Stoichiometry: Chemical Calculations Chemistry 120 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. Stoichiometry: Chemical Calculations Chemistry 120 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 Stoichiometry: Chemical Calculations Chemistry 120 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit Stoichiometry: Chemical Calculations Chemistry 120 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH4 SF6 NaCl Na2S2O3 Stoichiometry: Chemical Calculations Chemistry 120 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH4 SF6 NaCl Na2S2O3 12.0115 + 4 x 1.0079 Stoichiometry: Chemical Calculations Chemistry 120 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH4 12.0115 + 4 x 1.0079 SF6 32.066 + 6 x 18.9984 NaCl Na2S2O3 Stoichiometry: Chemical Calculations Chemistry 120 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH4 12.0115 + 4 x 1.0079 SF6 32.066 + 6 x 18.9984 NaCl 22.9898 + 35.453 Na2S2O3 2 x 22.9898 + 2 x 32.066 + 3 x 15.9994 Stoichiometry: Chemical Calculations Chemistry 120 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH4 12.0115 + 4 x 1.0079 SF6 32.066 + 6 x 18.9984 NaCl 22.9898 + 35.453 Na2S2O3 2 x 22.9898 + 2 x 32.066 + 3 x 15.9994 Stoichiometry: Chemical Calculations Chemistry 120 The molecular mass of a compound is the mass of each atom multiplied by the number of times it appears in the formula unit CH4 12.0115 + 4 x 1.0079 SF6 32.066 + 6 x 18.9984 NaCl 22.9898 + 35.453 Na2S2O3 2 x 22.9898 + 2 x 32.066 + 3 x 15.9994 Stoichiometry: Chemical Calculations Chemistry 120 The mole and atomic mass The mole is defined as the number of elementary entities as are present in 12.00 g of 12C. Numerically, this is equal to Avogadro’s Number 6.022 x 1023 Therefore, in 12.00 g of 12C there are 6.022 x 1023 ‘elementary entities’, in this case atoms. Stoichiometry: Chemical Calculations Chemistry 120 The mole and atomic mass Atomic masses, in atomic units, u, are defined relative to 12C. Therefore, The formula mass of an element or compound contains 1 mole, 6.022 x 1023, of particles Stoichiometry: Chemical Calculations Chemistry 120 Examples How many particles are there in 5 g of Na? Stoichiometry: Chemical Calculations Chemistry 120 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? Stoichiometry: Chemical Calculations Chemistry 120 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 Stoichiometry: Chemical Calculations Chemistry 120 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 (12C) And 1 mole = 6.022 x 1023 particles = number of particles in 12 g 12C Stoichiometry: Chemical Calculations Chemistry 120 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 Stoichiometry: Chemical Calculations Chemistry 120 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 1023) particles Na Stoichiometry: Chemical Calculations Chemistry 120 Examples How many particles are there in 5.0000 g of Na? 1.310 x 1023 atoms Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Chemistry 120 Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C4H10 Chemistry 120 Stoichiometry: Chemical Calculations Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: Atomic mass of C = 12.011g Atomic mass of H = 1.0079g C4H10 Chemistry 120 Stoichiometry: Chemical Calculations Chemistry 120 Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C4H10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u = 58.123 u Stoichiometry: Chemical Calculations Chemistry 120 Examples What is the mass of 0.23 mol of butane? Molecular formula of butane: C4H10 Atomic mass of C = 12.011g Atomic mass of H = 1.0079g Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u = 58.123 u Relative Molecular Mass of Butane = 58.123 g Stoichiometry: Chemical Calculations Chemistry 120 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 Stoichiometry: Chemical Calculations Chemistry 120 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 Stoichiometry: Chemical Calculations Chemistry 120 Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Sr(s) + Cl2(g) SrCl2(s) Stoichiometry: