Chemistry is concerned with the properties and the interchange of matter by reaction i.e. structure and change. In order to do this, we need to be able to talk about numbers of atoms. The key concept is the mole and the relationship between the mole and the mass of the atom. Each element has a distinct atomic mass based on the natural abundances of the various isotopes present. Atoms combine to form molecules in fixed proportions which are usually small integers for simple molecular or ionic compounds The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit
The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH 4 SF 6 NaCl Na 2 S 2 O 3 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH 4 12.0115 + 4 x 1.0079 SF 6 NaCl Na 2 S 2 O 3 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH 4 12.0115 + 4 x 1.0079 SF 6 32.066 + 6 x 18.9984 NaCl Na 2 S 2 O 3
The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH 4 12.0115 + 4 x 1.0079 SF 6 32.066 + 6 x 18.9984 NaCl 22.9898 + 35.453 Na 2 S 2 O 3 2 x 22.9898 + 2 x 32.066 + 3 x 15.9994 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH 4 12.0115 + 4 x 1.0079 SF 6 32.066 + 6 x 18.9984 NaCl 22.9898 + 35.453 Na 2 S 2 O 3 2 x 22.9898 + 2 x 32.066 + 3 x 15.9994 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH 4 12.0115 + 4 x 1.0079 SF 6 32.066 + 6 x 18.9984 NaCl 22.9898 + 35.453 Na 2 S 2 O 3 2 x 22.9898 + 2 x 32.066 + 3 x 15.9994
The mole and atomic mass The mole is defined as the number of elementary entities as are present in 12.00 g of 12 C. Numerically, this is equal to Avogadro s Number 6.022 x 10 23 Therefore, in 12.00 g of 12 C there are 6.022 x 10 23 elementary entities, in this case atoms. The mole and atomic mass Atomic masses, in atomic units, u, are defined relative to 12 C. Therefore, The formula mass of an element or compound contains 1 mole, 6.022 x 10 23, of particles Examples How many particles are there in 5 g of Na?
Examples How many particles are there in 5 g of Na? The particles are atoms how many atoms are there in 5 g of Na? Examples How many particles are there in 5 g of Na? The particles are atoms how many atoms are there in 5 g of Na? Atomic mass of Na = 22.9898 u Examples How many particles are there in 5 g of Na? The particles are atoms how many atoms are there in 5 g of Na Atomic mass of Na = 22.9898 u As 1 u = 1 / 12 x mass ( 12 C) And 1 mole = 6.022 x 10 23 particles = number of particles in 12 g 12 C
Examples How many particles are there in 5 g of Na? The particles are atoms how many atoms are there in 5 g of Na Atomic mass of Na = 22.9898 u Mass of 1 mole of Na = 22.9898 g Examples How many particles are there in 5.0000 g of Na? 22.9898 g Na = 1 mole Na Then 1 g Na = 1 mol Na 22.9898 5 x 1 g Na = 5 x 1 mol Na 22.9898 5 g Na = 0.2175 mol Na 5 g Na = 0.2175 x (6.022 x 10 23 ) particles Na Examples How many particles are there in 5.0000 g of Na? 1.310 x 10 23 atoms
Examples What is the mass of 0.23 mol of butane? Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C 4 H 10 Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C 4 H 10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g
Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C 4 H 10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g Molecular mass of C 4 H 10 = (4x12.011)+(10x1.0079)u = 58.123 u Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C 4 H 10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g Molecular mass of C 4 H 10 = (4x12.011)+(10x1.0079)u = 58.123 u Relative Molecular Mass of Butane = 58.123 g Examples What is the mass of 0.23 mol of butane? Relative Molecular Mass of Butane = 58.123 g 1 mole of butane = 58.123 g 0.23 x 1 mole of butane = 0.23 x 58.123 g
Examples What is the mass of 0.23 mol of butane? Relative Molecular Mass of Butane = 58.123 g 1 mole of butane = 58.123 g 0.23 x 1 mole of butane = 0.23 x 58.123 g 0.23 mole of butane = 13.368 g Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Sr (s) + Cl 2(g) SrCl 2(s) Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Sr (s) + Cl 2(g) SrCl 2(s) Physical state
Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Sr (s) + Cl 2(g) SrCl 2(s) Reactants Physical state Product Chemical Equations As matter cannot be created or destroyed in a chemical reaction, the total number of atoms on one side must be equal to the total number of atoms on the other. Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water
Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water Reactants: Cyclohexane, C 6 H 12 Oxygen, O 2 Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water Reactants: Cyclohexane, C 6 H 12 Oxygen, O 2 Products: Carbon Dioxide, CO 2 Water, H 2 O Chemical Equations Example Initially, we can write the reaction as C 6 H 12 + O 2 CO 2 + H 2 O
Chemical Equations Example Initially, we can write the reaction as C 6 H 12 + O 2 CO 2 + H 2 O This is NOT a correct equation there are unequal numbers of atoms on both sides Chemical Equations Example Initially, we can write the reaction as C 6 H 12 + O 2 CO 2 + H 2 O This is NOT a correct equation there are unequal numbers of atoms on both sides Reactants: 6 C, 12 H, 2 O Products: 1 C, 2 H, 3 O Balancing the equation C 6 H 12 + O 2 CO 2 + H 2 O
Balancing the equation C 6 H 12 + O 2 CO 2 + H 2 O 6 C, 12 H, 2 O 1 C, 2 H, 3 O 6 C on LHS means there must be 6 C on the RHS C 6 H 12 + O 2 6CO 2 + H 2 O 6 C, 12 H, 2 O 6 C, 2 H, 13 O 13 O on RHS means there must be 13 O on LHS C 6 H 12 + 13 / 2 O 2 6CO 2 + H 2 O 6 C, 12 H, 13 O 6 C, 2 H, 13 O Balancing the equation C 6 H 12 + 13 / 2 O 2 6CO 2 + H 2 O 6 C, 12 H, 13 O 6 C, 2 H, 13 O 12 H on RHS means there must be 12 H on LHS C 6 H 12 + 13 / 2 O 2 6CO 2 + 6H 2 O 6 C, 12 H, 13 O 6 C, 12 H, 18 O 18 O on RHS means there must be 18 H on LHS C 6 H 12 +9O 2 6CO 2 +6H 2 O 6 C, 12 H, 18 O 6 C, 12 H, 18 O The final balanced equation is C 6 H 12 +9O 2 6CO 2 +6H 2 O and the coefficients are known as the stoichiometric coefficients. These coefficients give the molar ratios for reactants and products This is a stoichiometric reaction one where exactly the correct number of atoms is present in the reaction
If cyclohexane were burnt in an excess of oxygen, the quantity of oxygen used would be the same although O 2 would be left over. Solutions and concentration If cyclohexane were burnt in an excess of oxygen, the quantity of oxygen used would be the same although O 2 would be left over. Solutions A solution is a homogenous mixture which is composed of two or more components the solvent - the majority component and one or more solutes - the minority components
Solutions Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent. Some are solids where both the solvent and the solute are solids. Brass is an example Solutions Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent. Some are solids where both the solvent and the solute are solids. Brass is an example Solutions Most common solutions are liquids where a solid, liquid or gas (the solute) is dissolved in the liquid solvent. Some are solids where both the solvent and the solute are solids. Brass NaCl is (s) an melts example Here copper is the solvent, zinc the solute. Cu Zn
Solutions Gas-Solid solution: Hydrogen in palladium Steel Solutions Common laboratory solvents are usually organic liquids such as acetone, hexane, benzene or ether or water. Water is the most important solvent. The oceans cover ~ ¾ of the surface of the planet and every cell is mainly composed of water. Solutions in water are termed aqueous solutions and species are written as E (aq). Solutions Aqueous Solutions Water is one of the best solvents as it can dissolve many molecular and ionic substances. The properties of solutions which contain molecular and ionic solutes are very different and give insight into the nature of these substances and solutions.
