One of the most important tasks every advanced student of chemistry must undertake is to develop a thorough working knowledge of electro valencies of common cations and anions. Without this knowledge, the student cannot hope to ever master the writing of chemical equations and, subsequently, the principles and applications of stoichiometry. The Table below lists some common ions which should be studied and committed to memory.

Table 2.1: Electro valencies

Cations

+1	+2	+3	+4
lithium Li⁺ sodium Na⁺ potassium K⁺ hydrogen H⁺ or H₃O⁺ copper (I) Cu⁺ silver Ag⁺ ammonium NH₄⁺	magnesium Mg²⁺ calcium Ca²⁺ strontium Sr²⁺ barium Ba²⁺ iron (II) Fe²⁺ copper (II) Cu²⁺ Nickel Ni²⁺ zinc Zn²⁺ tin (II) Sn²⁺ lead (II) Pb²⁺ manganese (II) Mn²⁺	aluminum Al³⁺ chromium Cr³⁺ iron (III) Fe³⁺	tin (IV) Sn⁴⁺ lead (IV) Pb⁴⁺

Anions

-1	-2	-3
fluoride F^- chloride Cl^- bromide Br^- iodide I^- hydride H^- hydroxide OH^- nitrate NO₃^- nitrite NO₂^- permanganate MnO₄^- chlorate ClO₃^- hydrogen sulfate HSO₄^- hydrogen sulfite HSO₃^- hydrogen sulfide HS^- hydrogen carbonate HCO₃^- dihydrogen orthophosphate H₂PO₄^-	oxide O^2- peroxide O₂^2- sulfate SO₄^2- sulfite SO₃^2- sulfide S^2- carbonate CO₃^2- chromate CrO₄^2- dichromate Cr₂O₇^2- thiosulfate S₂O₃^2- monohydrogen orthophosphate HPO₄^2-	nitride N^3- orthophosphate PO₄^3-

To write balanced chemical formula for particular substances ensure the overall positive charge of the cation balances that of the anion. Note that if more than one molecule-ion species is required to achieve electrical neutrality, brackets must be used.

Example 2.1

Write balanced chemical formula for the following ionic substances:

(a) magnesium chloride

(b) iron (II) sulfate

(d) potassium dichromate

(e) copper (II) nitrate

Solution

(a) Mg²⁺ + 2Cl^- = MgCl₂

(b) Fe²⁺ + SO₄^2- = FeCO₃

(d) 2K⁺ + Cr₂O₇^2- = K₂Cr₂O₇

(e) Cu²⁺ + 2NO₃^- = Cu(NO₃)₂

2.2 General types of Chemical reaction

A balanced chemical equation is one which shows the chemical formulae of both reactants and products and the ratio in which they react. The equation is balanced when the number of atoms of each element is equal on both sides of the equation. In the case of ionic and/or partial redox equations, the overall charge must also be balanced.

To assist in writing chemical equations it is useful to be able to recognise some general reaction types and then apply these “proforma” to the specific equation to be balanced. As with most general applications, these reaction types cannot be used in every situation and each equation needs to be considered individually.

1. Acid + metal hydroxide ----> salt + water

eg. HCl (aq) + NaOH (aq) ----> NaCl (aq) + H₂O (l)

2. Acid + metal oxide ----> salt + water

eg. H₂SO₄ (aq) + ZnO (s) ----> ZnSO₄ (aq) + H₂O (l)

3. Acid + metal carbonate ----> salt + carbon dioxide + water

eg. 2HCl (aq) + CaCO₃ (s) ----> CaCl₂ (aq) + CO₂(g) + H₂O (l)

4. Acid + metal sulfide ----> salt + hydrogen sulfide

eg. 2HNO₃ (aq) + FeS (s) ----> Fe(NO₃)₂ (aq) + H₂S (g)

5. Reactive metal + water ----> metal hydroxide + hydrogen

eg. 2Na (s) + 2H₂O(l) ----> 2NaOH (aq) + H₂ (g)

6. metal carbonate

metal oxide + carbon dioxide

eg. MgCO₃ (s)

MgO (s) + CO₂ (g)

7. nonmetal oxide + metal hydroxide ----> salt + water

eg. SO₂ (g) + 2NaOH (aq) ----> Na₂SO₃ (aq) + H₂O (l)

8. hydrocarbon + oxygen ----> carbon dioxide + water

eg. 2C₄H₁₀ (g) + 13O₂ (g) ----> 8CO₂ (g) + 10H₂O (l)

2.3 Writing Ionic equations

Ionic equations, as the name would suggest, are often used when ions are involved in reactions which occur in aqueous media. In these conditions, it is common for some ions to take no active role in the reaction and so they are classified as spectator ions. To recognise these ions it is necessary to establish those which remain in aqueous state throughout the reaction. The ions which do react with another substance and so change state are recorded in the ionic equation. A knowledge of the solubilities of some common ionic compounds is therefore necessary.

Table 2.2 - Solubilities of Ionic compounds

Ion	Soluble	Slightly soluble	Insoluble
sodium Na⁺ potassium K⁺ ammonium NH₄⁺ nitrate NO₃^- chloride Cl^- bromide Br^- sulfate SO₄^2- carbonate hydroxide sulfide	all all all all most most most Na₂CO₃, K₂CO₃, (NH₄)₂CO₃ NaOH, KOH, NH₄OH, Ba(OH)₂ Na₂S, K₂S, (NH₄)₂S, MgS, CaS, BaS	- - - - PbCl₂ PbBr₂ Ag₂SO₄, CaSO₄ - - -	- - - - AgCl AgBr, HgBr₂ BaSO₄, PbSO₄ most most most

To write a balanced ionic equation for a chemical reaction the following procedure may be adopted:

(i) write a balanced chemical equation, including states

(ii) re-write this equation listing all aqueous species as separate ions

(iii) write the ionic equation, showing only those species which change state

(iv) list the spectator ions as those which remain aqueous throughout, if required

Example 2.2

Write ionic equations for each of the following chemical reactions, listing spectator ions in each case:

(a) a solution of nitric acid is neutralised by dilute potassium hydroxide solution.

(b) a precipitate of silver bromide is formed when solutions of silver nitrate and copper (II) bromide are mixed.

(d) copper metal is displaced from a solution of copper (II) sulfate by tin granules.

Solution

(a) (i) HNO₃ (aq) + KOH (aq) ----> KNO₃ (aq) + H₂O (l)

(ii) H⁺ (aq) + NO₃^-(aq) + K⁺ (aq) + OH^- (aq) ----> K⁺ (aq) + NO₃^-(aq) + H₂O (l)

(iii) H⁺ (aq) + OH^- (aq) ----> H₂O (l)

Spectator ions are K⁺ (aq) and NO₃^-(aq)

(b) (i) AgNO₃ (aq) + CuBr₂ (aq) ----> 2AgBr (s) + Cu(NO₃)₂ (aq)

(ii) Ag⁺ (aq) + NO₃^-(aq) + Cu²⁺ (aq) + 2Br^- (aq) ----> 2AgBr (s) + Cu²⁺ (aq) + 2NO₃^-(aq)

(iii) Ag⁺ (aq) +Br^- (aq) ----> AgBr (s)

Spectator ions are Cu²⁺ (aq) and NO₃^-(aq)

(ii) Na⁺(aq) + HCO₃^-(aq) + H⁺(aq) + Cl^- (aq) ----> Na⁺(aq) + Cl^- (aq) + CO₂ (g) + H₂O(l)

(iii) HCO₃^-(aq) + H⁺(aq) ----> CO₂ (g) + H₂O(l)

Spectator ions are Na⁺(aq) and Cl^- (aq)

(d) (i) CuSO₄ (aq) + Sn (s) ----> Cu (s) + SnSO₄ (aq)

(ii) Cu²⁺ (aq) + SO₄^2- (aq) + Sn (s) ----> Cu (s) + Sn²⁺ (aq) + SO₄^2- (aq)

(iii) Cu²⁺ (aq) + Sn (s) ----> Cu (s) + Sn²⁺ (aq)

Spectator ion is SO₄^2- (aq)

2.4 Stoichiometric Processes

A thorough understanding of the various stoichiometric techniques is essential to all students of chemistry. Fortunately, they all follow the same basic process - once this is mastered, all stoichiometric problems can be solved as variations on a theme.

The process is as follows:

1. Write a balanced chemical equation for the reaction (or reactions) under consideration.

2. List all data given, including relevant units. These data may be masses, volumes and concentrations of aqueous solutions, pressures and temperatures of gases etc. etc. Remember to also write down the symbol of the unknown quantity!

3. Convert the data given to moles, using the relevant formulae.

ie.

4. Use the chemical equation to determine the mole ratio of the unknown quantity to the known quantity. This ratio enables calculation of the number of moles of the unknown quantity to be determined.

5. Finally, convert this number of mole back into the relevant units of the unknown. The problem is solved!

It is imperative that much practice of stoichiometric questions be conducted to ensure the student become very familiar and confident with the variety of problems that could be encountered.

2.4.1 Mass - mass stoichiometry

Example 2.3

5.6 g of sodium metal reacts vigorously with water to produce an alkaline solution of NaOH accompanied by the evolution of hydrogen gas and considerable amounts of heat. Calculate the mass of hydrogen gas evolved in this reaction.

Solution

2Na (s) + 2H₂O (l) ----> 2NaOH (aq) + H₂ (g)

m = 5.6g m = 5.0g

M = 23.0g.mol^-1 M = 2.00 g.mol^-1

n (Na) =

g = 0.243 mol

From the equation, n (H₂) =

x n (Na) = 0.122 mol

Mass of H₂ (g) = n x M = 0.122 x 2 = 0.224g

Example 2.4

Hydrogen sulfide gas was bubbled through 250.0ml of a solution containing 17.85g.L^-1 of lead (II) nitrate until all the lead was precipitated as lead sulfide.

(i) Write a balanced equation for the reaction.

(ii) Calculate the mass of lead sulfide precipitated.

(iii) Calculate the mass of hydrogen sulfide required.

Solution

(i) H₂S (g) + Pb(NO₃)₂ (aq) ----> PbS (s) + 2HNO₃ (aq)

(ii) m (Pb(NO₃)₂) =

x 17.85 = 4.463g

M (Pb(NO₃)₂) = 331.2

Þ n (Pb(NO₃)₂) =

mol

From the equation, n (PbS) = n (Pb(NO₃)₂) = 0.0135 mol

Þ mass of PbS = n x M = 0.0135 x 239.3 = 3.23g

(iii) From the equation, n (H₂S) = n (Pb(NO₃)₂) = 0.0135 mol

Þ mass of H₂S = n x M = 0.0135 x 34.1 = 0.460g

2.4.2 Mass - volume stoichiometry

Example 2.5

Calculate the mass of HgBr₂ precipitated when 15.00ml of 0.337 M Hg(NO₃)₂ solution is added to excess KBr solution.

Solution

The equation for the reaction is

Hg(NO₃)₂ (aq) + 2KBr (aq) ----> HgBr₂ (s) + 2KNO₃ (aq)

[Hg(NO₃)₂] = 0.337M m(HgBr₂) = ?

