species that results when a B–L base accepts a proton

6 a HCO3

–/SO42–: HF/F–

b HSO4–/SO4

2–: NH4+/NH3

c HF/F– : H3O+/H2O

Review exercise 8.3

1 a i HCl ii HCl

iii HCl b i H2SO4

ii H2SO4 iii H2SO4

2 a i NaOH ii NaOH





iii NaOH

b i Ba(OH)2 ii Ba(OH)2

iii Ba(OH)2 3 a 5 mol L–1 HCl

b 0.1 mol L–1 NaOH c 5 mol L–1 NH3

d 0.1 mol L–1 CH3COOH 4 a HBr(aq) → H+(aq) + Br–(aq)

b H2SO3(aq) ↔ HSO3–(aq) + H+(aq)

c RbOH(aq) → Rb+(aq) + OH–(aq)

d NaF(aq) ↔ Na+(aq) + F–(aq)

5 a small extent

b large extent c small extent

Review exercise 8.4

1 a S2–(aq) + H2O(l) ↔ HS–(aq) + OH–(aq)

b CO32–(aq) + H2O(l) ↔ HCO3

–(aq) + OH–(aq)

c NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)

d [Fe(H2O)6]3+(aq) + H2O(l) ↔ [Fe(OH)(H2O)5]2+(aq) + H3O+(aq)

e F–(aq) + H2O(l) ↔ HF(aq) + OH–(aq)

f HSO4–(aq) + H2O(l) ↔SO4

2–(aq) + H3O+(aq)

g ClO– + H2O↔ HClO + OH–

h CH3COO–(aq) + H2O(l) ↔ CH3COOH(aq) + OH–(aq) 2 Use Table 8.5.

a neutral b acidic due to reaction of NH4

+ ion with water

c basic due to reaction of ClO– ion with water d basic due to reaction of PO4

3– ion with water





e acidic due to reaction of Al3+ ion (actually will be aquated aluminium ion [Al(H2O)6]3+ with water)

f acidic due to reaction of H2PO4– ion with water

g neutral h basic due to reaction of CO3

2– ion with water

Chapter 8 Application and investigation 1 According to Arrhenius, acids are substances that release hydrogen ions in solution.

HCl(aq) completely ionises in water to produce H+ and Cl−, and therefore HCl(aq) is an Arrhenius acid. HCl(l) cannot release hydrogen ions and so is not an Arrhenius acid.

2 a HCl(aq) → H+(aq) + Cl–(aq)

b Ca(OH)2(aq) → Ca2+(aq) + 2OH–(aq)

c H2SO4(aq) → HSO4–(aq) + H+(aq)

HSO4–(aq) ↔ SO4

2–(aq) + H+(aq)

d H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

e Al(OH)3(aq) + 3HCl(aq) → AlCl3(aq) + 3H2O(l)

3 a 4

b 2 c 2

d CH2(COOH)2(aq) + 2NaOH(aq) → CH2(COONa)2(aq) + 2H2O(l) 4 a HCl(aq) + H2O(l) → H3O+(aq) + Cl–(aq)

b CH3COOH(aq) + H2O(l) ↔ H3O+(aq) + CH3COO–(aq)

c NH3(aq) + H2O(l) ↔ NH4+(aq) + OH–(aq)

d H2SO4(aq) + H2O(l) → H3O+(aq) + HSO4–(aq)

HSO4–(aq) + H2O(l) ↔ H3O+(aq) + SO4

2–(aq)

e HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

5 a ClO3–, S2–, NH3, OH–

b H2CO3, H2S, [Fe(H2O)6]3+, N2H5+

6 a As acid: HCO3–(aq) + H2O(l) ↔ CO3

2–(aq) + H3O+(aq)

As base: HCO3–(aq) + H2O(l) ↔ H2CO3(aq) + OH–(aq)





b The extent to which the HCO3– ion reacts as a base with water exceeds the degree

to which it reacts with water as an acid. (K value is greater for HCO3– as a base.)

Hence more OH– ions are produced than H3O+ ions.

7 a i H2C2O4/HC2O4–, H3O+/H2O

ii small extent

b i H2O/OH–, HCN/CN– ii small extent

c i CH3COOH/CH3COO– , HS–/S2– ii large extent

d i HCl/Cl–, HF/F– ii large extent

8 a HClO4(l) + H2O(l) → ClO4–(aq) + H3O+(aq)

b HCOOH(l) + H2O(l) ↔ HCOO–(aq) + H3O+(aq)

c LiOH(s) + H2O(l) → Li+(aq) + OH–(aq)

d N2H4(l) + H2O(l) ↔ N2H5+(aq) + OH–(aq)

9 a concentrated solution of a strong acid

b diluted solution of a strong acid





c concentrated solution of a weak acid

d diluted solution of a weak acid

10 a CH3COO–(aq) + H2O(l) ↔ CH3COOH(aq) + OH–(aq)

b The hydroxide ions produced raise the pH above 7. The ethanoic acid molecule formed has only a very small degree of dissociation.