Solutions Ionic Solutions An ionic substance, such as NaClO 4, contain ions in this case Na + and ClO 4-. The solid is held together through electrostatic forces between the ions. In water, the solid dissolves and the particles move away from each other and diffuse through the solvent. This process is termed Ionic Dissociation Solutions Ionic Solutions In an ionic solution, there are therefore charged particles the ions and as the compound is electrically neutral, then the solution is neutral. When a voltage is applied to the solution, the ions can move and a current flows through the solution. The ions are called charge carriers and whenever electricity is conducted, charge carriers are present. Solutions Molecular Solutions A molecular solution does not conduct electricity as there are no charge carriers present. The bonding in a molecule is covalent and involves the sharing of atoms and there is no charge separation.
Solutions Electrolytes A solute that, when dissolved, produces a solution that conducts is termed an electrolyte, which may be strong or weak. A strong electrolyte is one which is fully dissociated in solution into ions A weak electrolyte is one which is only partially dissociated. Solutions Moles and solutions When a substance is dissolved in a solvent, we relate the quantity of material dissolved to the volume of the solution through the concentration of the solution. The concentration is simply the number of moles of the material per unit volume: C = n V n = number of moles; V = volume of solvent Solutions Moles and solutions The units of concentration are: C = n = moles V L 3 and we define a molar solution as one which has 1 mole per liter. Alternatively, Concentration = Molarity = number of moles volume of solution
Solutions Example 4 g of Na 2 SO 4(s) is dissolved in 500 ml of water. What is the concentration of the solution? Solutions Example 4 g of Na 2 SO 4(s) is dissolved in 500 ml of water. What is the concentration of the solution? Formula mass of Na 2 SO 4(s) : Molar Atomic Mass of Na: 22.9898 gmol -1 Molar Atomic Mass of S: 32.064 gmol -1 Molar Atomic Mass of O: 15.9994 gmol -1 Solutions Example 4 g of Na 2 SO 4(s) is dissolved in 500 ml of water. What is the concentration of the solution? Formula mass of Na 2 SO 4(s) : (2 x 22.9898)+ 32.064+(4x15.9994)=142.041gmol -1 1 mole of Na 2 SO 4(s) = 142.041g 1 / 142.041 mole of Na 2 SO 4(s) = 1 g
Solutions Example 4 g of Na 2 SO 4(s) is dissolved in 500 ml of water. What is the concentration of the solution? 1 / 142.041 mole of Na 2 SO 4(s) = 1 g Therefore 4 g of Na 2 SO 4(s) = 4 / 142.041 mole = 2.82 x 10-2 mole Solutions Example 4 g of Na 2 SO 4(s) is dissolved in 500 ml of water. What is the concentration of the solution? 2.82 x 10-2 mole is therefore dissolved in 500 ml of water; So in 1 L, there are 2 x 2.82 x 10-2 mole Molarity of solution = 5.64 x 10-2 moll -1 Solutions Example The equation for the dissolution of Na 2 SO 4(s) is Na 2 SO 4(s) 2Na + 2- (aq) + SO 4 (aq) H 2 O So if we have 5.64 x 10-2 moll -1 Na 2 SO 4(s), we must have 1.13 x 10-1 moles Na + (aq) and 5.64 x 10-2 mol SO 2-4 (aq) as there are 2 Na cations for every sulfate ion
Solutions If we change the volume of the solution then we change the concentration: If the Na 2 SO 4 solution is diluted with 500ml of water, the concentration or molarity would be halved: 2.82 x 10-2 mole is therefore dissolved in 1000 ml of water Molarity = 2.82 x 10-2 moll -1 Solutions Dissolution on an atomic level. Solids are held together by very strong forces. NaCl (s) melts at 801 o C and boils at 1465 o C but it dissolves in water at room temperature. Solutions Dissolution on an atomic level. When we dissolve NaCl (s) in water we break the bonds between ions but make bonds between the ions and the water