V (Hg(NO₃)₂) = 15.00 ml

n (Hg(NO₃)₂) = CV = 0.337 x 0.015 = 5.055 x 10^-3 mol

From the equation, n (HgBr₂) = n (Hg(NO₃)₂) = 5.055 x 10^-3

Þ mass of H₂S = n x M = 5.055 x 10^-3 x 360.4 = 1.822g

Example 2.6

What mass of P₄O₁₀ can be produced by reacting 12.82g of P₄O₆ with 150 ml of 0.75M I₂ solution, and which reagent is in excess? The equation for the reaction is

5P₄O₆ (s) + 8I₂ (aq) ----> 4P₂I₄ (aq) + 3P₄O₁₀ (s)

Solution

Note: To solve stoichiometric problems where one reactant is in excess and we have to determine which it is, we must first calculate the number of moles of all reactants and then use the mole ratio from the equation to deduce which reagent is in excess and which is the limiting reagent. To calculate the amount of product formed we can only utilise the limiting reagent.

n (P₄O₆) =

mol n (I₂) = CV = 0.75 x 0.150 = 0.113 mol

From the equation, n (I₂) required to use up all of the P₄O₆is

x n (P₄O₆) = 0.0933 mol

ie. the I₂is in excess (by 0.113 - 0.0933 = 0.0197 mol)

In this case, the limiting reagent is P₄O₆ and so we can now calculate the n (P₄O₁₀) formed.

From the equation, n (P₄O₁₀) =

x n (P₄O₆) = 0.0350 mol

Þ mass of P₄O₁₀ = n x M = 0.0350 x 283.9 = 9.930g

2.5 The Gas Laws

There are several laws governing the (ideal) behaviour of gases which are listed below. Note that gases do not behave according to the “ideal” when the temperature becomes relatively low as the weak dispersion forces acting between particles can become of great enough significance for the gas to change state to the liquid.

The most commonly used of the Gas Laws in Year 12 chemistry is the

General Gas Equation PV = nRT

When using the General Gas equation, it is often useful to express Pressure in kPa and Volume in L. When these units are used, the Gas Constant, R = 8.314 JK^-1mol^-1

Under standard conditions of temperature and pressure this formula can be simplified to calculate the volume occupied by one mole of gas, called the molar volume; symbol V_M, where

and V_M = 22.4L at S.T.P. and 24.5L at S.L.C.

Note

S.T.P. stands for “standard temperature and pressure”, where T = 0 °C and P = 1.00atm. S.L.C. stands for “standard laboratory conditions”, where T = 25 °C and P = 1.00atm.

Some other useful laws governing the Ideal behaviour of gases are:

Boyle’s Law P a , where T and n remain constant.

Charles’ Law V a T , where V and n remain constant.

For calculation purposes, these two laws can be merged to produce the relation

, where n is constant

Units: Use of the correct Units is of critical importance when performing calculations involving gases (and all other calculations, of course).

Pressure: 1 atmosphere = 101.325 kPa (kNm^-2) = 101,325 Pa (Nm^-2) = 760 mmHg

Temperature: 0 °C = 273 K, where ‘Kelvin’ is the Absolute temperature scale.

Volume: 1000 ml (cm^-3) = 1.00L (dm^-3) = 0.001 kL (m³)

2.5.1 Volume stoichiometry

Example 2.7

A commonly available antacid preparation contains 2.32g of sodium bicarbonate (NaHCO₃) per 5g sample. Presuming all of the antacid reacts with the excess acid in the stomach, calculate the volume of carbon dioxide gas produced at 1.05 atm pressure and 37 °C.

Solution

The relevant equation is

NaHCO₃ (s) + 2HCl (aq) ----> 2NaCl (aq) + CO₂ (g) + H₂O (l)

n (NaHCO₃) =

mol

From the equation, n(NaHCO₃) = n(CO₂) = 0.0276 mol

PV = nRT Þ V =

Example 2.8

Nitroglycerine reacts explosively with oxygen gas in a highly exothermic reaction according to the equation

4C₃H₅(NO₃)₃ (l) ----> 12CO₂ (g) + 6N₂ (g) + O₂ (g) + 10H₂O (g)

Calculate the volume of gaseous products generated at 45 °C and 99.83 kPa pressure when 8.25g of nitroglycerine reacts completely.

Solution

n (C₃H₅(NO₃)₃) = mol

From the equation, n (gaseous products) = n (C₃H₅(NO₃)₃) = 0.263 mol

PV = nRT Þ V =

2.6 Summary/ Objectives

At the end of this Chapter you should

- learn the electro valencies of the common cations and anions as listed in Table 2.1

- be able to write balanced formulae of compounds and be able to accurately name them

- learn the general reaction types and be able to use them to help you to write balanced chemical equations

- be able to write balanced ionic equations having been given relevant solubility data. Recognise the existence of spectator ions.

- recognise that a balanced chemical equation gives the mole ratio of reactants to products for a reaction and that this equation can be used in stoichiometric calculations

- be familiar with the processes involved in performing stoichiometric calculations, including mass-mass, mass-volume and volume-volume types of problem

- learn and be able to use in context the formulae

and

- for problems involving solutions, be able to recognise and perform calculations in a range of concentration units; specifically molar and ppm

- for problems involving gases learn and be able to use in context the formula

, where n is constant

- learn and be able to use in context the General Gas Equation

- remember the correct units to employ when using and be able to interconvert units of temperature, pressure and volume

- learn the specific conditions of S.T.P. and S.L.C. and be able to use in its correct context the formula

where V_M = 22.4 L at S.T.P. and 24.5L at S.L.C.

- be able to perform calculations involving excess reagents, by using the mole ratio of the relevant chemical equation

Chapter 2 - Questions

1. Chemical and Ionic Equations

Q1. Write balanced equations, including states, for the following processes:

(a) When an electric current is passed through water it decomposes to its constituent elements.

(b) A dilute solution of hydrochloric acid dissolves a strip of magnesium to produce a solution of magnesium chloride and hydrogen gas.

(d) Propane undergoes complete combustion in excess oxygen to produce carbon dioxide and water.

(e) Aluminium metal reacts slowly with hydrochloric acid to produce a solution of aluminium chloride and hydrogen gas.

(f) Marble chips (calcium carbonate) dissolves in a solution of acetic acid to produce a solution of calcium acetate, carbon dioxide gas and water.

(g) When solutions of potassium iodide and lead (II) nitrate are mixed, a bright yellow precipitate of lead (II) iodide is formed.

(h) On standing, hydrogen peroxide spontaneously decomposes to water and oxygen gas.

(i) When carbon dioxide is bubbled through a solution of limewater (calcium hydroxide) a milky precipitate of calcium carbonate is formed.

(j) Solutions of silver nitrate and sodium chloride are mixed to produce a white precipitate of silver chloride.

Q2. Use your knowledge of general reaction types to write balanced chemical equations for each of the following reactions:

(a) solutions of nitric acid and sodium carbonate are mixed.

(b) sodium metal is added to water.

(d) solutions of strontium iodide and lithium sulphate are mixed.

(e) zinc metal is placed in a solution of copper sulphate.

(f) aluminium sulphide reacts with hydrochloric acid.

(g) calcium carbonate decomposes to calcium oxide upon strong heating.

(h) sulfur dioxide gas is bubbled through a solution of potassium hydroxide

(i) incomplete combustion of methane in a limited supply of air results in the formation of carbon monoxide gas.

(j) sodium hydrogen carbonate effervesces when reacted with a dilute solution of hydrochloric acid.

Q3. Write ionic equations for each of the following reactions:

(a) dilute acetic acid reacts with a solution of potassium hydroxide.

(b) Orthophosphoric acid reacts with a solution of barium hydroxide.

(d) powdered aluminium reacts readily with warm dilute sulfuric acid.

(e) solutions of tin (II) chloride and sodium hydroxide are mixed.

2. Mass-Mass stoichiometry

Q4. Calculate the mass of magnesium oxide formed when 2.431g of magnesium is burnt in excess oxygen.

Q5. Calculate the mass of carbon dioxide released when one litre of octane undergoes complete combustion in air according to the equation

2C₈H₁₈(l) + 25O₂(g) -----> 16CO₂(g) + 18H₂O (l)

Note: density of octane = 0.785g.cm^-3

Q6. Copper (II) nitrate decomposes on heating according to the equation

2Cu(NO₃)₂ (s) -----> 2CuO (s) + 4NO₂ (g) + O₂ (g)

After decomposition of a sample of copper nitrate, 3.6g of black copper oxide remains.

Calculate the original mass of copper nitrate required.

Q7. Assuming that limestone is 100% calcium carbonate, what mass of quicklime (calcium oxide) will be produced by the thermal decomposition of 500g of limestone?

Q8. Potassium chromate reacts with potassium chloride, in the presence of concentrated sulphuric acid, to form the deep red liquid chromyl chloride (Cr₂O₂Cl₂), of density 1.935 g.cm^-3. The equation for the reaction is as follows:

K₂Cr₂O₇ + 4KCl + 3H₂SO₄ ----> 2 Cr₂O₂Cl₂ + 3K₂SO₄ + 3H₂O

Calculate the volume of chromyl chloride liquid formed from 6.25g of potassium dichromate.

Q9. A 5.0 tonne charge of haematite ore (impure iron (III) oxide) undergoes reduction within a blast furnace according to the equation

Fe₂O₃ (s) + 3CO (g) ----> 2Fe (l) + 3CO₂ (g)

Calculate the percentage purity of the ore if 2.73 tonnes of iron is produced.

Q10. The sulfur content of coal may be estimated by igniting the sample of coal with Na₂CO₃ and then, by suitable treatment, converting the sulfur present to a precipitate of BaSO₄. If, in a particular analysis, a 2.00g sample of coal yields 0.493g of barium sulfate, what was the percentage by mass of sulfur in the coal?

Q11. A 1.233g sample of a mixture of NaBr and CdBr₂ is treated with an excess of silver nitrate solution, which precipitates all the Br^- as AgBr. A mass of 1.910g of AgBr is obtained. What is the percentage by mass of CdBr₂ in the sample?

Hint: Let m(NaBr) = x g, then m(CdBr₂) = (1.233 - x) g. This question is quite difficult!

3. Mass-Volume stoichiometry

Q12. A sample of 8.655g of sulfamic acid (H₂NSO₃H) is dissolved in water and the solution is carefully diluted to 100.00ml. What is the molarity of the solution?

Q13. Calculate the mass of zinc sulfate precipitated when 20.00ml of 0.45 M zinc nitrate solution reacts with excess sulfuric acid.

Q14. What mass of sodium hydrogen carbonate will react exactly with 25.00 ml of 1.55M nitric acid according to the equation

NaHCO₃ (s) + HNO₃ (aq) ----> NaNO₃ (aq) + H₂O (l) + CO₂ (g)

Q15. What volume of 0.20 M LiOH solution will exactly neutralise 35.9 ml of 0.11 M sulfuric acid solution?

Q16. The reaction between copper and concentrated nitric acid can be represented by the equation

Cu (s) + 4HNO₃ (aq) ----> Cu(NO₃)₂ (aq) + 2NO₂ (g) + 2H₂O(l)

What volume of 14M HNO₃ would be required to completely dissolve 8.4g of copper?

Q17. What mass of silver chloride will be precipitated when 25.0 ml of a solution containing 0.50 M silver nitrate is reacted with 30.0 ml of 0.35 M sodium chloride solution? Which reagent is in excess, and by how much (in grams)?

Q18. 45.0ml of 0.750 M AgNO₃ solution is added to 65.0ml of 1.375 M Cr(NO₃)₃ solution. Calculate the resultant concentration of nitrate ions in the mixture.

Q19. A student is required to make up 250.0 ml of 0.200M nitric acid solution from a stock supply of concentrated (14 M) acid. What volume of the concentrated acid will she need to use, and what volume of water will need to be added?