Chemical Calculations Chemistry 120 Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Sr(s) + Cl2(g) SrCl2(s) Physical state Stoichiometry: Chemical Calculations Chemistry 120 Chemical Equations These are formulæ which show the chemical change taking place in a reaction. Sr(s) + Cl2(g) Reactants SrCl2(s) Physical state Product Stoichiometry: Chemical Calculations Chemistry 120 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. Stoichiometry: Chemical Calculations Chemistry 120 Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water Stoichiometry: Chemical Calculations Chemistry 120 Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water Reactants: Cyclohexane, C6H12 Oxygen, O2 Stoichiometry: Chemical Calculations Chemistry 120 Chemical Equations Example Cyclohexane burns in oxygen to give carbon dioxide and water Reactants: Cyclohexane, C6H12 Oxygen, O2 Products: Carbon Dioxide, CO2 Water, H2O Stoichiometry: Chemical Calculations Chemistry 120 Chemical Equations Example Initially, we can write the reaction as C6H12 + O2 CO2 + H2O Stoichiometry: Chemical Calculations Chemistry 120 Chemical Equations Example Initially, we can write the reaction as C6H12 + O2 CO2 + H2O This is NOT a correct equation – there are unequal numbers of atoms on both sides Stoichiometry: Chemical Calculations Chemistry 120 Chemical Equations Example Initially, we can write the reaction as C6H12 + O2 CO2 + H2O 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 Stoichiometry: Chemical Calculations Chemistry 120 Balancing the equation C6H12 + O2 CO2 + H2O Stoichiometry: Chemical Calculations Balancing the equation C6H12 + O2 Chemistry 120 CO2 + H2O 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 C6H12 + O2 6CO2 + H2O 6 C, 12 H, 2 O 6 C, 2 H, 13 O 13 O on RHS means there must be 13 O on LHS 13 C6H12 + /2 O2 6 C, 12 H, 13 O 6CO2 + H2O 6 C, 2 H, 13 O Stoichiometry: Chemical Calculations Balancing the equation 13 Chemistry 120 C6H12 + /2 O2 6CO2 + H2O 6 C, 12 H, 13 O 6 C, 2 H, 13 O 12 H on RHS means there must be 12 H on LHS 13 C6H12 + /2 O2 6CO2 + 6H2O 6 C, 12 H, 13 O 6 C, 12 H, 18 O 18 O on RHS means there must be 18 H on LHS C6H12 +9O2 6 C, 12 H, 18 O 6CO2 + 6H2O 6 C, 12 H, 18 O Stoichiometry: Chemical Calculations The final balanced equation is C6H12 +9O2 Chemistry 120 6CO2 + 6H2O 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 Stoichiometry: Chemical Calculations Chemistry 120 If cyclohexane were burnt in an excess of oxygen, the quantity of oxygen used would be the same although O2 would be left over. Solutions and concentration Chemistry 120 If cyclohexane were burnt in an excess of oxygen, the quantity of oxygen used would be the same although O2 would be left over. Solutions Chemistry 120 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 Chemistry 120 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 Chemistry 120 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 Chemistry 120 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 NaCl melts (s) solute are solids. Brass is an example Here copper is the solvent, zinc the solute. Zn Cu Solutions Chemistry 120 Gas-Solid solution: Hydrogen in palladium Steel Solutions Chemistry 120 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 Chemistry 120 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 Chemistry 120 Ionic Solutions An ionic substance, such as NaClO4, contain ions – in this case Na+ and ClO4-. 