Solutions Dissolution on an atomic level. When we dissolve NaCl (s) in water we break the bonds between ions but make bonds between the ions and the water The ions are hydrated or solvated in solution and these bonds between solvent and solute make the dissolution energetically possible If something does not dissolve then the energetics are wrong for it do do so. Solutions Solubility rules All ammonium and Group I salts are soluble. All Halides are soluble except those of silver, lead and mercury (I) All Sulfates are soluble except those of barium and lead. All nitrates are soluble. Everything else is insoluble The Exam
Solutions Solutions are homogenous mixtures in which the majority component is the solvent and the minority component is the solute Solutions are normally liquid but solutions of gases in solids and solids in solids are known. Ionic compounds dissolve in water to give conducting solutions they are electrolytes Solutions Electrolytes are either strong or weak depending on the degree of dissociation in solution Molecular solutions do not conduct as molecules do not dissociate in solution The concentration or molarity of a solution is defined by C = n = moles V L 3 and the units are moll -1 or moldm -3 Solutions When ionic substances dissolve, bonds between particles in the solid break and bonds between the solvent and the ions are made There are general rules for the solubilities of ionic compounds
Reactions in Solution Reactions in solution include Acid base reactions Precipitation reactions Oxidation- reduction reactions Reactions in Solution Reactions and equilibria Reactions are often written as proceeding in one direction only with an arrow to show the direction of the chemical change, reactants to products. Not all reactions behave in this manner and not all reactions proceed to completion. Even those that do are dynamic. Reactions in Solution: Acid - Base A saturated solution of NaI is placed in contact with Na 131 I (s), which is radioactive. Na + I - I - Na + I - I - Na + Na + I - I - I - I - I - Na + Na + Na + Na + Na + I - Na + I - I - Na + Na + Na + Na + I - I - I - I - Na + I - Na + Na + NaI (aq) NaI * (s)
Reactions in Solution: Acid - Base I - Na + I - I - Na + I - I - Na + Na + I - I - I - I - Na + Na + Na + Na + Na + I - Na + I - I - Na + Na + Na + Na + I - I - I - I - Na + I - Na + Na + NaI (aq) NaI * (s) A saturated solution of NaI is placed in contact with Na 131 I (s), which is radioactive. After time, the activity in the solution is measured and... Reactions in Solution: Acid - Base Radioactivity is found in the solution, even though the concentration of I - (aq) has not changed. I - Na + I - I - Na + I - I - Na + Na + I - I - I - I - Na + Na + Na + Na + Na + I - Na + I - I - Na + Na + Na + Na + I - I - I - I - Na + I - Na + Na + I - Na + I - I - Na + I - I - Na + Na + I - I - I - I - Na + Na + Na + Na + Na + I - Na + I - I - Na + Na + Na + Na + I - I - I - I - Na + I - Na + Na + Reactions in Solution: Acid - Base The equilibrium here is composed of two reactions: Na 131 I (s) H 2 O Na + (aq) + 131 I - (aq) Na + (aq) + I - (aq) H 2 O NaI (s) So we write NaI (s) H 2 O Na + (aq) + I - (aq)
Reactions in Solution: Acid - Base Such reactions are termed equilibria and all chemical reactions are equilibria. The symbol for an equilibrium is a double-headed arrow Forward reaction Reactions in Solution: Acid - Base Such reactions are termed equilibria and all chemical reactions are equilibria. The symbol for an equilibrium is a double-headed arrow Forward reaction + Reactions in Solution: Acid - Base Such reactions are termed equilibria and all chemical reactions are equilibria. The symbol for an equilibrium is a double-headed arrow Forward reaction + Reverse reaction