4. Gas Volume calculations

Q20. Calculate the volume occupied by 4.4g of carbon dioxide gas at Standard Laboratory Conditions. (Note: V_m = 24.5 L at S.L.C.)

Q21. Hydrogen peroxide decomposes over time to produce oxygen gas and water. What volume of O₂ (g) at S.T.P. will be produced by the complete decomposition of 200ml of 6.0 M H₂O₂ solution.

Q22. An oxygen cylinder of internal volume 6.65L contains oxygen gas under a pressure of 16.7 MPa at a temperature of 28°C.

(i) Calculate the mass of oxygen in the cylinder.

(ii) What volume would this gas occupy at 22°C and 1.13 atm. pressure?

Q23. Propane is commonly used as the principle fuel in BBQ gas bottles. Given that a 9.0 kg gas bottle contains 85% propane which undergoes complete combustion according to the equation

C₃H₈ (l) + 5O₂ (g) ----> 3CO₂ (g) + 4H₂O (l)

Calculate the volume of CO₂ produced at 35°C and 98.6 kPa pressure.

Q24. Calculate the number of atoms of neon gas in a fluorescent tube of volume 3.21 L and pressure 560 Pa at a temperature of 26°C.

Q25. In the preparation of nitrogen monoxide by the reaction

3Hg (l) + 8HNO₃ (aq) ----> 3Hg(NO₃)₂ (aq) + 2NO (g) + 4H₂O (l)

24.7 g of mercury was treated with 180 ml of 6.0 M nitric acid. Calculate

(a) Which reagent is in excess and by how much (in grams)?

(b) What volume of NO is formed at S.T.P. ? (Note: V_m = 22.4 L at S.T.P.)

Q26. A gaseous compound of carbon and hydrogen contains 92.26% carbon. When 1.373g of the compound is collected at a temperature of 25°C and a pressure of 99.485 kPa it is found to occupy a volume of 1.308 L. What is the molecular formula of the compound?

Chapter Two Solutions

Q4. 4.031g Q5. 2424g Q6. 8.494g

Q7. 280g Q8. 4.54 ml Q9. 78.1% Q10. 3.39% Q11. 20.4% Q12. 0.893 M

Q13. 1.453g Q14. 3.255g Q15. 39.5 ml Q16. 37.8 ml

Q17. AgNO₃ is in excess by 0.340g, m(AgCl) = 1.505g Q18. 2.744 M Q19. 3.57 ml,

246.43 ml Q20. 2.45 L Q22. (i) 1420g, (ii) 950.6 L Q23. 1.355 x 10⁴ L

Q24. 4.355 x 10²⁰ Q25 a) HNO₃ is in excess by 47.35g, b) 1.839 L Q26. CH, C₂H₂

The Reactivity of Metals

When an iron nail is left in copper sulphate solution a reaction occurs, the iron removes the copper from solution. This is a displacement reaction.

The iron is more reactive than the copper. We know from experience that different metals have varying abilities to react with other substances.

Eg.

Metals can be put in a reactivity list which is called the reactivity series of metals or the activity series. Table 18.1

Reactivity of metals and the periodic table.

The metals in groups I and II are very reactive. Metals tend to act as reductants. The reactivity of metals tends to decrease across the periodic table and down the group.

We can use the reactivity series to predict redox reactions.

ELECTROCHEMICAL CELLS

Redox reactions that occur spontaneously can be used to produce energy. The energy can be released as heat, light, electricity or sound depending on the set-up of the displacement reaction. Eg. Iron nail in copper sulfate solution produces heat energy.

If the reactions are separated the electrons can be forced to move through wires to create an electrical current. This is the set-up in batteries. Such set-ups are called ELECTROCHEMICAL CELLS. Their function is to be an energy converter, changing chemical energy into electrical energy.

A simple electrochemical cell consists of:

Two half-cells, containing two electrodes (anode and cathode) and two electrolytes
A conducting wire
A salt bridge, containing another electrolyte.

In half-cells the more reactive metal will always give its electrons to the ions of the less reactive metal. The ions of the less reactive metal will gain electrons from the other.

· THE ELECTRODE AT WHICH OXIDATION OCCURS IS CALLED THE ANODE.

THE ELECTRODE AT WHICH REDUCTION OCCURS IS CALLED THE CATHODE

Competition for Electrons.

From the reactivity series of metals the higher the metal is on the list, the more reactive the metal and the stronger the ability to act as a reductant. Therefore the higher metal will release its electrons more readily and undergoes an oxidation reaction.

If there was competition between two ions for electrons then the ions lower on the reactivity series will more readily accept the electrons.

Electrons flow to the half-cell that most readily accepts electrons.

The ELECTROCHEMICAL SERIES OF METALS

The relative abilities of all half-cells to attract electrons can be ranked in an ELECTROCHEMICAL SERIES OF METALS as shown in table 18.2.

In an electrochemical series the half-equations are written in the direction in which reduction occurs. Strongest oxidant at the top left and strongest reductant bottom right.

IN GENERAL THE STRONGER REDUCTANT WILL ALWAYS REACT WITH THE STRONGER OXIDANT.

Corrosion

Corrosion is the oxidation of a metal that causes the metal to be damaged as a result of chemical change.

Two types of corrosion:

Dry corrosion occurs when a metal reacts directly with oxygen. Eg burning magnesium ribbon in oxygen.
Wet corrosion occurs when a metal reacts directly with oxygen in the presence of water.

The corrosion of iron is commonly known as rusting.

Rusting starts at a ‘stress’site.

Fe (s) Fe²⁺ + 2e^-

The oxygen accepts electrons and is reduced.

O₂(g) + 2H₂0 + 4e^- 4OH^- (aq)

The overall equation for the reaction is

2Fe(s) + O₂(g) + 2H₂O(l) 2Fe(OH)₂(s)

4Fe(OH)₂(s) + O₂(g) + 2H₂O(l) 4Fe(OH)₃(s)

Effects of corrosion.

Loss of strength
Cannot conduct
Corroded pipes leak
Products of corrosion flake off and cause blockages
Corroded metals expand may cause nuts and bolts to jam machinery.

Three main methods that we can use to protect iron and steel from corrosion.

Surface protection –paint, plastic, grease and metal coatings.
Alloying
Electrochemical protection which involves placing a more reactive metal in electrical contact with the metal that needs protection.

REDOX CHEMISTRY

4.1 Introduction

The corrosion of iron, the production of small-scale energy sources, the electroplating of silver and gold onto base metals, the production of aluminium and iron - all of these and so many more are examples of this critically important class of chemical reactions. Redox reactions are frequently observed in both large-scale industrial processes and complex biochemical reactions occurring within cells. Much of the technology upon which we have become increasingly reliant depends on light, highly efficient, small-scale energy sources, more commonly known as “batteries”. As is the case with acid/base chemistry our understanding of the mechanisms of redox reactions is relatively recent, although this type of reaction has been known and exploited since the Iron Age.

4.2 Definitions of oxidation and reduction

As the name would suggest, oxidation reactions were originally defined as any reaction in which a substance accepts oxygen. Reduction was defined as that process whereby a substance loses oxygen in a chemical reaction. This definition was later expanded to include any reaction whereby hydrogen is gained (reduction) or lost (oxidation). Neither of these definitions are suitable in all situations however, and so the formal definition of redox reactions with which the VCE student must be familiar is as follows:

Oxidation reactions involve the loss of electrons.

Reduction reactions involve the gain of electrons.

Note that the terms “oxidation” and “reduction” are referring to processes which may occur in chemical reactions. The terms “oxidant” and “reductant” are reserved for actual chemical species which undergo these reactions. The oxidant, or oxidising agent, is the chemical which oxidises another substance: to achieve this end, it must (by definition) force the other species to lose electrons. In other words, the oxidant must gain electrons. Following similar logic, a reductant, or reducing agent, is the chemical which reduces another substance: it must (by definition) force the other species to gain electrons by donating its own. And so, we have derived another most important definition which should be committed to memory:

An oxidant (oxidising agent) accepts electrons and is itself reduced.

A reductant (reducing agent) donates electrons and is itself oxidised.

An important corollary can be drawn between redox reactions and acid/base systems at this stage. Just as acids and bases always function together, so too must oxidants and reductants; an oxidant cannot accept electrons unless there is a reductant available to donate them, and vice versa. Furthermore, when an oxidant accepts electrons, the product of this reaction must, by definition, be able to donate electrons in the reverse reaction. Just as with acid/base systems, the existence of conjugate redox pairs is noted as a fundamental aspect of redox chemistry.

4.3 The determination and use of Oxidation numbers.

It is not always easy to determine at first glance whether or not a particular reaction is redox in nature and if it is, which chemical species are acting as oxidant and reductant. A very useful tool to assist in this regard is the “oxidation number” - an assigned number that indicates the number of electrons that an atom appears to have lost or gained from its normal complement when it is combined with other atoms. Note that oxidation numbers have no real meaning, they are merely a useful device to help us recognise redox reactions.

The rules for assigning oxidation numbers are as follows:

i) the oxidation number of atoms in neutral elements is always zero. eg. He, O₂, Br₂, Fe, Cu all have oxidation numbers of zero.

ii) The oxidation number of atoms in simple ions is the same as the charge on the ion: eg. H⁺, Cl^-, Zn²⁺, Cr³⁺, O^2- have oxidation numbers of +1, -1, +2, +3 and -2 respectively.

iii) The oxidation number of oxygen is nearly always -2, the only common exception being -1 in the case of the peroxide ion, O₂^2-

iv) The oxidation number of hydrogen is usually +1, the exceptions being the metal hydrides, such as NaH, KH and CaH₂, where hydrogen has an oxidation number of -1.

v) The sum of the oxidation numbers of all the atoms in a compound or ion is equal to the charge on that compound or ion. In the case of a neutral compound, the sum of the oxidation numbers must be zero.

vi) The oxidation number of the most electronegative element in a compound is always assigned the negative value.

Example 4.1

Calculate the oxidation number of the underlined species for each of the following:

a) SO₂ b) Fe₂O₃ c) S₈

d) MnO₄^- e) VO₃^2- f) Cr₂O₇^2-

g) OCl^- h) H₂O₂ i) CuHSO₄

Solution

a) +4 b) +3 c) 0

d) +7 e) +4 f) +6

g) +2 h) -1 i) +1

Once the oxidation numbers of species involved in chemical reactions have been determined it is relatively simple to determine whether a reaction is redox in nature or not, as change in oxidation number always occurs in a redox reaction. Furthermore, if the oxidation number of a particular atom increases, that atom has undergone oxidation; if it has decreased, the atom has undergone reduction.

Oxidation involves an increase in oxidation number.

Reduction involves a decrease in oxidation number

When examining a chemical reaction so as to determine which species acts as oxidant and which as reductant we must first determine the oxidation number of each constituent atom and establish which have changed. From the rules above, it can be seen that the oxidant will have undergone a decrease in oxidation number as it undergoes reduction. Similarly, the reductant will have undergone an increase in oxidation number as it undergoes oxidation.