11 a neutral

b basic PO43–(aq) + H2O(l) ↔ HPO4

2–(aq) + OH–(aq)

c basic CO32–(aq) + H2O(l) ↔ HCO3

–(aq) + OH–(aq)

d acidic NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)

e acidic [Cr(H2O)6]3+(aq) + H2O(l) ↔ [Cr(OH)(H2O)5]2+(aq) + H3O+(aq)

f basic SO42–(aq) + H2O(l) ↔ HSO4

–(aq) + OH–(aq)

g basic CN–(aq) + H2O(l) ↔ HCN(aq) + OH–(aq)

h acidic NH4+(aq) + H2O(l) ↔ NH3(aq) + H3O+(aq)

12

Acid Base Neutralisation

Arrhenius produces H+ ions in water

produces OH– ions in water

H+ + OH–↔H2O

Brønsted–Lowry proton donor proton acceptor NH3(aq) + H2O↔NH4+

+ OH– proton transfer





13 The reaction: HCl(g) + NH3(g) → NH4Cl(s) can be viewed as a Brønsted–Lowry acid–base reaction because a proton is transferred. However, the reaction is not happening in solution and there are no hydroxide ions released, so the reaction cannot be viewed as an Arrhenius acid–base reaction. When nitric acid is mixed with sulfuric acid, for the manufacture of nitroglycerine, the reaction that occurs is

H2SO4 + HNO3 ↔ HSO4– + H2NO3

+ This is a B–L acid–base reaction because a proton is transferred but it is not an Arrhenius acid–base reaction.

14 Investigation

CHAPTER 9: HYDROGEN ION CONCENTRATION AND THE pH SCALE

Review exercise 9.1

1 a KW = [H+] [OH–]

1.0 × 10–14 = [H+] × 1.6 × 10–7 [H+] = 6.2 × 10–8 mol L–1

b [H+] is less than 1 × 10–7 mol L–1 ∴ pool is slightly basic

2 a [H+] = 5.0 × 10–3 mol L–1 [NO3

–] = 5.0 × 10–3 mol L–1 [OH–] = 1 × 10–14 ÷ 5.0 × 10–3 = 2 × 10–12 mol L–1

b [H+] = 1.5 mol L–1 [Cl–] = 1.5 mol L–1

[OH–] = 1 × 10–14 ÷ 1.5 = 6.7 × 10–15 mol L–1

c [K+] = 0.25 mol L–1

[OH–] = 0.25 mol L–1 [H+] = 4 × 10–14 mol L–1

d [Ba2+] = 6.0 × 10–2 mol L–1

[OH–] = 0.12 mol L–1 [H+] = 8.3 × 10–14 mol L–1





Review exercise 9.2

1 a pH = –log[H+] = –log[5.0 × 10–1] = 0.30

b [H+] = 1 × 10–14 ÷ 0.0065 = 1.54 × 10–12 mol L–1

pH = –log[1.54 × 10–12]

= 11.81 c pH = –log[3.6 × 10–3] = 2.44

d [H+] = 1 × 10–14 ÷ (6.5 × 10–4 × 2) = 7.7 × 10–12 mol L–1

pH = –log[7.7 × 10–12]

= 11.1

2 a pH = 3.50; [H+] = 3 × 10–4 mol L–1; [OH–] = 3.16 × 10–11 mol L–1

b pH = 11.90; [H+] = 1.3 × 10–12 mol L–1; [OH–] = 7.9 × 10–3 mol L–1

c pH = 0.80; [H+] = 0.16 mol L–1; [OH–] = 6.3 × 10–14 mol L–1

3 pH = 7 so [H+] = 1 × 10–7

pH = 5 so [H+] = 1 × 10–5

∴ pH changes by factor of 100

Review exercise 9.3 1 a HClO2

b HClO2 c HClO2, HF, HCN

d No. Complete ionisation of 0.1 mol L–1 gives pH = 1.

Review exercise 9.4

1 a A buffer is a species that resists changes to pH despite small additions of acid or base.

b A buffer contains approximately equal amounts of a weak acid and its conjugate base (some are a mixture of weak bases and their conjugate acids).