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 Chemistry 120 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 Chemistry 120 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 Chemistry 120 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 Chemistry 120 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 Chemistry 120 Moles and solutions The units of concentration are: C = n = moles V L3 and we define a molar solution as one which has 1 mole per liter. Alternatively, Concentration = Molarity = number of moles volume of solution Solutions Chemistry 120 Example 4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution? Solutions Chemistry 120 Example 4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution? Formula mass of Na2SO4(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 Chemistry 120 Example 4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution? Formula mass of Na2SO4(s): (2 x 22.9898)+ 32.064+(4x15.9994)=142.041gmol-1 1 mole of Na2SO4(s) = 142.041g 1/ 142.041 mole of Na2SO4(s) = 1 g Solutions Chemistry 120 Example 4 g of Na2SO4(s) is dissolved in 500 ml of water. What is the concentration of the solution? 1/ 142.041 mole of Na2SO4(s) = 1 g Therefore 4 g of Na2SO4(s) = 4/142.041 mole = 2.82 x 10-2 mole Solutions Chemistry 120 Example 4 g of Na2SO4(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 Chemistry 120 Example The equation for the dissolution of Na2SO4(s) is + Na2SO4(s) 2Na (aq) + 2SO4 (aq) H2O So if we have 5.64 x 10-2 molL-1 Na2SO4(s), we must have 1.13 x 10-1 moles Na+(aq) and 5.64 x 10-2 mol SO42-(aq) as there are 2 Na cations for every sulfate ion Solutions Chemistry 120 If we change the volume of the solution then we change the concentration: If the Na2SO4 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 Chemistry 120 Dissolution on an atomic level. Solids are held together by very strong forces. NaCl(s) melts at 801oC and boils at 1465 oC but it dissolves in water at room temperature. Solutions Chemistry 120 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 Chemistry 120 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 Chemistry 120 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 Chemistry 120 Solutions • Chemistry 120 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 Chemistry 120 • 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 L3 and the units are molL-1 or moldm-3 Solutions • Chemistry 120 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 Chemistry 120 Reactions in Solution Chemistry 120 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. Chemistry 120 Reactions in Solution: Acid - Base A saturated solution of NaI is placed in contact with Na131I(s), which is radioactive. Na+ II- Na+ + Na I I Na+ I I I + + Na Na+ - Na I I+ + Na Na Na+ I- + I Na Na+ Na + II- Na+ Na+ II- Na+ - NaI(aq) Na I- I- I- Na+ Na + + Na I I Na+ I I I + + Na Na+ - Na I I+ + Na Na Na+ I- + I Na Na+ I- - I Na+ + NaI*(s) I- Na+ - I- I- I- Na+ Na + INa+ Chemistry 120 Reactions in Solution: Acid - Base A saturated solution of NaI is placed in contact with Na131I(s), which is radioactive. Na+ II- Na+ + Na I I Na+ I I I + + Na Na+ - Na I I+ + Na Na Na+ I- + I Na Na+ Na + II- Na+ - Na+ After time, Na+ II-theI- activity in I- Na+ Na IIthe I- solution Na - INa + I- - I Na+ Na+ + NaI(aq) + Na Na+ - Na I measured I+ .......... + and + Na Na I- I+ I- Na+ NaI*(s) I- I Na+ Na+ Na+ I- I- I- Na+ Na + INa+ is Chemistry 120 Reactions in Solution: Acid - Base Radioactivity is found in the solution, even though the concentration of I-(aq) has not changed. Na+ II- Na+ + Na I I Na+ I I I + + Na Na+ - Na I I+ + Na Na Na+ I- + I Na Na+ Na + II- Na+ - I- Na+ Na + Na I I Na+ I I I + + Na Na+ - Na I I+ + Na Na Na+ I- + I Na Na+ Na I- I+ Na+ II- Na+ I- - I Na+ + I- Na+ - I- I- I- Na+ Na + INa+ Reactions in Solution: Acid - Base Chemistry 120 The equilibrium here is composed of two reactions: Na131I(s) Na+(aq) + I-(aq) H2O Na+(aq) + 131I-(aq) NaI(s) H2O So we write NaI(s) H2O + Na (aq) + I (aq) Reactions in Solution: Acid - Base Chemistry 120 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 Chemistry 120 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 Chemistry 120 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 Chemistry 120 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 Chemistry 120 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 Chemistry 120 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, H3O+(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 Chemistry 120 