Example 4.2

By assigning oxidation numbers determine which species is acting as the oxidant and which the reductant in the following redox reactions:

a) Mg (s) + 2HCl (aq) ----> MgCl₂ (aq) + H₂ (g)

b) 2Fe³⁺ (aq) + Sn²⁺ (aq) ----> 2Fe²⁺ (aq) + Sn⁴⁺ (aq)

c) 2Al₂O₃ (l) + 4C (s) ----> 4Al (l) + 3CO₂ (g)

d) 5S^2- (aq) + 8MnO₄^- (aq) + 24H⁺ (aq) ----> 8Mn²⁺ (aq) + 12H₂O (l) + 5SO₄^2- (aq)

e) Zn (s) + 2MnO₂ (s) + 2NH₄⁺ (aq) ----> Zn²⁺ (aq) + Mn₂O₃ (s) + 2NH₃ (aq) + 12H₂O (l)

Solution a) Oxidant is HCl as the oxidation number of hydrogen decreases from +1 to zero. Reductant is Mg as the oxidation number of magnesium increases from zero to +2. b) Oxidant is Fe³⁺ as the oxidation number of iron decreases from +3 to +2. Reductant is Sn²⁺as the oxidation number of tin increases from +2 to +4. c) Oxidant is Al₂O₃ as the oxidation number of aluminium decreases from +3 to zero. Reductant is C as the oxidation number of carbon increases from zero to +4 d) Oxidant is acidified MnO₄^- as the oxidation number of manganese decreases from +7 to +2. Reductant is S^2- as the oxidation number of sulfur increases from -2 to +6. e) Oxidant is acidified MnO₂ as the oxidation number of manganese decreases from +4 to +3.

Reductant is Zn as the oxidation number of zinc increases from zero to +2.

4.4 Writing Partial Redox equations

There are a number of occasions where you will be required to write a balanced partial or full ionic equation for a redox reaction, having been given only the formulae of the reactant and product species. Whilst this is very simple in the case of monatomic species, more complex polyatomic species require more care and attention to detail.

The following procedure may be adopted to generate balanced partial redox equations:

(i) Balance the equation with respect to all atoms except hydrogen and oxygen

(ii) Balance the oxygen atoms by adding water to the relevant side

(iii) Balance the hydrogen atoms by adding H⁺ (aq) ions to the relevant side

(iv) Balance the overall charge by adding electrons to the relevant side

Example 4.3

Write balanced partial redox equations for each of the following:

a) Cu²⁺ (aq) ----> Cu (s)

b) Cl₂ (g) ----> Cl^- (aq)

c) SO₂ (g) ----> SO₄^2- (aq)

d) MnO₄^- (aq)----> Mn²⁺ (aq)

e) CH₃CH₂OH (aq) ----> CH₃CHO (aq)

Solution

a) Cu²⁺ (aq) ----> Cu (s)

Balance charge by adding 2 electrons to the left-hand side:

Cu²⁺ (aq) + 2e ----> Cu (s)

b) Cl₂ (g) ----> Cl^- (aq)

Balance chlorine atoms:

Cl₂ (g) ----> 2Cl^- (aq)

Balance charge by adding 2 electrons to the right-hand side:

Cl₂ (g) + 2e ----> 2Cl^- (aq)

c) SO₂ (g) ----> SO₄^2- (aq)

Balance oxygen atoms by adding 2 water molecules to the left-hand side:

SO₂ (g) + 2H₂O (l) ----> SO₄^2- (aq)

Balance hydrogen atoms by adding 4 H⁺ (aq) ions to the right-hand side:

SO₂ (g) + 2H₂O (l) ----> SO₄^2- (aq) + 4H⁺(aq)

Balance charge by adding 2 electrons to the right-hand side:

SO₂ (g) + 2H₂O (l) ----> SO₄^2- (aq) + 4H⁺(aq) + 2e

d) MnO₄^- (aq) ----> Mn²⁺ (aq)

Balance oxygen atoms by adding 4 water molecules to the right-hand side:

MnO₄^- (aq) ----> Mn²⁺ (aq) + 4H₂O (l)

Balance hydrogen atoms by adding 8 H⁺ (aq) ions to the right-hand side:

MnO₄^- (aq) + 8H⁺ (aq) ----> Mn²⁺ (aq) + 4H₂O (l)

Balance charge by adding 5 electrons to the left-hand side:

MnO₄^- (aq) + 8H⁺ (aq) + 5e ----> Mn²⁺ (aq) + 4H₂O (l)

e) CH₃CH₂OH (aq) ----> CH₃CHO (aq)

Balance hydrogen atoms by adding 2H⁺ (aq) ions to the right-hand side:

CH₃CH₂OH (aq) ----> CH₃CHO (aq) + 2H⁺ (aq)

Balance charge by adding 2 electrons to the right-hand side:

CH₃CH₂OH (aq) ----> CH₃CHO (aq) + 2H⁺ (aq) + 2e

4.5 Writing balanced Ionic equations for Redox reactions

To write a balanced overall ionic equation for a redox reaction, first generate partial equations for the oxidation and the reduction reactions and then balance these with respect to electrons. The two partial equations can then be added and any reagents shown on both sides of the equation can be cancelled.

Example 4.4

Write a balanced overall ionic equation for the reaction whereby copper metal is formed when a piece of zinc metal is placed in a solution of copper (II) sulfate solution.

Solution

The relevant partial equations are

Cu²⁺ (aq) + 2e ----> Cu (s) and

Zn (s) ----> Zn²⁺ (aq) + 2e

When the two equations are added the electrons cancel out, as required. The overall ionic equation for this reaction is:

Cu²⁺ (aq) + Zn (s) ----> Cu (s) + Zn²⁺ (aq)

Example 4.5

Write a balanced overall ionic equation for the reaction whereby ethanol is oxidised to ethanoic acid by acidified potassium dichromate. The dichromate ion is itself reduced to the chromium (III) ion by this reaction.

Solution

The relevant partial equations are

Cr₂O₇^2- (aq) + 14H⁺ (aq) + 6e ----> 2Cr³⁺ (aq) + 7H₂O (l) and

CH₃CH₂OH (aq) + H₂O (l) ----> CH₃COOH (aq) + 4H⁺ (aq) + 4e

Note that we need to multiply the first equation by 2 and the second by 3 in this instance in order to balance the number of electrons on both sides of the equation:

2Cr₂O₇^2- (aq) + 28H⁺ (aq) + 12e ----> 4Cr³⁺ (aq) + 14H₂O (l) and

3CH₃CH₂OH (aq) + 3H₂O (l) ----> 3CH₃COOH (aq) + 12H⁺ (aq) + 12e

The partial equations can now be added:

2Cr₂O₇^2- (aq) + 28H⁺ (aq) + 12e + 3CH₃CH₂OH (aq) + 3H₂O (l)

----> 4Cr³⁺ (aq) + 14H₂O (l) + 3CH₃COOH (aq) + 12H⁺ (aq) + 12e

This equation can now be simplified to produce the overall ionic equation:

2Cr₂O₇^2- (aq) + 16H⁺ (aq) + 3CH₃CH₂OH (aq) ---> 4Cr³⁺ (aq) + 11H₂O (l) + 3CH₃COOH (aq)

4.6 Volumetric analysis involving redox reactions

The fundamental principles of volumetric analysis are unchanged, whether the reagents involved are acid/base or redox in nature. The most commonly used primary redox standards employed in the school laboratory are acidified potassium permanganate (H⁺/ MnO₄^- ) and acidified potassium dichromate (H⁺ /Cr₂O₇^2- ). A significant advantage of using the permanganate ion is that no indicator is required as the permanganate ion has an intense purple colour while its conjugate, Mn²⁺,has no colour. Note that the common acid/base indicators, such as phenolphthalein and methyl red, are not suitable for redox titrations; specific redox indicators (such as starch) are required.

Some examples of volumetric analyses involving redox reactions are

(i) the determination of alcohol (or SO₂) content in wine

(ii) the determination of acetyl salicylate in an aspirin tablet

(iii) the determination of ascorbic acid in a Vitamin C tablet

(iv) the determination of available chlorine in pool chlorine

It is likely that the student will have performed one or more of these exercises as part of the Product Analysis work requirement.

4.7 Summary/ Objectives

At the end of this Chapter you should

- recall the definitions of oxidation and reduction as the loss of electrons and gain of electrons respectively

- recognise that an oxidant (or oxidising agent) accepts electrons and a reductant (or reducing agent) donates electrons

- recall that oxidants and reductants must function as conjugate pairs, just as must acids and their conjugate bases

- learn the rules for assigning oxidation numbers and be able to apply these elements in particular compounds

- use oxidation numbers to determine whether a particular species has undergone oxidation (increase in oxidation number) or reduction (decreases in oxidation number)

- learn the rules for writing balanced partial redox equations and be able to apply these rules as required

- be able to write a balanced full ionic equation for a redox reaction by first writing balanced partial redox equations (always ensure electrons ‘cancel out’ on both sides)

- recognise the role of acidic media in many important oxidant species (such as permanganate, MnO₄^-, and dichromate, Cr₂O₇^2-)

- recall applications of redox chemistry to analytical procedures, such as gravimetric and volumetric analyses (egs. determination of ascorbic acid content in a Vitamin C tablet, determination of bleach content in a commercially available laundry bleach)

Chapter 4 questions

Q1. Calculate the oxidation number of the underlined element for each of the following species:

a) O₂ b) SO₃ c) Fe(NO₃)₃

d) MnO₄^- e) CrO₂ f) Na₂B₄O₇

g) Zn(BrO₃)₂ h) Ga₂(CO₃)₃ i) PbI₄

j) SnSO₄ k) H₃PO₄ l) Mg₃(AsO₄)₂

Q2. Calculate the oxidation number of sulfur in each of the following species:

a) SO₂ b) H₂SO₄ c) H₂S

d) SO₃^2- e) S₂ f) S₂O₃^2-

Q3. Calculate the oxidation number of nitrogen in each of the following species:

a) NO b) NO₂ c) N₂

d) HNO₂ e) HCN f) N₂O₄

g) NO₃^- h) N₂O i) NH₃

Q.4 For each of the following redox reactions, state which species is acting as the oxidant and which as the reductant:

a) Cl₂(g) + Zn (s) ----> Zn²⁺ (aq) + 2Cl^- (aq)

b) 2Li (s) + 2H₂O (l) ----> 2LiOH (aq) + H₂ (g)

c) Fe₂O₃ (s) + 3H₂ (g) ----> 2Fe (s) + 3H₂O (l)

d) 2H₂S (g) + SO₂(g) ----> 3S (s) + 2H₂O (l)

e) HNO₂ (aq) + H₂O₂ (aq) ----> HNO₃ (aq) + H₂O (l)

Q5. Write partial ionic equations for each of the following:

a) iodide ions are oxidised to iodine

b) nitric acid is reduced to nitrogen dioxide

c) vanadate ions (VO₃^2-) are reduced to vanadium (III) ions

d) sulfur is oxidised to thiosulfate ions (S₂O₃^2-)

e) methanol (CH₃OH) is oxidised to methanal (HCHO)

f) sulfite ions (SO₃^2-) are reduced to hydrogen sulfide (H₂S)

Q6. Write overall ionic equations for the following redox reactions:

a) A strip of magnesium metal displaces silver metal from a solution of AgNO₃ and is itself dissolved.

b) Concentrated nitric acid reacts with silver metal to produce Ag⁺ ions and nitrogen dioxide gas, among other products

c) When sulfur dioxide gas is bubbled through an acidified solution of potassium dichromate, the SO₂ is oxidised to SO₄^2- and the Cr₂O₇^2- is reduced to Cr³⁺ ions.

d) A solution of potassium permanganate is decolourised by the addition of excess oxalic acid (H₂C₂O₄). CO₂ and Mn²⁺ are among the products of the reaction.

e) Hydrogen peroxide spontaneously decomposes over a period of time to produce water and oxygen gas

Q7. The reaction which occurs within the common Leclanché dry cell may be represented by

2MnO₂ (s) + Zn (s) + NH₄⁺ (aq) ----> Mn₂O₃ (s) + Zn²⁺ (aq) + 2NH₃ (aq)

(i) Which species is acting as the oxidant and which as the reductant?