2 a Buffered; contains approximately equal amounts of the weak acid HSO4– and its

conjugate base SO42–.

b Not buffered; HNO3 is a strong acid.

c Buffered; contains approximately equal amounts of the weak acid HClO and its conjugate base ClO–.

3 a HPO4–(aq) + H3O+(aq) ↔ H2PO4

–(aq) + H2O(l) When excess hydrogen ions are added, the system shifts to the right and thus

decreases the H+ concentration.

b H2PO4+(aq) + OH–(aq) ↔ HPO4

–(aq) + H2O(l) When excess hydroxide is added, the system shifts to the right and thus decreases the OH– concentration.

4 The concentration of the acid and conjugate base ions is limited. If too much H+ or OH– is added then the system is unable to provide sufficient acid or conjugate base particles to counteract their effect.

5 Lactic acid in the blood produces lactate ions and hydrogen ions. The blood’s buffering system H2CO3(aq) + H2O(l) ↔ HCO3

–(aq) + H3O+(aq) responds by combining the extra hydrogen ions with the hydrogencarbonate ions, thus shifting the system to the left and decreasing their concentration.


1 a [H+] = 0.5 mol L–1 [OH–] = 1 × 10–14 ÷ 0.5

= 2 × 10–14 mol L–1 pH = –log[0.5] = 0.30

b [OH–] = 3.0 × 10–3 × 2 = 6.0 × 10–3 mol L–1

[H+] = 1 × 10–14 ÷ 0.006

= 1.67 × 10–12 mol L–1 pH = –log[1.67 × 10–12] = 11.8

2 a [H+] = 1.0 × 10–3 mol L–1





[OH–] = 1 × 10–14 ÷ 0.001 = 1 × 10–11 mol L–1

b [H+] = 3.9 × 10–3 mol L–1 [OH–] = 2.51 × 10–12 mol L–1

c [H+] = 3.2 × 10–9 mol L–1 [OH–] = 3.2 × 10–6 mol L–1

d [H+] = 7.9 × 10–13 mol L–1 [OH–] = 1.3 × 10–2 mol L–1

3 a i [H+] = 0.20 mol L–1

[NO3–] = 0.20 mol L–1

[HNO3] = 0 mol L–1

ii pH = –log10[0.20] = 0.70

b As HNO2 is a weak acid, it will not dissociate completely. Therefore [H+] will be less than 0.20 mol L–1 and pH will be higher than 0.70.

4 H2CO3, H3PO4, HClO2, H2SO3, HCl 5 Investigation

6 Investigation

CHAPTER 10: VOLUMETRIC ANALYSIS AND ESTERIFICATION

Review exercise 10.1

1 a NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

b 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

c 2NaOH(aq) + H2CO3(aq) → Na2CO3(aq) + H2O(l)

d 3NaOH(aq) + H3PO4(aq) → Na3PO4(aq) + H2O(l)

2 a 2HCl(aq) + Ca(OH)2(aq) → CaCl2(aq) + 2H2O(l)

b 3HCl(aq) + Al(OH)3(s) → AlCl3(aq) + 3H2O(l)





c 2HCl(aq) + MgO(s) → MgCl2(aq) + H2O(l)

d 2HCl(aq) + Na2CO3(s) → 2NaCl(aq) + CO2(g) + H2O(l)

3 Neutralisation is the result of an acid–base interaction. In Brønsted–Lowry terms this refers to proton transfer from the acid to the base.

4 Neutralisation results in the formation of a new bond as the proton bonds to its base. Bond formation is an exothermic process, and thus neutralisation is an exothermic process.

5 a Add solid sodium hydrogencarbonate until the ‘fizzing’ stops. The reaction produces carbon dioxide, so when the fizzing stops, this indicates that the neutralisation is complete.

b Rinse with copious amounts of water. A base should not be used, as the neutralisation reaction is exothermic—i.e. it releases a lot of heat.


1 2OH–(aq) + H2SO4(aq) → 2H2O(l) + SO42–(aq)

Calculate moles of OH– n(OH–) = cV

= 1.90 × 0.02368 = 0.045 moles

∴ moles of H2SO4 : n(H2SO4) = 0.5n(OH–) (from equation)

= 0.0225 moles

∴ concentration of H2SO4

n(H2SO4) = cV 0.0225 = c × 0.005

c = 4.50 mol L–1

2 NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) Calculated moles of HCl

n = cV = 0.02052 × 0.952

= 0.019535 moles moles of HCl = moles NaOH (from equation)





∴ concentration of NaOH = Vn

= 0.019535 / 0.005

c = 3.91 mol L–1 NaOH 3 a Pipettes are more accurately calibrated than a measuring cylinder; this accuracy is

important in volumetric analysis. b If rinsed with water, any drops left in the pipette or burette will alter the

concentration of the solution and this will affect the results. Rinsing with the solution will not have this effect.

c Volume may change slightly once the solid is dissolved; the customary unit of concentration is mol/L, where the volume refers to litres of solution, not water.