A strong acid is a compound which dissolves and dissociates completely in water or reacts with water to give hydronium ions, H3O+(aq) H3O+(aq) + Cl-(aq) HCl(g) H2O - 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: H3O+(aq) + Cl-(aq) HCl(g) H2O Chemistry 120 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) Na+(aq) + OH-(aq) NaOH(s) H2O 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 MeCO2H Chemistry 120 H3O+(aq) + MeCO2-(aq) H2O 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 MeCO2H Recombination H3O+(aq) + MeCO2-(aq) H2O H3O+(aq) + MeCO2-(aq) H2O MeCO2H Chemistry 120 Reactions in Solution: Acid - Base We write the reaction for acetic acid, MeCO2H, as an equilibrium to include the ionization and recombination. H3O+(aq) + MeCO2-(aq) Ionization MeCO2H Recombination H3O+(aq) + MeCO2-(aq) MeCO2H H2O H2O MeCO2H H3O+(aq) + MeCO2-(aq) H2O As the amount of ionization and recombination are the same, the concentrations of the ions and the molecular form are constant Chemistry 120 Reactions in Solution: Acid - Base In solution, weak acids establish an equilibrium between the un-ionized or molecular form and the ionized form: MeCO2H un-ionized molecular form + H3O (aq) + MeCO2 (aq) H2O ionized Chemistry 120 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 + H3O - (aq) + Br (aq) H2O ionized Chemistry 120 Reactions in Solution: Acid - Base Acids with more than one ionizable hydrogen are termed Polyprotic The common polyprotic acids are H3PO4 Phosphoric acid H2SO4 Sulfuric acid Chemistry 120 Reactions in Solution: Acid - Base Polyprotic acids can ionize more than once H3PO4 H2SO4(aq) HSO4-(aq) HPO42-(aq) H2O H2O H2O H3O+(aq) + HSO4-(aq) H3O+(aq) + PO42-(aq) H3O+(aq) + PO42-(aq) Each proton is ionizable and dihydrogen phosphate (H2PO4-(aq)) the anions, and hydrogen phosphate (HPO42-(aq)) both act as acids, though H3PO4 is a weak acid. Chemistry 120 Reactions in Solution: Acid - Base Polyprotic acids can ionize more than once H3PO4 H3PO4(aq) H2PO4-(aq) HPO42-(aq) H2SO4 H2SO4(aq) HSO4-(aq) H2O H2O H2O H2O H2O H3O+(aq) + H2PO4-(aq) H3O+(aq) + HPO42-(aq) H3O+(aq) + PO42-(aq) H3O+(aq) + HSO4-(aq) H3O+(aq) + PO42-(aq) Chemistry 120 Reactions in Solution: Acid - Base In contrast, H2SO4 is a strong acid and hydrogen sulfate (HSO4-(aq)) is also a strong acid. H2SO4(aq) HSO4 (aq) H2O H2O H3O+(aq) + HSO4-(aq) H3O+(aq) + PO42-(aq) Reactions in Solution: Acid - Base Strong or weak? Chemistry 120 All acids can be assumed to be weak except the following: HCl(aq) hydrochloric acid HBr(aq) hydrobromic acid HI(aq) hydriodic acid HClO4(aq) perchloric acid HNO3(aq) nitric acid H2SO4(aq) sulfuric acid Chemistry 120 Reactions in Solution: Acid - Base Hydrogens attached to carbon are not ionizable in water Acetic acid, MeCO2H (or CH3CO2H) has the H structure O H H O H Reactions in Solution: Acid - Base Chemistry 120 Only the hydrogen attached to oxygen is ionized in aqueous solution H H O H H O H H2O O H O + H O H The methyl hydrogens are NOT ionizable in aqueous solution. H H Reactions in Solution: Acid - Base Chemistry 120 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 2 Li the solid. 3 Strong bases are therefore the hydroxides of the group I and II metals 3 Na11 Mg1 2 4 K 19 Ca 20 5 Rb 37 Sr 38 6 Cs 55 Ba 56 Reactions in Solution: Acid - Base Chemistry 120 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 N H3C N CH 3 CH 3 Trimethylamine H2O H3C CH 3 + OH- CH 3 Trimethylammonium Chemistry 120 Reactions in Solution: Acid - Base Neutralization reactions and titrations Hydroxide and hydronium ions will react to form water. + H3O (aq) + OH (aq) 2H2O(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. Chemistry 120 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 Chemistry 120 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. Chemistry 120 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) + H2SO4(aq) Ca2SO4(s) + H2O(l) 2Ca(OH)2(s) + H2SO4(aq) Ca2SO4(s) + 2H2O(l) Chemistry 120 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) + H2SO4(aq) Ca2SO4(s) + H2O(l) Is this balanced? 