(ii) What mass of MnO₂ would be required to react completely with 15.0g of zinc?

Q8. Calculate the mass of powdered zinc metal that would be required to displace all of the Ag⁺ ions from 5.0 L of waste photographic emulsion solution as silver metal, according to the equation

2Ag⁺ (aq) + Zn (s) ----> 2Ag (s) + Zn²⁺ (aq)

Presume [Ag⁺] = 120 ppm and d (waste solution) = 1.00 g.cm^-3

Q9. 18.22 ml of a solution of potassium permanganate exactly reacts with 0.605 g of nickel metal; Mn²⁺ and Ni²⁺ being among the products. Calculate the concentration of the KMnO₄ solution.

Q10. 8.27 ml of 2.25 M sulfuric acid precisely oxidises 15.0g of an impure sample of iron; SO₂ and Fe²⁺ being among the products. Presuming iron is the only substance undergoing oxidation, what is the percentage purity of iron in the sample?

Q11. A 0.2817g sample of iron ore was dissolved in acid and all of the Fe present was converted to Fe²⁺ ions. The resulting Fe²⁺ solution was titrated with 0.01864 M K₂Cr₂O₇ solution, a titre volume of 21.24 ml being required to reach endpoint; Fe³⁺ and Cr³⁺being among the products. Calculate the percentage of iron in the iron ore sample.

Q12. A student wishes to verify the mass of active constituent (ascorbic acid, C₆H₈O₆) in a 250 mg tablet of Vitamin C. She weighs the tablet and records the mass as 0.344g. She then dissolves the tablet in about 50 ml of deionised water in a conical flask. This solution is then titrated against a 0.0528 M solution of iodine, using starch as indicator. A titre volume of 27.18 ml was required to reach endpoint.

The equation for this reaction may be written as:

C₆H₄O₂(OH)₄ (aq) + I₂ (aq) ----> C₆H₄O₄(OH)₂(aq) + 2H⁺(aq) + 2I^- (aq)

(a) Calculate the mass of ascorbic acid in the tablet (in mg).

(b) What possible functions may the substances perform which make up the remainder of the tablet’s mass?

Q13. Household bleach contains the strong oxidant ‘sodium hypochlorite’, NaOCl, as the active ingredient. The concentration of NaOCl is generally known as the ‘available chlorine’ and is often expressed as the mass of active ingredient per unit volume of solution, or w/v.

A student wishes to verify the available chlorine content in a sample of commercially available bleach, quoted as ‘40g/L available chlorine’.

To achieve this aim he pipettes 25.00ml of the bleach into a 250.0 ml standard flask and makes up to the mark with deionised water. He then pipettes three 20.00 ml aliquots of this diluted solution into separate conical flasks and adds about 10 ml of acidified potassium iodide solution. He notes that the solution in the flask immediately becomes dark brown, as the available chlorine oxidises the iodide ions to iodine, according to the equation:

OCl^- (aq) + 2I^- (aq) + 2H⁺ (aq) -----> I₂ (aq) + Cl^- (aq) + H₂O (l)

He then titrates this solution with a standard solution of 0.0513 M sodium thiosulfate solution for each flask; an average titre volume of 9.57 ml being required. The relevant equation is:

I₂ (aq) + 2S₂O₃^2- (aq) ----> S₄O₆^2- (aq) + 2I^- (aq)

(i) Calculate the no. of mol of S₂O₃^2- in each flask.

(ii) Calculate the no. of mol of I₂reduced by the S₂O₃^2- ions in each flask.

(iii) Deduce the no. of mol of OCl^-in the aliquot of diluted solution.

(iv) Calculate the no. of mol of OCl^-in the original sample of bleach.

(v) Thus deduce the no. of mol of chlorine atoms (as Cl^-) in the original sample of bleach.

(vi) Calculate the mass of available chlorine in the original sample, and thus determine the w/v ratio (Presume d (bleach solution) = 1.00 g.ml^-1).

Q14. A scientist wishes to determine the percentage by volume of alcohol in a bottle of Chilean wine. She dispenses a 20.00 ml sample of the wine into a 250.0 ml standard flask and makes it up to the mark with deionised water. She then withdraws a 20.00 ml aliquot of this mixture and adds to it precisely 20.00 ml of 0.05279 M K₂Cr₂O₇ and approximately 10 ml of 8M H₂SO₄. After heating this acidified mixture carefully for 10 minutes all of the alcohol in the wine has been oxidised to ethanal by the excess dichromate, according to the equation:

3CH₃CH₂OH (aq) + Cr₂O₇^2- (aq) + 8H⁺ (aq) ----> CH₃CHO (aq) + 2Cr³⁺ (aq) + 7H₂O (l)

The solution is then allowed to cool, about 2g of potassium iodide is added and the excess Cr₂O₇^2- oxidises the iodide ions to iodine, according to the equation:

Cr₂O₇^2- (aq) + 14H⁺ (aq) + 6I^- (aq) ----> 2Cr³⁺ (aq) + 7H₂O (l) + 3I₂ (aq)

Finally, the iodine thus formed is titrated against a standard solution of 0.1091 M sodium thiosulfate solution, an average titre of 7.72 ml being required to reach endpoint. The relevant equation is:

I₂ (aq) + 2S₂O₃^2- (aq) ----> S₄O₆^2- (aq) + 2I^- (aq)

(i) Calculate the no. of mol of S₂O₃^2- in the titre.

(ii) Calculate the no. of mol of I₂reduced by the S₂O₃^2- ions in each flask.

(iii) Calculate the no. of mol of Cr₂O₇^2- in excess.

(iv) Calculate the no. of mol of Cr₂O₇^2- required to oxidise the ethanol in the aliquot.

(v) Calculate the no. of mol of ethanol in the aliquot.

(vi) Calculate the mass of ethanol in the original sample.

(vii) Given that the density of ethanol = 0.785 g.ml^-1, determine the % volume of alcohol (% v/v) in the wine.

Solutions to Chapter 4 questions

Q1. a) 0 b) +6 c) +3 d) +7 e) +4 f) +3

g) +5 h) +3 i) +4 j) +2 k) +5 l) +5

Q2. a) +4 b) +6 c) -2 d) +4 e) 0 f) +2

Q3. a) +2 b) +4 c) 0 d) +3 e) -3 f) +4

g) +5 h) +1 i) -3

Q4. a) oxidant is Cl₂, reductant is Zn

b) oxidant is H₂O, reductant is Li

c) oxidant is Fe₂O₃, reductant is H₂

d) oxidant is SO₂, reductant is H₂S

e) oxidant is H₂O₂, reductant is HNO₂

Q5. a) 2I^- (aq) ----> I₂ (aq) + 2e

b) HNO₃ (aq) + H⁺(aq) + e ----> NO₂ (g) + H₂O (l)

c) VO₃^2- (aq) + 6H⁺(aq) + e ----> V³⁺ (aq) + 3H₂O (l)

d) 2S (s) + 3H₂O (l) ----> S₂O₃^2- (aq) + 6H⁺(aq) + 4e

e) CH₃OH (aq) ----> HCHO (aq) + 2H⁺(aq) + 2e

f) SO₃^2- (aq) + 8H⁺(aq) + 6e ----> H₂S (g) + 3H₂O (l)

Q6. a) Mg (s) + 2Ag⁺ (aq) ----> Mg²⁺ (aq) + 2Ag (s)

b) HNO₃ (aq) + H⁺(aq) + Ag (s) ----> NO₂ (g) + Ag⁺ (aq) + H₂O (l)

c) Cr₂O₇^2- (aq) + 2H⁺(aq) + 3SO₂ (g) ----> 2Cr³⁺ (aq) + H₂O (l) + 3SO₄^2- (aq)

d) 2MnO₄^-(aq) + 6H⁺(aq) + 5H₂C₂O₄ (aq) ----> 2Mn²⁺ (aq) + 8H₂O (l) + 10CO₂ (g)

e) 2H₂O₂ (aq) ----> 2H₂O (l) + O₂ (g)

Q7. (i) oxidant is MnO₂, reductant is Zn

(ii) 39.9g

Q8. 0.182g Q9. 0.226 M Q10. 69.3% Q11.47.1%

Q12. a) 253 mg b) flavour, colour, binding agents Q13. (i) 4.909 x 10^-4 mol

(ii) 2.455 x 10^-4 mol (iii) 2.455 x 10^-4 mol (iv) 0.0307 mol (v) 0.0307 mol

(vi) 1.088g, 43.5g.L^-1 Q14. (i) 8.423 x 10^-4 mol (ii) 4.211 x 10^-4 mol

(iii) 1.404 x 10^-4 mol (iv) 9.154 x 10^-4 mol (v) 0.0343 mol (vi) 1.580g

(vii) 10.06%

ACIDS AND BASES

3.1 Introduction

The existence of acids and bases has been recognised since Ancient times, though definitions of their properties and reactivity are much more recent. The Romans knew how to make soap: by boiling animal fats with “caustic soda”, what we would now describe as sodium hydroxide. In the Middle Ages, alchemists described chemical substances in terms of their physical properties, such as colour, taste and smell. Hence, the unfortunate person who first described sulfuric acid was moved to describe the substance as “Oil of Vitriol”.

In a more modern context, we are aware of this important class of chemicals from such significant examples as the amino acids (building blocks of proteins), ascorbic acid (Vitamin C) - an important and widely used food preservative, citric acid (the principle component of citrus fruits that imparts the characteristically tart taste to these fruits), and antacid preparations (typically containing such compounds as Na₂CO₃, NaHCO₃ and Al(OH)₃) used to ease the indigestion caused by excess stomach acid. In Western societies, the serious problem of “acid rain” (mainly caused by the hydrolysis reactions of the oxides of sulfur, nitrogen and carbon with rainwater) has yet to be effectively addressed.

3.2 Definition of acid-base function

The most useful way in which the behaviour of this class of compounds may be defined, as far as the VCE chemistry student is concerned, is in terms of the BrÆnsted/ Lowry theory.

A BrÆnsted/ Lowry acid is defined as any ion or molecule that can donate a proton (H⁺).

A BrÆnsted/ Lowry base is defined as any ion or molecule that can accept a proton (H⁺).

Note that acids and bases always function together; an acid cannot donate a proton unless there is a base available to accept it, and vice versa. Furthermore, when an acid donates a proton, the product of this reaction must, by definition, be able to accept a proton in the reverse reaction. This concept of conjugicity, or the existence of conjugate acid-base pairs, is really of critical importance to a thorough understanding of the chemistry of acids and bases.

Students should also recognise that a proton (H⁺) in aqueous solution will attach itself to a water molecule (due to the polarity of water) to generate the species H₃O⁺, generally known as the ‘hydronium ion’:

H₂O (l) + H⁺ (aq) ----> H₃O⁺ (aq)

Example 3.1

For each of the following acid/base reactions, link the conjugate acid/base pairs:

a) HCl (aq) + H₂O (l) ----> H₃O⁺ (aq) + Cl^- (aq)

b) H₂SO₄ (aq) + H₂O (l) ----> HSO₄^- (aq) + H₃O⁺ (aq)

c) NH₃ (g) + H₂O (l) ----> NH₄⁺ (aq) + OH^- (aq)

d) H₃O⁺ (aq) + OH^- (aq) -----> 2 H₂O (l)

Solution

acid/base base/acid

a) HCl (aq) / Cl^- (aq) H₂O (l) / H₃O⁺ (aq)

b) H₂SO₄ (aq) / HSO₄^- (aq) H₂O (l) / H₃O⁺ (aq)

c) H₂O (l) / OH^- (aq) NH₃ (g) / NH₄⁺ (aq)

d) H₃O⁺ (aq) / H₂O (l) OH^- (aq) / H₂O (l)

Note that in each of the examples the members of the conjugate pair differ by only a H⁺ and that one half of the pair is always on the left-hand side of the equation, the other on the right-hand side.