Reactions in Solution: Acid - Base Such reactions are termed equilibria and all chemical reactions are equilibria. The symbol for an equilibrium is a double-headed arrow Forward reaction + = Reverse reaction Reactions in Solution: Acid - Base Equilibria are important in the chemistry of acids and bases Strong acids and bases are completely ionized But... Weak acids and bases are not. Reactions in Solution: Acid - Base The Arrhenius definition of acid and bases are: an acid is a compound which dissolves in water or reacts with water to give hydronium ions, H 3 O + (aq) a base is a compound which dissolves in water or reacts with water to give hydroxide ions, OH - (aq) Svante Arrhenius (1859 1927)
Reactions in Solution: Acid - Base A strong acid is a compound which dissolves and dissociates completely in water or reacts with water to give hydronium ions, H 3 O + (aq) HCl (g) H 3 O + (aq) + Cl - (aq) H 2 O - the double arrow implies that the reaction can go both ways it is an equilibrium. As a strong acid, the reaction is completely on the RHS: HCl (g) H 2 O H 3 O + (aq) + Cl - (aq) Reactions in Solution: Acid - Base A strong base is a compound which dissolves and dissociates completely in water or reacts with water to give hydroxide ions, OH - (aq) NaOH (s) H 2 O Na + (aq) + OH - (aq) Again, we could write this reaction as an equilibrium with a double headed arrow, but the base is strong and the reaction is completely over to the right hand side. Reactions in Solution: Acid - Base In a reaction such as MeCO 2 H H 3 O + - (aq) + MeCO 2 (aq) H 2 O we write the reaction as going from LHS to RHS. Chemical reactions run both ways, so in this reaction, there are two reactions present: Ionization MeCO 2 H H 3 O + - (aq) + MeCO 2 (aq) H 2 O Recombination H 3 O + (aq) + MeCO 2 - (aq) H 2 O MeCO 2 H
Reactions in Solution: Acid - Base We write the reaction for acetic acid, MeCO 2 H, as an equilibrium to include the ionization and recombination. Ionization MeCO 2 H H 2 O H 3 O + (aq) + MeCO 2 - (aq) Recombination H 3 O + (aq) + MeCO 2 - (aq) H 2 O MeCO 2 H MeCO 2 H H 2 O H 3 O + (aq) + MeCO 2 - (aq) As the amount of ionization and recombination are the same, the concentrations of the ions and the molecular form are constant Reactions in Solution: Acid - Base In solution, weak acids establish an equilibrium between the un-ionized or molecular form and the ionized form: MeCO 2 H un-ionized molecular form H 2 O H 3 O + (aq) + MeCO 2 - (aq) ionized Reactions in Solution: Acid - Base In solution, strong acids are completely ionized and even though there is an equilibrium, it lies entirely on the RHS and recombination is negligible: HBr (g) un-ionized molecular form H 2 O H 3 O + (aq) + Br - (aq) ionized
Reactions in Solution: Acid - Base Acids with more than one ionizable hydrogen are termed Polyprotic The common polyprotic acids are H 3 PO 4 H 2 SO 4 Phosphoric acid Sulfuric acid Reactions in Solution: Acid - Base Polyprotic acids can ionize more than once H 3 PO 4 H 2 SO 4(aq) H2 O H 3 O + (aq) + HSO 4 - (aq) HSO 4 - (aq) H2 O H 3 O + (aq) + PO 4 2- (aq) HPO 4 2- (aq) H2 O H 3 O + (aq) + PO 4 2- (aq) Each proton is ionizable and the anions, dihydrogen phosphate (H 2 PO - 4 (aq) ) and hydrogen phosphate (HPO 2-4 (aq) ) both act as acids, though H 3 PO 4 is a weak acid. Reactions in Solution: Acid - Base Polyprotic acids can ionize more than once H 3 PO 4 H 3 PO 4(aq) H2 O H 3 O + (aq) + H 2 PO 4 - (aq) H 2 PO 4 - (aq) H2 O H 3 O + (aq) + HPO 4 2- (aq) HPO 4 2- (aq) H2 O H 3 O + (aq) + PO 4 2- (aq) H 2 SO 4 H 2 SO 4(aq) H2 O H 3 O + (aq) + HSO 4 - (aq) HSO 4 - (aq) H2 O H 3 O + (aq) + PO 4 2- (aq)
Reactions in Solution: Acid - Base In contrast, H 2 SO 4 is a strong acid and hydrogen sulfate (HSO - 4 (aq) ) is also a strong acid. H 2 SO 4(aq) H2 O H 3 O + (aq) + HSO 4 - (aq) HSO 4 - (aq) H2 O H 3 O + (aq) + PO 4 2- (aq) Reactions in Solution: Acid - Base Strong or weak? All acids can be assumed to be weak except the following: HCl (aq) HBr (aq) HI (aq) HClO 4(aq) HNO 3(aq) H 2 SO 4(aq) hydrochloric acid hydrobromic acid hydriodic acid perchloric acid nitric acid sulfuric acid Reactions in Solution: Acid - Base Hydrogens attached to carbon are not ionizable in water Acetic acid, MeCO 2 H (or CH 3 CO 2 H) has the structure H O H H O H