2Ca(OH)2(s) + H2SO4(aq) Ca2SO4(s) + 2H2O(l) Chemistry 120 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) + H2SO4(aq) Ca2SO4(s) + H2O(l) Is this balanced? No 2Ca(OH)2(s) + H2SO4(aq) Ca2SO4(s) + 2H2O(l) Chemistry 120 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) + H2SO4(aq) CaSO4(s) + H2O(l) Is this balanced? No 2Ca(OH)2(s) + H2SO4(aq) CaSO4(s) + 2H2O(l) Chemistry 120 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: CO32-(aq) carbonate CO2(g) HCO3-(aq) hydrogen carbonate CO2(g) S2-(aq) sulfide H2S(g) HS-(aq) hydrogen sulfide H2S(g) SO32-(aq) sulfite SO2(g) HSO3-(aq) hydrogen sulfite SO2(g) Solution Reactions: Oxidation-Reduction Chemistry 120 Reactions between elements and between compounds often involves the exchange of electrons. Mg(s) + Cl2(g) • MgCl2(s) Magnesium and chlorine are in their elemental states and react together so that magnesium forms Mg2+ and chlorine forms Cl• The overall product is neutral - MgCl2(s) has no net charge (even though it is ionic). Solution Reactions: Oxidation-Reduction Mg(s) + Cl2(g) Chemistry 120 MgCl2(s) In this reaction, Mg has lost two electrons: 2+ Mg + 2e Mg ½Cl2 has gained an electron: 1 - - /2Cl2 + e Cl - Solution Reactions: Oxidation-Reduction Chemistry 120 The oxidation state of magnesium has changed from zero to +2 2+ Mg Mg - + 2e The oxidation state of ½Cl2 (Cl) has changed from zero to -1 1 /2Cl2 + e - Cl - Solution Reactions: Oxidation-Reduction Chemistry 120 The oxidation state of magnesium has changed from zero to +2 Mg Oxidation state zero 2+ Mg - + 2e Oxidation state two Solution Reactions: Oxidation-Reduction Chemistry 120 The oxidation state of magnesium has changed from zero to +2 2+ Mg Mg Oxidation state zero - + 2e Oxidation state two The oxidation state of ½Cl2 (Cl) has changed from zero to -1 1 /2Cl2 + e Oxidation state zero - Cl - Oxidation state two Solution Reactions: Oxidation-Reduction Mg(s) + Cl2(g) Chemistry 120 MgCl2(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 - Solution Reactions: Oxidation-Reduction Mg(s) + Cl2(g) Chemistry 120 MgCl2(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 Solution Reactions: Oxidation-Reduction Mg(s) + Cl2(g) Chemistry 120 MgCl2(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 Solution Reactions: Oxidation-Reduction Chemistry 120 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 Solution Reactions: Oxidation-Reduction Chemistry 120 In an oxidation-reduction reaction, or redox reaction, there MUST be and oxidation part and a reduction part. Solution Reactions: Oxidation-Reduction Chemistry 120 In an oxidation-reduction reaction, or redox reaction, there MUST be and oxidation part and a reduction part. WHY? Solution Reactions: Oxidation-Reduction Chemistry 120 If the reduction and oxidation portions of the reaction do not balance then • Charges will not balance overall Solution Reactions: Oxidation-Reduction Chemistry 120 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 Solution Reactions: Oxidation-Reduction Chemistry 120 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 Solution Reactions: Oxidation-Reduction Chemistry 120 Initially, Oxidation implied reaction with oxygen Reduction implied the liberation of a metal from it’s ore - usually by reaction with carbon or air Solution Reactions: Oxidation-Reduction Chemistry 120 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) Solution Reactions: Oxidation-Reduction Chemistry 120 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: Be2+, Mg2+, Ca2+, Sr2+, Ba2+ +2 Group IIIA: B3+, Al3+, Ga3+ +3 Solution Reactions: Oxidation-Reduction Chemistry 120 The mono-atomic anions can be treated in the same way: Element Group VIIA: F-, Cl-, Br-, IGroup VIA: O2-, S2-, Group VA: N3-, P3-, Ox. State -1 -2 -3 Solution Reactions: Oxidation-Reduction Chemistry 120 For an ion the oxidation state is the charge assignment of the charge requires some thought Q: What is the