3.3 Strength of acids and bases

We know from experience that some acids and bases are much stronger than others, but how can we explain this phenomenon? The answer lies in the extent of the reaction of these substances - a concept that will be discussed in much more detail when we study “Chemical Equilibrium”.

Three strong acids with which we are familiar are hydrochloric, sulfuric and nitric acids. Each of these inorganic acids are formed from the reaction of covalently bonded molecules with water in ionisation reactions:

HCl (g) + H₂O (l) ----> H₃O⁺ (aq) + Cl^- (aq)

H₂SO₄ (l) + H₂O (l) ----> HSO₄^- (aq) + H₃O⁺ (aq)

HNO₃ (l) + H₂O (l) ----> H₃O⁺ (aq) + NO₃^- (aq)

As these are all strong acids we can deduce that the reaction proceeds to a large extent; that effectively all of the molecules react with water to produce large numbers of hydrogen ions.

Acetic (ethanoic) acid, on the other hand, is a typical example of a weak acid; this ionisation reaction proceeds to only a relatively small extent and so not as many H⁺ ions are formed:

CH₃COOH (l) + H₂O (l) ----> CH₃COO^- (aq) + H₃O⁺ (aq)

The extent of the reaction of bases explains their relative strength in a similar way.

Examples of strong bases are NaOH, KOH and Mg(OH)₂. In each of these cases, the ionic compound undergoes complete dissociation to produce large numbers of hydroxide ions (OH^-). It is the hydroxide ion which functions as a strong base:

NaOH (s) Na⁺ (aq) + OH^- (aq)

KOH (s) K⁺ (aq) + OH^- (aq)

Mg(OH)₂ (s) Mg²⁺ (aq) + 2OH^- (aq)

A weak base, such as ammonia, ionises with water to only a small extent:

NH₃ (g) + H₂O (l) ----> NH₄⁺ (aq) + OH^- (aq)

3.4 Polyprotic and amphiprotic substances

Some substances, such as sulfuric acid, are capable of donating more than one proton. In the case of sulfuric acid, each molecule of H₂SO₄ is able to donate two protons - we say it is a “diprotic acid”. The successive ionisation reactions of H₂SO₄are given below:

H₂SO₄ (l) + H₂O (l) ----> HSO₄^- (aq) + H₃O⁺ (aq)

HSO₄^- (aq) + H₂O (l) ----> SO₄^2- (aq) + H₃O⁺ (aq)

Example 3.2

Write the successive ionisation reactions of the polyprotic acid H₃PO₄ reacting with water.

Solution

H₃PO₄ (l) + H₂O (l) ----> H₂PO₄^- (aq) + H₃O⁺ (aq)

H₂PO₄^- (aq) + H₂O (l) ----> HPO₄^2- (aq) + H₃O⁺ (aq)

HPO₄^2- (aq) + H₂O (l) ----> PO₄^3- (aq) + H₃O⁺ (aq)

An important property of some substances is their ability to function as either acids or bases, depending on the other substance with which they are reacting. Perhaps the most important example of this class of compound is water:

eg. H₂O (l) + H₂S (g) ----> H₃O⁺ (aq) + S^2- (aq)

H₂O (l) + CO₃^2- (aq) ----> OH^-(aq) + HCO₃^- (aq)

Example 3.3

Write relevant equations to show how the hydrogen sulfite ion, HSO₃^- , acts as an amphiprotic substance:

HSO₃^- (aq) + H₂O (l) ----> SO₃^2- (aq) + H₃O⁺ (aq)

HSO₃^- (aq) + H₂O (l) ----> H₂SO₃ (aq) + OH^-(aq)

In summary,

A polyprotic acid is one which is capable of donating more than one proton.

A polyprotic base is one which is capable of accepting more than one proton.

An amphiprotic (amphoteric) substance can act as either an acid or a base.

3.5 K_W and the pH scale

The pH scale was devised to allow a quantitative measure of the relative acidity or basicity of particular solutions. It is important to recognise that pH is a mathematical scale based on logarithms, so that a difference of one unit on the pH scale correlates with a 10 fold change in relative acidity or basicity.

The scale is based on the fact that water undergoes self-ionisation to a very small extent. We have already seen that water is an amphiprotic substance and it has been known for some time that even very pure water conducts electricity to a small extent. These two facts taken in conjunction with each other allow us to see the validity of the reaction below:

H₂O (l) + H₂O (l) Û H₃O⁺ (aq) + OH^-(aq)

It can be experimentally proven that the product of the concentrations of H₃O⁺ and OH- ions in any aqueous solution remains a constant value at a specified temperature. This product is known as the “Ionisation Constant of water”, symbol K_W, where

K_W = [H₃O⁺][ OH^-] = 10^-14 at 25 °C

In neutral solution, [H₃O⁺] = [OH^-] = 10^-7 M at 25 °C

In acidic solutions, [H₃O⁺] > [OH^-]

ie. [H₃O⁺] > 10^-7 M and [OH^-] < 10^-7 M at 25 °C

In basic solutions, [H₃O⁺] < [OH^-]

ie. [H₃O⁺] < 10^-7 M and [OH^-] > 10^-7 M at 25 °C

Note that the Ionisation constant expression forms the basis of the pH scale, as

pH = -log₁₀[H⁺] = -log₁₀[H₃O⁺] , or

[H⁺] = [H₃O⁺] = 10^-pH

ie. a solution of pH <7 is acidic, pH = 7 is neutral and pH >7 is basic.

Example 3.4

Calculate the pH of each of the following solutions:

a) [H₃O⁺] = 0.010 M b) [HCl] = 1.0 x 10^-4 M

c) [H₂SO₄] = 5.0 x 10^-3 M d) [H₃O⁺] = 10^-3.8 M

Solution

a) [H₃O⁺] = 0.010 M = 10^-2 M

As [H₃O⁺] = 10^-pH, pH = 2

b) [HCl] = [H₃O⁺] = 1.0 x 10^-4 M

As [H₃O⁺] = 10^-pH, pH = 4

c) [H₂SO₄] = 5 x 10^-3 M Þ [H₃O⁺] = 2 x 5.0 x 10^-3 M = 1.0 x 10^-2 M

As [H₃O⁺] = 10^-pH, pH = 2

d) [H₃O⁺] = 10^-3.8 M, so pH = 3.8

Example 3.5

Calculate the pH of each of the following solutions:

a) [OH^-] = 10^-6 M b) [KOH] = 1.0 M

c) [Mg(OH)₂] = 0.00050 M

Solution

a) [H₃O⁺][ OH^-] = 10^-14and [OH^-] = 10^-6 M

Þ [H₃O⁺] = = 10^-8 M

Þ pH = 8

b) [H₃O⁺][ OH^-] = 10^-14and [KOH] = [OH^-] = 1.0 M = 10⁰ M

Þ [H₃O⁺] = = 10^-14 M

Þ pH = 14

c) [Mg(OH)₂] = 0.00050 M = 5 x 10^-4 M

Þ [OH^-] = 2 x 5.0 x 10^-4 M = 1.0 x 10^-3 M

[H₃O⁺][ OH^-] = 10^-14and [OH^-] = 1.0 x 10^-3 M

Þ [H₃O⁺] = = 10^-11 M

Þ pH = 11

Example 3.6

25.0 ml of 8.0 M nitric acid is diluted with 175.0 ml of distilled water. What is the pH of the resultant solution?

Solution

n (HNO₃) = n (H₃O⁺) = C.V = 8.0 x = 0.200 mol

Total volume of solution = 25.0 ml + 175.0 ml = 200.0 ml = 0.200 L

Þ [HNO₃] = [H₃O⁺] = = 1.00 M

As [H₃O⁺] = 10^-pH, pH = 0

3.6 Volumetric Analysis

3.6.1 Primary standards

One of the most valuable experimental tools available to the analytical chemist is volumetric analysis: the analysis of amounts of compounds in solution. Many of the experiments you will undertake early in Year 12 chemistry will involve this process.

In essence, volumetric analysis involves the accurate measurement of a known volume of a

standard solution of precisely known concentration, delivered by a pipette. This solution is reacted with the unknown solution delivered by a burette; the endpoint being recognised by a suitable indicator.

Only a relatively small number of chemicals are suitable for use in the production of primary standard solutions. A primary standard must satisfy the following criteria:

i) it must be highly pure and of known chemical formula

ii) it must remain stable over time and not react with moisture or gases from the atmosphere

iii) it should be readily available and relatively inexpensive, and

iv) it should have a relatively high molar mass (to minimise error)

Some examples of acid/base primary standards are anhydrous sodium carbonate (Na₂CO₃), sodium oxalate (Na₂C₂O₄) and potassium hydrogen phthalate (KC₈H₅O₄).

Any solution to be used in volumetric analysis which does not meet the criteria for being classified as a primary standard must first be standardised immediately before use. Solutions of this nature are classified as secondary standards.

3.6.2 The experimental procedure

To produce a primary standard solution, an accurately known mass of reagent is measured and delivered into a volumetric flask (while there are a wide range of sizes available, the most commonly used in this context has a volume of 250.0 ml). A pipette is used to measure a known volume (known as an aliquot) of this solution to the conical flask in readiness for reaction. The pipette must first be prepared by washing with water and then rinsing with the primary standard solution.

A few drops of a suitable indicator is then usually added. An indicator is a substance which changes colour at a particular pH (common examples are litmus, methyl orange and phenolphthalein).

The burette is prepared in a similar manner to the pipette; first washed with water and then rinsed with the unknown solution. This unknown is then slowly added to the flask until a colour change is noted; endpoint has been reached. The volume of the unknown solution delivered is known as the titre volume.

Note that the exact point at which acid and base are neutralised is known as the equivalence point. The endpoint is defined as the point exactly between the colours shown by the indicator in its acid form and its base form. In the case of methyl orange, which changes from yellow in basic solution to red in acid, endpoint occurs at the intermediate ‘orange’ colour. Generally, the equivalence point and the endpoint can be made to approximate each other very closely by carefully choosing an indicator which changes colour at appropriate pH.

Example 3.7

A student is required to accurately determine the concentration of a solution of approximately 0.1 M hydrochloric acid, using the supplied anhydrous sodium carbonate to prepare a standard solution. He weighed 1.311 g of Na₂CO₃ into a 250.0 ml volumetric flask, dissolved it in a little distilled water and then made the solution up to the mark. After suitably preparing both the pipette and burette, he delivered three 20.00 ml aliquots of the solution into conical flasks and added three drops of methyl orange indicator to each flask. He then titrated the Na₂CO₃ against the hydrochloric acid solution, recording an average titre volume of 18.87 ml being required to reach endpoint.

a) Calculate the exact concentration of the HCl solution.

b) Had the student neglected to rinse the pipette with the Na₂CO₃solution, what would have been the effect on the calculated value of the HCl concentration?

c) Had the student neglected to rinse the burette with the HClsolution, what would have been the effect on the calculated value of its concentration?