4 a No. moles Na2CO3 = 0.250 × 0.1 = 0.025 moles

moles = Mm

so 0.25 = 99.105m

m = 2.65 g of Na2CO3 is dissolved in water in a 250 mL volumetric flask with solution made up to the calibrated mark

b n = cV

nH2C2O4.2H2O = 0.5 × 0.015

= 7.5 × 10–3 moles

n = Mm

7.5 × 10–3 = 07.126m

m = 0.95 g oxalic acid is dissolved in water in a 500 mL volumetric flask with the solution made up to the calibrated mark

5 a n = Mm





= 07.126

3162.0

= 0.0025 moles

n = cV

0.0025 = c × 0.25

c = 0.010 mol L–1 H2C2O4.2H2O

b 2NaOH(aq) + H2C2O4(aq) → Na2C2O4(aq) + 2H2O(l) n(H2C2O4) = cV

n = 0.010 × 0.02061

= 2.068 × 10–4 moles H2C2O4

n(OH–) = 2 × nH2C2O4 (from equation)

n(OH–) = 4.136 × 10–4 mol n(NaOH) = cV

4.136 × 10–4 = c × 0.020

c = 0.0207 mol L–1 NaOH

c NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) n(NaOH) = cV

n = 0.0207 × 0.020

= 4.14 × 10–4 mol NaOH volume of HCl used = 20.26 – 1.83

= 18.43 mL = 0.01843 L

nHCl = nNaOH from equation

n(HCl) = 4.14 × 10–4 moles

nHCl = cV

4.14 × 10–4 = c × 0.01843

c = 2.24 × 10–2 HCl mol L–1

Review exercise 10.3 1 a i NaF(aq)





ii basic

iii phenolphthalein b i KNO3(aq)

ii neutral iii methyl orange/phenolphthalein

c i NaCl(aq) ii acidic

iii methyl orange 2 i a Phenolphthalein; as a strong acid/strong base titration and endpoint will be

around 7. b Colourless in acid changing to pink as base added, with pale pink at

equivalence point. ii a Methyl orange; as strong acid/weak base titration and endpoint will be less

than pH 7. b Yellow in base and changing to orange at equivalence point, and to red as

point is passed. iii a Bromothymol blue; weak acid/strong base titration and endpoint will be

greater than pH 7. b Yellow in acid changing to blue as base added, with pale yellow at

equivalence point. 3 This is a weak acid/strong base titration with an equivalence point above pH 7. If

methyl orange was used as an indicator, a colour change would be observed at around pH 3.1–4.4 and an equivalence point assumed, so the volume of hydroxide recorded would be too low.

4 a Running this strong base into the weak acid.





b Running this weak base into the strong acid.

c Running this strong base into the strong acid.


1 a n = 71.22V =

71.225.2

= 0.11 mol H2

b n = Mm

0.11 = 016.2m

= 0.22 g hydrogen gas

2 a n = Mm =

320.48

= 1.5 mol





n = 71.22V

1.5 = 71.22V

v = 34.07 L at 0°C and 100 kPa

b n = 79.24V

1.5 = 79.24V

v = 37.19 L at 25°C and 100 kPa 3 Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

1 2 1 2

a nH2 79.24V

n = 79.245.2

= 0.1 moles H2

From equation nZn = 0.1 (nH2)

= 0.1 moles

nZn = Mm

0.1 = 4.65m

= 6.54 g Zn required

b From equation nHCl = 2 × nH2

= 2 × 0.1 = 0.2 moles

nHCl = cV

0.2 = 1.5 × V

V = 0.13 L HCl required





4 CaCO3(s) + 2HNO3(aq) → Ca(NO3)2(aq) + CO2(g) + H2O(l)

1 2 1 1 1

a nCaCO3 = Mm =

1.10010

n = 0.10 moles

From equation nCaCO3 = nCO2

∴ nCO2 = 0.10 moles

nCO2 = 71.22V

0.1 = 71.22V

V = 2.271 L CO2 at 0°C and 100 kPa

b nCO2 = 79.24V

0.1 = 79.24V

V = 2.48 L CO2 at 25°C and 100 kPa

c From equation nHNO3 = 2 × nCaCO3

∴ nHNO3 = 2 × 0.1 = 0.2 moles nHNO3 = cV

0.2 = 1.00 × V V = 0.2 L HNO3 required

d From equation nCaCO3 = nCa(NO3)2

∴ nCa(NO3)2 = 0.10 moles

nCa(NO3)2 = Mm

0.1 = 1.164

m

m = 16.41 g of Ca(NO3)2 produced






1 a 1-propanol b 1-butanol

c 2-butanol 2 a

b

c

3 a pentanoic acid

b 5-methylhexanoic acid 4 a





b

5 They form more H-bonds because they have a C=O group as well as an −OH group, making two H-bonds possible between two molecules.