oxidation state in MnO4-? Q: What is the oxidation state in SO4-? To answer these questions , we use some basic rules which count the valence electrons in a species Solution Reactions: Oxidation-Reduction Chemistry 120 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 Solution Reactions: Oxidation-Reduction Example 1: Sodium fluoride, NaF Chemistry 120 Solution Reactions: Oxidation-Reduction Chemistry 120 Example 1: Sodium fluoride, NaF The formula unit is neutral, so the oxidation numbers must sum to zero. Solution Reactions: Oxidation-Reduction Chemistry 120 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 Solution Reactions: Oxidation-Reduction Chemistry 120 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 Solution Reactions: Oxidation-Reduction Example 2: What is the oxidation state of P in PO43-? Chemistry 120 Solution Reactions: Oxidation-Reduction Example 2: What is the oxidation state of P in PO43-? The charge on this ion is 3- Chemistry 120 Solution Reactions: Oxidation-Reduction Example 2: What is the oxidation state of P in PO43-? The charge on this ion is 3Oxygen has an oxidation number of -2 Chemistry 120 Solution Reactions: Oxidation-Reduction Chemistry 120 Example 2: What is the oxidation state of P in PO43-? The charge on this ion is 3Oxygen has an oxidation number of -2 With four oxygens present, the total oxidation number of the oxygens is 4 x -2 = -8 Solution Reactions: Oxidation-Reduction Chemistry 120 Example 2: What is the oxidation state of P in PO43-? The charge on this ion is 3Oxygen 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 Solution Reactions: Oxidation-Reduction Chemistry 120 Example 2: What is the oxidation state of P in PO43-? The charge on this ion is 3Oxygen 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 Solution Reactions: Oxidation-Reduction Chemistry 120 Example 2: What is the oxidation state of P in PO43-? So, P has an oxidation state of +5 as (+5) + (-8) = -3 Remember that this does NOT imply that PO43- is ionic, just that the oxidation state of P is +5 Solution Reactions: Oxidation-Reduction Chemistry 120 Example 3: What is the oxidation state of Fe in Fe3O4? Solution Reactions: Oxidation-Reduction Chemistry 120 Example 3: What is the oxidation state of Fe in Fe3O4? The formula unit, Fe3O4, is neutral so all the oxidation numbers must sum to zero Solution Reactions: Oxidation-Reduction Chemistry 120 Example 3: What is the oxidation state of Fe in Fe3O4? The formula unit, Fe3O4, is neutral so all the oxidation numbers must sum to zero O has an oxidation state of -2, -2 x4 = 8 Solution Reactions: Oxidation-Reduction Chemistry 120 Example 3: What is the oxidation state of Fe in Fe3O4? The formula unit, Fe3O4, 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. Solution Reactions: Oxidation-Reduction Chemistry 120 Example 3: What is the oxidation state of Fe in Fe3O4? 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. Fe3O4 is actually Fe2O3.FeO, where the oxidation states are + 3 and +2. The average is 1/3[(2 x 3) + 2] Solution Reactions: Oxidation-Reduction Chemistry 120 Redox reactions CuO(s) + H2(g) Cu(s) + H2O(l) This reaction is the reduction of Copper (II) Oxide with hydrogen gas to give copper metal and water. Solution Reactions: Oxidation-Reduction Chemistry 120 Redox reactions CuO(s) + H2(g) Cu(s) + H2O(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? Solution Reactions: Oxidation-Reduction CuO(s) + H2(g) H2 is the elemental form, Oxidation state = 0 Cu2+: oxidation state +2 O2-: oxidation state -2 Chemistry 120 Cu(s) + H2O(l) Solution Reactions: Oxidation-Reduction CuO(s) + H2(g) Chemistry 120 Cu(s) + H2O(l) Cu is the elemental form, Oxidation state = 0 H: oxidation state +1 O: oxidation state -2 Solution Reactions: Oxidation-Reduction Chemistry 120 Redox reactions CuO(s) + H2(g) Cu(s) + H2O(l) The oxidation state of Cu has changed from +2 to 0 - it has been reduced The oxidation state of H2 has changed from 0 to +1 - it has been oxidized