Solution

a) The equation for the reaction is as follows:

Na₂CO₃(aq) + 2 HCl (aq) ----> 2NaCl (aq) + H₂O (l) + CO₂ (g)

n (Na₂CO₃) = mol

Þ n (Na₂CO₃) in aliquot = x 10^-4 mol

From the equation, n (HCl) = 2 x n (Na₂CO₃) = 1.979 x 10^-3 mol

V (HCl) = 18.87 ml = 0.01887 L

Þ [HCl] =

b) Had the student not rinsed the pipette with the solution there would still be traces of water adhering to the pipette and so the actual volume of Na₂CO₃ delivered would be less than 20.00 ml. As a consequence, a lesser volume of HCl would be required to neutralise it and the concentration of HCl calculated would be greater than its true value, as C=.

c) Had the student not rinsed the burette with HCl as required, it would have been diluted slightly by the traces of water adhering to the inside of the glassware. As a consequence, more HCl would be required to neutralise the aliquot of Na₂CO₃ and the concentration of HCl calculated would be lesser than its true value.

3.6.3 Back titration

Occasionally, direct titration of one reactant against another does not produce satisfactory results. The most common reasons for this problem are a poor colour change occurring when weak acid or base solutions are being titrated, resulting in accurate determination of endpoint not being possible; and a slow rate of reaction making direct titration unfeasible.

In such cases an excess of reactant is deliberately added to ensure complete reaction and then titration is carried out to determine how much excess has been used. Such a procedure is known as a “back titration”.

Example 3.8

A laboratory technician wishes to accurately determine the percentage by mass of available nitrogen (present as the ammonium ion) in a sample of commercially available lawn fertiliser. To achieve this aim, she accurately weighs 1.399 g of fertiliser into a 250.0 ml volumetric flask and makes it up to the mark with distilled water. 20.00 ml aliquots of this fertiliser solution are then reacted with 20.00 ml of 0.1022 M NaOH solution, according to the ionic equation

NH₄⁺ (aq) + OH^- (aq) ----> NH₃ (g) + H₂O (l)

To ensure the reaction proceeds to completion the solution is boiled for approximately 10 minutes until no further evolution of ammonia can be measured. The excess OH^- is then titrated against 0.108 M HCl solution, the average titre required being 11.26 ml. The relevant equation for this reaction is:

HCl (aq) + NaOH (aq) ----> NaCl (aq) + H₂O (l)

Calculate the percentage by mass of nitrogen in the fertiliser.

Solution

Step 1. As excess NaOH was added to the sample of fertiliser in the 20.00 ml aliquot we must first determine how much NaOH was actually consumed in the reaction with the NH₄⁺ contained in the fertiliser aliquot. This is achieved by determining the difference between the amount of NaOH present initially and that in excess after the reaction has proceeded.

n (NaOH) present initially = C.V.

= 2.044 x 10^-3 mol

The excess NaOH is reacted with 0.108 M HCl, the titre volume required being 11.26 ml.

n (NaOH) in excess = n (HCl) in titre volume = C.V.

= 1.216 x 10^-3 mol

Þ n(NaOH) consumed in reaction = 2.044 x 10^-3 - 1.216 x 10^-3

= 8.28 x 10^-4 mol

Step 2. Presuming that the ammonium ion is the only acid present, the no. of mol of NH₄⁺ present in the aliquot of fertiliser is the same as the no. of mole of NaOH consumed, according to the equation

NH₄⁺ (aq) + OH^- (aq) ----> NH₃ (g) + H₂O (l)

Þ n(NH₄⁺) in aliquot of fertiliser solution = 8.28 x 10^-4 mol

Step 3. A 20.00 ml aliquot of the fertiliser solution was pipetted from the 250.0 ml volumetric flask, so the no. of mol of NH₄⁺ in the original sample is given by

n(NH₄⁺) in volumetric flask = n(NH₄⁺) in original fertiliser sample

= 8.28 x 10^-4 mol

= 1.035 x 10^-2 mol

Step 4. We can now determine the mass of NH₄⁺ in the fertiliser sample and consequently the percentage by mass of nitrogen in the sample:

mass of NH₄⁺ = n x M

= 1.035 x 10^-2 x 18

= 0.1863 g

% mass of nitrogen =

= 13.32%

3.7 Summary/ Objectives

At the end of this Chapter you should

- recall the definition of the terms acid (as a proton donor) and base (as proton acceptor)

- recognise that acids and bases always function as conjugate pairs

- understand the difference between the terms ‘ionisation’ (as applies to a covalently bonded molecule) and dissociation (as it applies to an ionically bonded compound)

- recall that the strength of an acid or base is dependent on the extent of the reaction whereby it donates or accepts protons

- memorise two examples each of strong and weak acids and bases

- be able to define the terms ‘polyprotic’ and ‘amphiprotic’ and recall at least two examples of each

- understand the basis of the pH scale as being the equilibrium that exists between water and the products of its ionisation, H₃O⁺ and OH^-

- recall the definition of pH as

pH = -log₁₀[H₃O⁺]

- be able to apply the rule above to calculate the pH of acidic or basic solutions

- given the pH of a solution, be able to determine the concentration of H₃O⁺ and OH^- ions in solution

- recognise the terms volumetric analysis, primary and secondary standard, aliquot, titre, endpoint, equivalence point, indicator

- recall the criteria which must be met by a primary standard to be used in a volumetric analysis. Memorise two examples of primary standards

- understand the role of indicators as weak acids or bases which change colour at a specific pH

- be familiar with the ways in which glassware, such as standard flasks, pipettes, burettes, volumetric flasks; must be prepared prior to the experimental performance of a volumetric analysis

- be able to deduce the effect, if any, of not preparing the above-mentioned glassware correctly before the volumetric analysis is performed

- be able to perform calculations involving volumetric analyses from relevant data provided, including both direct and back titrations

- recognise possible sources of experimental error arising from volumetric analyses and how these errors may alter the calculated result.

Chapter 3 Questions

Q1. Define the following terms:

a) acid b) base

c) polyprotic d) amphiprotic (amphoteric)

e) pH f) indicator

g) volumetric analysis h) primary standard

i) secondary standard j) equivalence point

k) endpoint l) titration

m) titre n) aliquot

o) ionisation p) dissociation

Q2. Write the conjugate base of the following acids:

a) HCl b) HNO₃

c) H₂SO₄ d) NH₄⁺

e) H₂O f) HOCl

g) CH₃COOH h) H₂PO₄^-

i) HSO₃^- j) HCN

Q3. Write the conjugate acid of the following bases:

a) OH^- b) NH₃

c) Cl^- d) H₂O

e) HPO₄^2- f) PO₄^3-

g) HS^- h) CO₃^2-

i) O^2- j) NH₂^-

Q4. For each of the following reactions, state whether the first named species is acting as an acid, a base, or neither:

a) H₂SO₄ (aq) + 2NaOH (aq) ----> Na₂SO₄ (aq) + 2H₂O (l)

b) BaCl₂ (aq) + 2KOH (aq) ----> Ba(OH)₂ (s) + 2KCl (aq)

c) 3NH₃ (aq) + H₃PO₄ (aq) ----> (NH₄)₃PO₄ (aq)

d) 2HCl (aq) + Zn (s) ----> ZnCl₂ (aq) + H₂ (g)

e) Ca(OH)₂ (aq) + CO₂ (g) ----> Ca(HCO₃)₂ (aq)

Q5. Write equations to show the successive ionisations of the triprotic acid H₃PO₄ reacting with water.

Q6. State whether each of the following species undergoes ionisation or dissociation when added to water:

a) Mg(OH)₂ (s) b) KI (s)

c) HCl (g) d) H₂O (l)

e) CuSO₄ (s) f) HNO₃ (l)

g) NaOH (s) h) Na₂Cr₂O₇ (s)

i) CH₃COOH (l) j) (NH₄)₂CO₃ (s)

Q7. 250.0 ml of 0.0400 M limewater solution is prepared and through it is bubbled 75 cm^-3 of CO₂ (g) at S.T.P. Calculate the mass of calcium carbonate formed, according to the equation

Ca(OH)₂ (aq) + CO₂ (g) ----> CaCO₃ (s) + H₂O (l)

Q8. 120.0 ml of 0.566 M nitric acid solution is exactly neutralised by 85.5 ml potassium hydroxide solution. What is the concentration of the KOH solution?

Q9. What volume of 2.088 M sulfuric acid is required to exactly neutralise a solution made up of 100.0 g of sodium hydroxide pellets dissolved in 320.0 ml of water?

Q10. If V ml of 0.20 M NaOH is mixed with 2V ml of 0.80 M NaOH, what volume of 0.45 M H₂SO₄ would be required to effect neutralisation?

Q11. What volume of water would need to be added to 100.0ml of 0.354 M NaOH to make the resulting solution 0.250 M NaOH?

Q12. What volume of 0.0641 M HClO₄ solution would be required to make up 200.0 ml of 0.0250 M solution?

Q13. A laboratory technician wishes to make up 5.00 L of 0.450 M nitric acid solution from a stock bottle of 14 M acid. What volume of the concentrated acid will he require? What safety precautions should he adopt when diluting the concentrated acid?

Q14. Calculate the pH of each of the following solutions:

a) 0.010 M HNO₃ b) 0.10 M HCl

c) 1 x 10^-4M HNO₃ d) 0.0050 M H₂SO₄

e) [H⁺] = 1 x 10^-13M f) [H₂SO₄] = 5 x 10^-6M

Q15. Calculate the pH of the solution which results from bubbling 2.45 L of hydrogen chloride gas at S.L.C. through 100 ml of water.

Q16. Calculate the pH of the solution which results from the dilution of 80.0 ml of 14 M HNO₃ with 1040 ml of distilled water.

Q17. 5.89g of potassium hydroxide pellets are added to 28.00 ml of 0.625 M sulfuric acid solution and the mixture is thoroughly stirred. Is the resultant solution acidic, basic or neutral?

Q18. Calculate the pH of each of the following solutions:

a) [OH^-] = 0.10 M b) [NaOH] = 1 x 10^-5M

c) [KOH] = 1.0 M d) [Mg(OH)₂] = 5 x 10^-4M

e) [OH^-] = 10^-8.3M f) [LiOH] = 10 M

Q19. 2.00g of solid NaOH is carefully dissolved in 500 ml of water. Calculate the pH of the resultant solution.

Q20. Why can solid NaOH not be used to prepare a primary standard solution?

Q21. In order to accurately determine the concentration of a solution of potassium hydroxide, a student intends to titrate it against 20.00 ml aliquots of a standard solution of hydrochloric acid. Which one, or more, of the following procedures, is incorrect?

a) the pipette is first washed with water and then rinsed with the KOH solution.

b) the pipette is first washed with water and then rinsed with the HCl solution.

c) the burette is washed with water and then filled with the HCl solution.

Q22. A commonly used primary standard acid in analytical chemistry is potassium hydrogen phthalate (KC₈H₅O₄) - a monoprotic acid of molar mass 204.1 g.mol^-1. In a titration exercise it is found that a sample of 1.305g of potassium hydrogen phthalate requires 17.28 ml of sodium hydroxide solution to neutralise it. What is the molarity of the NaOH solution?