6 a methyl butanoate b ethyl methanoate

7 a

b

8 a

b





c

d

e

f

9 a C3H7OH + CH3COOH

€

conc. H2SO4" → " " " " " CH3COOC3H7 + H2O

The ester is propyl ethanoate.

b C5H11OH + C3H7COOH

€

conc. H2SO4" → " " " " " C3H7COOC5H11 + H2O

The ester is pentyl butanoate.

10 a H2SO4 acts as a catalyst. b Refluxing prevents volatile reactants from escaping before equilibrium is reached

and allows the reaction to be carried out at temperatures lower than would normally be possible.


1 a 2HCl(aq) + CuO(s) → CuCl2(aq) + H2O(l)





b 3H2SO4(aq) + 2Fe(OH)3(s) → Fe2(SO4)3(aq) + 6H2O(l)

c 2HNO3(aq) + CaCO3(s) → Ca(NO3)2(aq) + CO2(g) + H2O(l)

d HCl(aq) + NaHCO3(aq) → NaCl(aq) + H2O(l) + CO2(g)

2 a i Na+(aq), SO42–(aq), OH–(aq), H+(aq), H2O(l)

ii H+(aq), Na+(aq), OH–(aq) all at 2 mol L–1

b i H+(aq), SO42–(aq), H2O(l)

ii H+(aq)

c i Na+(aq), SO42–(aq), H2O(l)

ii Na+(aq), SO42–(aq)

3 a i Pipette: used to place an accurate volume of solution into a conical flask. ii Burette: used to deliver variable volumes into a conical flask.

iii Conical flasks: used to mix two solutions. b i citric acid

ii NaOH iii deionised water

4 Sodium hydroxide cannot be made as a primary standard and hence must be standardised by titration with an acid of accurately known concentration.

5 a A good primary standard has a relatively high molecular weight to minimise weighing errors, does not react with any component of the atmosphere such as oxygen, moisture or carbon dioxide, and is available in a very pure form.

b Investigation

6 a nNaHCO3 = cV = 2 × 1

= 2 moles

n = Mm

2 = 008.84m

m = 168 g

b nNa2CO3 = cV

n = 0.05 × 0.5





= 0.025 moles

n = Mm

0.025 =

m = 2.65 g c nBa(OH)2.8H2O = cV

n = 2 × 10–3 × 2

= 0.004

n = Mm

m = 0.004 × 315.444 = 1.26 g d nH2C2O4.2H2O = cV

n = 0.150 × 0.25 = 0.0375 moles

n = Mm

m = 0.0375 × 126.068 = 4.728 g

7 a NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(l) nHNO3 = cV

= 0.0961 × 0.02

= 1.9 × 10–3 From equation nNaOH = nHNO3

= 1.9 × 10–3

nNaOH = cV

1.9 × 10–3 = 0.118 × V V = 1.62 × 10–2 L





b 2NaOH(aq) + H2SO3(aq) → Na2SO3(aq) + 2H2O(l) nH2SO3 = cV

= 0.0531 × 0.02 = 1.062 × 10–3 moles

nNaOH = 2 × nH2SO3 from equation

2 × 1.062 × 10–3 = 0.118 × V V = 18 × 10–3 L

c 2NaOH(aq) + H2C2O4.2H2O(s) → Na2C2O4(aq) + H2O(l) nH2C2O4.2H2O

= Mm

=

126134.0

= 1 × 10–3 moles

2 × n(oxalic acid) = n(NaOH)