Reactions in Solution: Acid - Base Only the hydrogen attached to oxygen is ionized in aqueous solution H H H O O H H 2 O H H H O O + H O H H The methyl hydrogens are NOT ionizable in aqueous solution. Reactions in Solution: Acid - Base Strong bases are those which ionize in solution of react to generate hydroxide ion. The common strong bases are those which already contain the OH - ion in the solid. 2 Strong bases are therefore the hydroxides of the group I and II metals 3 4 5 6 Li 3 Na11 K 19 Rb 37 Cs 55 Mg1 2 Ca 20 Sr 38 Ba 56 Reactions in Solution: Acid - Base Weak bases are the majority and are usually amines and ammonia. These react with water and deprotonate it, forming hydroxide ion and an ammonium ion: H H 3 C N CH 3 CH 3 H 2 O H 3 C N CH 3 CH 3 + OH - Trimethylamine Trimethylammonium
Reactions in Solution: Acid - Base Neutralization reactions and titrations Hydroxide and hydronium ions will react to form water. H 3 O + (aq) + OH - (aq) 2H 2 O (l) From the stoichiometry of the balanced equation, the hydroxide and hydronium react in a 1:1 ratio. We can therefore neutralize a known concentration of base or acid with the same quantity of acid or base. This is an example of a titration. Reactions in Solution: Acid - Base Neutralization reactions and titrations We use an indicator to determine the acidity or basicity of a solution: An indicator is a compound which changes color strongly at a certain level of acidity. Reactions in Solution: Acid - Base Neutralization reactions and titrations We add acid or base the titrant - to a solution of unknown concentration containing a few drops of the indicator solution. When the solution is still acid, no color change occurs; when the indicator changes color, we know the equivalence point the point where the acidity or basicity has been neutralized. By knowing the concentration and the volume of the titrant, we can calculate the concentration of the of the unknown solution.
Reactions in Solution: Acid - Base Decompostion in acid A solid base, such as Ca(OH) 2(s), will dissolve with reaction in an acid. The anion, hydroxide, reacts with the acid to form the calcium salt of the acid: Ca(OH) 2(s) + H 2 SO 4(aq) Ca 2 SO 4(s) + H 2 O (l) Reactions in Solution: Acid - Base Decompostion in acid A solid base, such as Ca(OH) 2(s), will dissolve with reaction in an acid. The anion, hydroxide, reacts with the acid to form the calcium salt of the acid: Ca(OH) 2(s) + H 2 SO 4(aq) Is this balanced? Ca 2 SO 4(s) + H 2 O (l) Reactions in Solution: Acid - Base Decompostion in acid A solid base, such as Ca(OH) 2(s), will dissolve with reaction in an acid. The anion, hydroxide, reacts with the acid to form the calcium salt of the acid: Ca(OH) 2(s) + H 2 SO 4(aq) Is this balanced? No Ca 2 SO 4(s) + H 2 O (l)
Reactions in Solution: Acid - Base Decompostion in acid A solid base, such as Ca(OH) 2(s), will dissolve with reaction in an acid. The anion, hydroxide, reacts with the acid to form the calcium salt of the acid: Ca(OH) 2(s) + H 2 SO 4(aq) Is this balanced? 2Ca(OH) No 2(s) + H 2 SO 4(aq) CaSO 4(s) + H 2 O (l) CaSO 4(s) + 2H 2 O (l) Reactions in Solution: Acid - Base Decompostion in acid Some anions also decompose in acid. These are usually anions which are derived from gases which are not soluble in water: CO 2-3 (aq) carbonate CO 2(g) HCO 3 - (aq) hydrogen carbonate CO 2(g) S 2- (aq) sulfide H 2 S (g) HS - (aq) hydrogen sulfide H 2 S (g) SO 3 2- (aq) sulfite SO 2(g) HSO 3 - (aq) hydrogen sulfite SO 2(g) Reactions between elements and between compounds often involves the exchange of electrons. Mg (s) + Cl 2(g) MgCl 2(s) Magnesium and chlorine are in their elemental states and react together so that magnesium forms Mg 2+ and chlorine forms Cl - The overall product is neutral - MgCl 2(s) has no net charge (even though it is ionic).