Q23. A 40.00 ml sample of vinegar containing acetic acid was diluted to 250.0ml in a standard flask. A 20.00 ml aliquot of this solution required 27.35 ml of 0.0942 M NaOH solution to reach endpoint. Calculate the percentage by mass of pure acetic acid in the original vinegar sample. [Note: presume d (vinegar) = 1.00 g.cm^-3]

Q24. A student carries out a volumetric analysis to determine the concentration of hydrochloric acid in a sample of “spirits of salts” (a cleaning agent used to remove cement from bricks). She carefully pours about 5 ml of the liquid into a 250.0 ml standard flask and determines its mass to be 5.117g. Distilled water is slowly added to the flask and the solution is agitated to ensure uniform mixing. 25.00 ml aliquots of this solution are then pipetted into three separate volumetric flasks and 2-3 drops of methyl orange indicator is added to each. Titration against a standard solution of 0.117 M Na₂CO₃ solution is carried out; the average titre volume required to reach endpoint being 17.83 ml. Calculate the concentration of HCl in the “spirits of salts” in

(a) g.L^-1 (b) Molar concentration

[Note: presume d (spirits of salts) = 1.00 g.cm^-3]

Q25. A student wishes to determine the percentage by mass of available nitrogen (present as the ammonium ion) in a sample of commercially available lawn fertiliser. To achieve this aim, he accurately weighs 1.556 g of fertiliser into a 250.0 ml standard flask and makes it up to the mark with distilled water. 20.00 ml aliquots of this fertiliser solution are then reacted with 25.00 ml of 0.1061 M NaOH solution, according to the ionic equation

NH₄⁺ (aq) + OH^- (aq) ----> NH₃ (g) + H₂O (l)

The excess OH^- is then titrated against 0.119 M HCl solution, the average titre required being 12.07 ml.

Calculate the percentage by mass of nitrogen in the fertiliser.

Solutions to Chapter 3 questions

Q2. a) Cl^- b) NO₃^- c) HSO₄^- d) NH₃ e) OH^- f) OCl^-

g) CH₃COO^- h) HPO₄^2- i) SO₃^2- j) CN^-

Q3. a) H₂O b) NH₄⁺ c) HCl d) H₃O⁺ e) H₂PO₄^- f) HPO₄^2-

g) H₂S h) HCO₃^- i) OH^- j) NH₃

Q4. a) acid b) neither c) base d) neither e) base

Q5. H₃PO₄ (aq) + H₂O (l) ----> H₂PO₄^- (aq) + H₃O⁺ (aq)

H₂PO₄^- (aq) + H₂O (l) ----> HPO₄^2- (aq) + H₃O⁺ (aq)

HPO₄^2- (aq) + H₂O (l) ----> PO₄^3- (aq) + H₃O⁺ (aq)

Q6. a) dissociation b) dissociation c) ionisation d) ionisation

e) dissociation f) ionisation g) dissociation h) dissociation

i) ionisation j) dissociation

Q7. 0.335 g Q8. 0.794 M Q9. 0.59 L Q10. 2 V ml

Q11. 41.6 ml Q12. 78.00 ml Q13. 161 ml Q14. a) 2 b) 1

c) 4 d) 2 e) 13 f) 5 Q15. 0 Q16. 0 Q17. alkaline

Q18. a) 13 b) 9 c) 14 d) 11 e) 5.7 f) 15 Q19. 13

Q20. NaOH is hydroscopic and reacts with CO₂ from atmosphere

Q21. (b) is correct Q22. 0.370 M Q23. 4.83 % Q24. a) 297 g.L^-1

b) 8.15 M Q25. 17.58 %

The Reactivity of Metals

When an iron nail is left in copper sulfate solution a reaction occurs, the iron removes the copper from solution. This is a displacement reaction.

The iron is more reactive than the copper. We know from experience that different metals have varying abilities to react with other substances.

Eg.

Metals can be put in a reactivity list which is called the reactivity series of metals or the activity series. Table 18.1

Reactivity of metals and the periodic table.

The metals in groups I and II are very reactive. Metals tend to act as reductants. The reactivity of metals tends to decrease across the periodic table and down the group.

You can use the reactivity series to predict redox reactions.

ELECTROCHEMICAL CELLS

A simple electrochemical cell consists of:

Two half-cells, containing two electrodes (anode and cathode) and two electrolytes
A conducting wire
A salt bridge, containing another electrolyte.

In half-cells the more reactive metal will always give its electrons to the ions of the less reactive metal. The ions of the less reactive metal will gain electrons from the other.

THE ELECTRODE AT WHICH OXIDATION OCCURS IS CALLED THE ANODE.

THE ELECTRODE AT WHICH REDUCTION OCCURS IS CALLED THE CATHODE.

Competition for Electrons.

If there was competition between two ions for electrons then the ions lower on the reactivity series will more readily accept the electrons.

Electrons flow to the half-cell that most readily accepts electrons.

The ELECTROCHEMICAL SERIES OF METALS

The relative abilities of all half-cells to attract electrons can be ranked in an ELECTROCHEMICAL SERIES OF METALS as shown in table 18.2.

In an electrochemical series the half-equations are written in the direction in which reduction occurs. Strongest oxidant at the top left and strongest reductant bottom right.

IN GENERAL THE STRONGER REDUCTANT WILL ALWAYS REACT WITH THE STRONGER OXIDANT.

Corrosion

Corrosion is the oxidation of a metal that causes the metal to be damaged as a result of chemical change.

Two types of corrosion:

Dry corrosion occurs when a metal reacts directly with oxygen. Eg burning magnesium ribbon in oxygen.
Wet corrosion occurs when a metal reacts directly with oxygen in the presence of water.

The corrosion of iron is commonly known as rusting.

Rusting starts at a ‘stress ‘site.

Fe (s) Fe²⁺ + 2e^-

The oxygen accepts electrons and is reduced.

O₂(g) + 2H₂0 + 4e^- 4OH^- (aq)

The overall equation for the reaction is

2Fe(s) + O₂(g) + 2H₂O(l) 2Fe(OH)₂(s)

4Fe(OH)₂(s) + O₂(g) + 2H₂O(l) 4Fe(OH)₃(s)

Effects of corrosion.

Loss of strength
Cannot conduct
Corroded pipes leak
Products of corrosion flake off and cause blockages
Corroded metals expand may cause nuts and bolts to jam machinery.

Three main methods that we can use to protect iron and steel from corrosion.

· Surface protection –paint, plastic, grease and metal coatings.

· Alloying

· Electrochemical protection which involves placing a more reactive metal in electrical contact with the metal that needs protection.

Reactions of Alkanes

· Since Alkanes are Saturated Hydrocarbons, they do not readily react. However, they can be made to react under certain conditions, so giveuseful products or energy output.

· Alkanes will react with Oxygen if they are given sufficient Activation Energy. This will result in a highly Exothermic reaction, producing Carbon Dioxide and Water, which makes Alkanes very useful as fuels.

· When Alkanes C₄-C₆ are heated to 150°C with a Platinum of Aluminium Oxide Catalyst, Isomerism occurs. The reaction produces BranchedAlkanes, and so is useful in improving Octane Number.

· Reforming is the process of increasing the amount of Cycloalkanes and Hydrocarbons containing Benzene Rings, to improve Octane Number. This is done with Alkanes in the Naphtha Fraction (C₆-C₁₀) at 500°C with a Platinum or Aluminium Oxide Catalyst. Hydrogen is recycled through the mixture to reduce 'coking'.

· Cracking is the process of breaking longer chained Alkanes down into smaller Alkanes and Alkenes, sometimes for Polymer manufacture, and sometimes to improve Octane Number.

· Steak Cracking involves heating Alkanes from the Naphtha and Kerosene Fractions (C₆-C₁₆) to 900°C without a catalyst and using Steam as adiluent to reduce 'coking'. This is used in the manufacture of Polymers.

· Catalytic Cracking takes feedstock from longer chained Alkanes in the Gas Oil Fraction (C₁₄-C₂₀) and heating to 500°C with a Zeolite. This can produce Branched and Cyclic Hydrocarbons and is used to improve Octane Number.

Alkenes

· Alkenes are a Homologous Series composed of unsaturatedHydrocarbons, since they contain at least one C=C Double Bond. They have a general formula C_nH_2n. Since they are unsaturated, they are quite reactive.

· Alkenes are named with an 'ene' suffix. Often, in naming Alkenes with more than three Carbons, the position of the Double Bond is identified. For example, Penta-2-ene means the double bond starts on Carbon 2.

· If an Alkenes has more than one Double Bond, this is shown in the name, and the position of all the double bonds are identified. For example, Hex-1,2,4-triene represents a 6 Carbon Alkene with Double Bonds on Carbons 1, 2 and 4.

· Like Alkanes, Alkenes can form Branched Structures, and Cyclic Structures, called Cycloalkenes.

Electrophilic Addition Mechanism

· The Double Bond in and Alkene is a region of high density negative charge. Positive Ions or molecules with a partially positively charged region may be attracted to this Double Bond, and act as Electrophiles, accepting a pair of Electrons from it.

· The Electrophonic Addition Mechanism describes the way in which some molecules are thought to react with the Double Bond in an Alkene. One example of such a reaction is the reaction with Bromine, which is a testfor Unsaturation.

· The Bromine molecule is thought to become Polarised by the negative charge of the Double Bond, so that electrons are pushed to the other end of the Bromine molecule. This means that the Bromine atom nearest the Double Bond becomes slightly positive.

· This Bromine atom then reacts with the Double Bond like an Electrophile, accepting a pair of electrons and bonding to one of the Carbon atoms. This leaves a negatively charged Bromine Ion, and a positively charged Carbon atom in the Alkene, called a Carbocation.

· The Bromine Ion reacts with the Carbocation so that it bonds with it. This mechanism produces Dibromoethene.

· This is the mechanism for numerous reactions with Alkenes. Water reacts with Alkenes, slitting into OH- and H⁺ Ions, forming an Alcohol when bonded to the Alkene. Hydrogen will react like this as well. This process is known as Hydrogenation, and occurs at 150°C, 5atm, with a Platinum orNickel Catalyst.

· If an Alkene is reacted with Bromine in the presence of Chlorine atoms, some of the products will contain a Chlorine atom instead of one of the Bromine atoms. This gives evidence in support of the mechanism for Electrophilic Addition.

Bond Enthalpies

· In a compound, atoms are bonded together by electromagnetic forces. Each bond between atoms has a specific bond energy - the energy required to break it - called the Bond Dissociation Enthalpy. For example, a C=O bond has a bond energy of 805KJmol^-1. Most bond enthalpies are average enthalpies, the arithmetic mean of the energy required to break the bond in all (most) compounds.

· The greater the energy of the bond, the shorter the bond length. Between two atoms, there are attractive forces between the positive nuclei and the negative electrons, but there is also a repulsive force between the two positive nuclei, which increases as the atoms get closer together. So it can be seen that there is an equilibrium point reached whereby the distance is such that the energy is greatest and the forces of attraction and repulsion are equal.

· The change in energy of a reaction can be calculated by adding the bond enthalpies of all the bonds in the reactants and subtracting the bond enthalpies of all the products. For example, the reaction N₂H₄ + O₂ → N₂ + 2H₂O has an energy change of -581.1 KJmol^-1. The calculated value may differ from the ΔH_r^O because some compounds are not in their standard states and average bond energies are used, rather than ones specific to the compounds in the reaction.

Universal Learning House

Electro valencies and Chemical formula

Corrosion

O₂(g) + 2H₂0 + 4e^- 4OH^- (aq)

Corrosion

O₂(g) + 2H₂0 + 4e^- 4OH^- (aq)

Reactions of Alkanes

Alkenes

Electrophilic Addition Mechanism

Bond Enthalpies

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Bond Enthalpies

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