n = cV

2 × 10–3 = 0.118 × V V = 16.95 × 10–3 L

8 a H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

nNaOH = cV = 0.5 × 0.02 = 0.01 moles

nH2SO4 = 0.5nNaOH = 0.005 moles H2SO4

n = cV

V = 202.0005.0

= 2.48 × 10–2 L

b H2SO4(aq) + Ba(OH)2 → BaSO4(aq) + 2H2O(l) nBa(OH)2 = 0.02 × 0.165

= 0.0033 moles





nBa(OH)2 = nH2SO4

∴ nH2SO4 = 0.0033 = 0.202 × V V = 1.63 × 10–2 L

c Na2CO3(s) + H2SO4(aq) → Na2SO4(aq) + CO2(g) + H2O(l) nNa2CO3 =

Mm

=

106343.0

= 3.24 × 10–3 moles nNa2CO3 = nH2SO4

nH2SO4: 3.24 × 10–3 = 0.202 × V V = 1.60 × 10–2 L

d total nOH– = nOH– in Ca(OH)2 + nOH– in NaOH

nCa(OH)2 = Mm

=

1.740746.0

= 1 × 10–3 moles moles OH– = 1 × 10–3 × 2

= 2 × 10–3 moles

nNaOH = Mm

=

40247.0

= 6.2 × 10–3 moles OH– Total nOH– = 2 × 10–3 + 6.2 × 10–3

= 8.18 × 10–3 moles

nH2SO4 = nOH– × 0.5 = 4.08 × 10–3 moles

n = cV





= 4.08 × 10–3 = 0.202 × V VH2SO4 = 2.02 × 10–2 L

9 a HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) nHCl = cV

= 0.121 × 0.02565

= 3.1 × 10–3 moles nHCl = nNaOH

nNaOH = cV

3.1 × 10–3 = c × 0.020 c = 0.16 mol L–1

b Ba(OH)2(aq) + 2HCl(aq) → BaCl2(aq) + 2H2O(l) nHCl = cV

= 0.0503 × 0.01523

= 7.66 × 10–4 moles nBa(OH)2 = 0.5nHCl

= 0.5 × 7.66 × 10–4

= 3.83 × 10–4 moles Ba(OH)2 = 3.83 × 10–4 ÷ 0.02

= 1.92 × 10–2 mol L–1

10 a H2SO4(aq) + CuO(s) → CuSO4(aq) + H2O(l)

nCuO = Mm =

5.7905.1

= 0.013 moles nCuO = nH2SO4

nH2SO4 = 0.13 moles = 2.0 × V V = 6.6 × 10–3 L

b 1 mole H2SO4 → 1 mole CuSO4.5H2O

∴ nCuSO4.5H2O = 0.013 moles





n = Mm

0.013 = 6.249

m

m = 3.24 g 11 Investigation

12 A known mass of the tablets could be dissolved in 20 mL of water. This solution could be titrated against HCl of known concentration to determine the concentration of NaHCO3.

13 nNaOH = cV

= 0.119 × 0.02053

= 2.4 × 10–3 moles

∴ n vitamin C = 2.44 ×10–3 moles

n = Mm

2.44 × 10–3 = 176m

m = 0.4294 g

% in capsule = 100450.04294.0

×

= 95.4%

14 a NaOH + HCl → NaCl + H2O 1 1

nHCl = cV

= 0.125 × 0.01759

= 0.0022 moles nNaOH = nHCl

= 0.0022 moles

c = 020.00022.0

=Vn

= 0.11 mol L–1





NaOH + CH3COOH → NaCH3COO + H2O nNaOH = cV

= 0.11 × 0.03812

= 4.2 × 10–3 moles nNaOH = nCH3COOH

= 4.2 × 10–3

nCH3COOH = 4.2 × 10–3

= Mm

4.2 × 10–3 = 60m

m = 0.25 g in 5.00 mL sample vinegar

b % mass = 0.25/5.00 × 100 = 5% ethanoic acid in vinegar

15 NH3(aq) + HCl(aq) → NH4Cl(aq) nHCl = cV

= 0.110 × 0.02462

= 2.7 × 10–3 moles

∴ nNH3 = 2.7 × 10–3 = cV

2.7 × 10–3 = c × 0.02 c = 0.135 mol L–1 ammonia

Number of moles NH3 present in 250 mL

n = cV

= 0.135 × 0.25

= 0.339 moles

n = Mm





0.339 = 17m

m = 0.575 g ammonia in 9.97 g

% mass ammonia in household ammonia

= 10097.9575.0

×

= 5.77%

16 Na2CO3 + H2O + 2HCl(aq) → 2NaCl(aq) + 2H2O(l) + CO2(g) nHCl = cV

= 0.104 × 0.02520

= 2.62 × 10–3 moles nNa2CO3 = 0.5nHCl

nNa2CO3 = 0.5 × 2.62 × 10–3

= 1.31 × 10–3 moles

nNa2CO3 = Mm

1.31 × 10–3 = 106m

m = 0.1389 g Mass of water in washing soda = 0.374 – 0.1389

= 0.235 g H2O

nH2O = Mm

n = 18235.0

n = 0.0131 moles

Ratio of nNa2CO3 : nH2O = 1.31 × 10–3 : 1.31 × 10–2

= 1 : 10

∴Number of molecules of water of crystallisation = 10 i.e. Na2CO3.10H2O





17 The pH at the equivalence point is below 7 due to the reaction of NH4+ produced at the

equivalence point reacting with H2O to form H3O+. Phenolphthalein changes colour above 7, and therefore it would not be appropriate.