Mg (s) + Cl 2(g) MgCl 2(s) In this reaction, Mg has lost two electrons: Mg Mg 2+ + 2e - ½Cl 2 has gained an electron: 1 / 2 Cl 2 + e - Cl - The oxidation state of magnesium has changed from zero to +2 Mg Mg 2+ + 2e - The oxidation state of ½Cl 2 (Cl) has changed from zero to -1 1 / 2 Cl 2 + e - Cl - The oxidation state of magnesium has changed from zero to +2 Mg Mg 2+ + 2e - Oxidation state zero Oxidation state two
The oxidation state of magnesium has changed from zero to +2 Mg Mg 2+ + 2e - Oxidation state zero Oxidation state two The oxidation state of ½Cl 2 (Cl) has changed from zero to -1 1 / 2 Cl 2 + e - Cl - Oxidation state zero Oxidation state two Mg (s) + Cl 2(g) MgCl 2(s) In this reaction, Mg has been oxidized - the oxidation state of Mg has increased - and Cl has been reduced - the oxidation state of Cl has decreased - Mg (s) + Cl 2(g) MgCl 2(s) In this reaction, Mg has been oxidized - the oxidation state of Mg has increased - and Cl has been reduced - the oxidation state of Cl has decreased - Chlorine is an oxidant or an oxidizing agent
Mg (s) + Cl 2(g) MgCl 2(s) In this reaction, Mg has been oxidized - the oxidation state of Mg has increased - and Cl has been reduced - the oxidation state of Cl has decreased - Chlorine is an oxidant or an oxidizing agent Magnesium is a reductant or a reducing agent Oxidation and reduction reactions are extremely common and involves the formal interchange of electrons between atoms. The Oxidation State and Oxidation Number are key concepts in the discussion of these reactions In an oxidation-reduction reaction, or redox reaction, there MUST be and oxidation part and a reduction part.
In an oxidation-reduction reaction, or redox reaction, there MUST be and oxidation part and a reduction part. WHY? If the reduction and oxidation portions of the reaction do not balance then Charges will not balance overall If the reduction and oxidation portions of the reaction do not balance then Charges will not balance overall the Mass Balance - the number of atoms on both sides - will not balance
If the reduction and oxidation portions of the reaction do not balance then Charges will not balance overall the Mass Balance - the number of atoms on both sides - will not balance Mass and energy will therefore be created or destroyed Initially, Oxidation implied reaction with oxygen Reduction implied the liberation of a metal from it s ore - usually by reaction with carbon or air The modern definition is based on the changes in oxidation numbers and the actual charges or the formal charges on atoms and ions (including polyatomic ions) (In organic chemistry, oxidation is still based on reaction with oxygen and reduction in the addition of hydrogen)
Some rules for redox reactions All elements have an oxidation state of zero The oxidation state of simple mono-atomic cations is the charge on the ion: Element Ox. State Group IA: Li +, Na +, K +, Rb +, Cs + +1 Group IIA: Be 2+, Mg 2+, Ca 2+, Sr 2+, Ba 2+ +2 Group IIIA: B 3+, Al 3+, Ga 3+ +3 The mono-atomic anions can be treated in the same way: Element Ox. State Group VIIA: F -, Cl -, Br -, I - -1 Group VIA: O 2-, S 2-, -2 Group VA: N 3-, P 3-, -3 For an ion the oxidation state is the charge - assignment of the charge requires some thought Q: What is the oxidation state in MnO 4-? Q: What is the oxidation state in SO 4-? To answer these questions, we use some basic rules which count the valence electrons in a species
All Group IA cations have an oxidation state of +1 Hydrogen has an oxidation state of +1, rarely -1 Fluorine always has an oxidation state of -1 The sum of the oxidation states must always equal the charge on the species Example 1: Sodium fluoride, NaF Example 1: Sodium fluoride, NaF The formula unit is neutral, so the oxidation numbers must sum to zero.
Example 1: Sodium fluoride, NaF The formula unit is neutral, so the oxidation numbers must sum to zero. All group 1A cations have an oxidation number (or are in oxidation state) +1 Example 1: Sodium fluoride, NaF The formula unit is neutral, so the oxidation numbers must sum to zero. All group 1A cations have an oxidation number (or are in oxidation state) +1 Fluoride must have an oxidation state of -1 Example 2: What is the oxidation state of P in PO 3-4?
Example 2: What is the oxidation state of P in PO 3-4? The charge on this ion is 3- Example 2: What is the oxidation state of P in PO 3-4? The charge on this ion is 3- Oxygen has an oxidation number of -2 Example 2: What is the oxidation state of P in PO 3-4? The charge on this ion is 3- Oxygen has an oxidation number of -2 With four oxygens present, the total oxidation number of the oxygens is 4 x -2 = -8
Example 2: What is the oxidation state of P in PO 3-4? The charge on this ion is 3- Oxygen has an oxidation number of -2 With four oxygens present, the oxidation number of all the oxygens is 4 x -2 = -8 The balance of the oxidation states must equal -3 Example 2: What is the oxidation state of P in PO 3-4? The charge on this ion is 3- Oxygen has an oxidation number of -2 With four oxygens present, the oxidation number of all the oxygens is 4 x -2 = -8 The balance of the oxidation states must equal -3 So, P has an oxidation state of +5 as (+5) + (-8) = -3 Example 2: What is the oxidation state of P in PO 3-4? So, P has an oxidation state of +5 as (+5) + (-8) = -3 Remember that this does NOT imply that PO 3-4 is ionic, just that the oxidation state of P is +5
Example 3: What is the oxidation state of Fe in Fe 3 O 4? Example 3: What is the oxidation state of Fe in Fe 3 O 4? The formula unit, Fe 3 O 4, is neutral so all the oxidation numbers must sum to zero Example 3: What is the oxidation state of Fe in Fe 3 O 4? The formula unit, Fe 3 O 4, is neutral so all the oxidation numbers must sum to zero O has an oxidation state of -2, -2 x4 = 8
Example 3: What is the oxidation state of Fe in Fe 3 O 4? The formula unit, Fe 3 O 4, is neutral so all the oxidation numbers must sum to zero O has an oxidation state of -2, -2 x4 = 8 The oxidation states of Fe must balance this number, so the oxidation state of Fe is 8 / 3 - a fractional oxidation state. Example 3: What is the oxidation state of Fe in Fe 3 O 4? The oxidation state of Fe is 8 / 3 This is the average over all Fe in the solid and does not represent the charges on the ions. Fe 3 O 4 is actually Fe 2 O 3.FeO, where the oxidation states are + 3 and +2. The average is 1 / 3 [(2 x 3) + 2] Redox reactions CuO (s) + H 2(g) Cu (s) + H 2 O (l) This reaction is the reduction of Copper (II) Oxide with hydrogen gas to give copper metal and water.
Redox reactions CuO (s) + H 2(g) Cu (s) + H 2 O (l) This reaction is the reduction of Copper (II) Oxide with hydrogen gas to give copper metal and water. What are the oxidation states of the reactants and products? CuO (s) + H 2(g) Cu (s) + H 2 O (l) H 2 is the elemental form, Oxidation state = 0 Cu 2+ : oxidation state +2 O 2- : oxidation state -2 CuO (s) + H 2(g) Cu (s) + H 2 O (l) Cu is the elemental form, Oxidation state = 0 H: oxidation state +1 O: oxidation state -2
Redox reactions CuO (s) + H 2(g) Cu (s) + H 2 O (l) The oxidation state of Cu has changed from +2 to 0 - it has been reduced The oxidation state of H 2 has changed from 0 to +1 - it has been oxidized