18 a i Na+, CH3COO–, H2O, CH3COOH, OH– ii basic

iii phenolphthalein b i KCl(aq), H2O

ii neutral iii bromothymol blue

c i NH4+, NO3

–, NH3, H3O+ ii acidic

iii methyl orange d i BaCl2(aq), H2O

ii neutral iii bromothymol blue

e i H2CO3(aq), NaCl, H3O+, HCO3–

ii acidic

iii methyl orange

19 a i nO2 = 71.22V

0.500 = 71.22V

V = 11.36 L O2 at 0ºC and 100 kPa

ii nO2 = 79.24V

0.500 = 79.24V

V = 12.4 L O2 at 25ºC and 100 kPa

b i nN2 = 71.22V

= 71.222





= 0.09 moles N2 at 0ºC and 100 kPa

ii nN2 = 79.24V

= 79.2410

= 0.40 moles N2 at 25ºC and 100 kPa

20 a i nCl2 = 71.22V

= 71.223

= 0.13 moles

nCl2 = Mm

0.13 = 71m

m = 9.5 g Cl2 at 0ºC and 100 kPa

ii nCl2 = 79.24V

79.243

= 0.12 moles

n = Mm

0.12 = 71m

m = 8.7 g Cl2 at 25ºC and 100 kPa

b i nCO2 = Mm

= 4455

= 1.25 mol





n = 71.22V

1.25 = 71.22V

V = 28.39 L CO2 at 0ºC and 100 kPa ii nCO2 = 1.25 mol

n = 79.24V

1.25 = 79.24V

V = 30.99 L CO2 at 25ºC and 100 kPa

21 a nFeS2 = Mm

= 1201000

= 8.33 mol

From equation, FeS2:O2 equals 1:2.75

nO2 = 2.75 × nFeS2

= 2.75 × 8.33

= 22.92 mol O2

nO2 = 71.22V

22.92 = 71.22V

V = 520.5 L O2 at 0ºC and 100 kPa

b i nSO2 = 2 × nFeS2 = 2 × 8.33 = 16.67 moles

At 0°C and 101.3 kPa:





16.67 = 71.22V

V = 378.6 L SO2 produced at 0ºC and 100 kPa

ii At 25°C and 101.3 kPa:

16.67 = 79.24V

V = 413.3 L SO2 produced at 25ºC and 100 kPa

22 a CH3CH2CH3; CH3CH2CH2OH; CH3CH2COOH

b propane –41°C

1-propanol 97°C

propanoic acid 141°C

Propane has only weak dispersion forces between molecules ∴ lowest boiling point. 1-propanol has strong H-bonds between molecules but propanoic acid has stronger intermolecular forces due to two H-bonds between acid molecules.

23 a

b

24 a Reflux ethanol with butanoic acid with concentrated H2SO4 as a catalyst as an

esterification process

b Reflux pentanol with propanoic acid with concentrated H2SO4 as a catalyst as an esterification process

Refluxing is necessary so that none of the volatile components of the reaction are lost while this slow, low-yielding reaction proceeds.

25 Investigation





Module 2

REVIEW 1 A

2 D 3 D

4 B 5 D

6 A 7 B

8 C 9 A

10 B 11 a CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

b nNaOH = 0.095 × 0.02 = 1.9 × 10–3 moles

Average = 3

0.198.189.18 ++

= 18.9 mL

nCH3COOH = 1.9 × 10–3 = c × 0.0189 c = 0.1005 mol L–1 acetic acid

c c1V1 (original) = c2V2 (volumetric flask)

c × 0.02 = 0.1005 × 0.2 c = 1.005 mol L–1

d Phenolphthalein; pH at end point will be above 7. e Distilled water. The number of moles of vinegar added will not be affected by

water in the flask.

12 NH3(g) + H2O(l) → NH4+(aq) + OH–(aq)

The OH– causes the solution to be basic.

NH4+(aq) + H2O(l) → NH3(aq) + H3O+(aq)

The H3O+ causes the solution to be acidic.





13 a HCl is a strong acid which dissociates completely and therefore has high conductivity and low pH. Acetic acid is weak and therefore only slightly dissociates, causing a higher pH and lower conductivity.

b The CH3COO– reacts with H2O to form some OH– ions. c Greater number of ions are present in the solution in Beaker 3.

14 a Bacteria can decompose organic matter to produce H2S, which then oxidises to SO2. Burning of fossil fuels and smelting of sulfide ores results in sulfur compounds which are oxidised to SO2.

b 2CuFeS2(s) + 5O2(g) + 2SiO2(s) → 2Cu(l) + 4SO2(g) + 2FeSiO3(l)

c i SO3(g) + H2O(l) → H2SO4(aq) sulfuric acid

ii Lakes and waterways become acidic, killing aquatic life. Soil can become acidic, affecting plant growth. Limestone can be dissolved by the action of acid rain. These are all of concern if we want to protect our environment.

iii Student opinion

15 a Both molecules have an −OH group but ethanoic acid contains a C=O group, which increases the intermolecular forces. Even though 1-butanol has more dispersion forces than acetic acid, acetic acid has more H-bonding and so the intermolecular forces are very nearly the same in each substance.

b i CH3COOH(aq) + C4H9OH(l) → CH3COOC4H9(aq) + H2O(l)

Catalyst is conc. H2SO4. ii H2SO4 acts as a catalyst.

c Refluxing allows the reaction mixture to be boiled without losing the volatile reactants and increases the time over which this quite slow reaction can occur.

d Acetic acid would be more soluble than 1-butanol. Compound X would be least soluble.





e

16 a i Equilibrium shifts to the left and more CO2(g) is produced

∴ the drink goes flat. ii More space above the liquid means that pressure is lower and CO2(aq)

→ CO2(g) and the drink will be flat. b The soft drink could be carefully poured into a beaker and gently heated. If

bubbles of CO2 are observed forming in the solution (which they should be), this would indicate that the equilibrium is shifting to the left (CO2(g) ↔ CO2(aq)). As the addition of heat favours the reverse reaction, it follows that this reaction must be endothermic. Therefore the forward reaction must be exothermic.

Alternatively:

Take an unopened bottle of soft drink. Immediately upon opening it, insert a thermometer (at the same initial temperature as the drink) and observe any change in temperature. Bubbles of CO2 gas should be observed forming in the solution, indicating that the equilibrium is shifting to the left (as written). If a temperature decrease is recorded (which it should), this will indicate that the reverse reaction is endothermic, as heat is being absorbed from the surroundings (the drink itself). Therefore the forward reaction must be exothermic.

17 a An acid loses an H+ to become its conjugate base, e.g. C6H5N2O5(aq) → C6H4N2O5

–(aq). b Yellow. The CN– would react with the H3O+, causing the equilibrium to shift to

the right.

CN–(aq) + H3O+ → H2O(l) + HCN(aq)





c i Yes: strong acid–strong base titration. pH = 7 at end point and the pH changes from 11 to 3 very rapidly.

ii Yes: weak base–strong acid; pH < 7 at end point.

iii No: weak acid–strong base; pH > 7 at end point. 18 a [H+] = 0.12 mol L–1

pH = –log10[H+] = –log10[0.12] pH = 0.92

b Mg(OH)2(aq) + 2HCl(aq) → MgCl2(aq) + 2H2O(l) The antacid reacts with excess acid to produce salt and water.

19 a Curve 1

b The acetate ion reacts with water to form OH– and CH3COOH, raising the pH at equivalence point. There is no such reaction in the NaCl(aq) produced at the equivalence point where pH = 7.

c Starting pH is higher for acetic acid due to less dissociation.

20 a CH3COOH(aq) ↔ CH3COO–(aq) + H+(aq) If CH3COO– is added then the equilibrium will shift left, resulting in less H+ and

a higher pH.

Also CH3COO–(aq) + H2O(l) ↔ CH3COOH(aq) + OH–(aq). The presence of OH– ions raises the pH.

b Buffering systems result from the action of a weak acid in approximately equal concentration with its conjugate base. Acetic acid is a weak acid and the acetate ions from the sodium acetate would be close in concentration to the acetic acid. Thus the resulting solution could act as a buffering system.

21 n (SO2) = 2 × n (CuFeS2) = 2 × mass (CuFeS2) ÷ molar mass (CuFeS2) Vol (SO2) = 2 × mass (CuFeS2) ÷ molar mass (CuFeS2) × molar volSATP

= 2 × 500 × 103 ÷ 183.54 × 24.79 = 135066 L 22 a Decreased pressure would cause the equilibrium to shift to the left. This would

result in the formation of CaCO3(s) and CO2(g). b Change in temperature. Increased temperature would cause this exothermic

reaction to shift to the left to counteract the change. This would cause a decrease in [Ca2+].

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