OCR A Level Chemistry A Acid-base & Redox Reactions PDF

Head to www.savemyexams.com for more awesome resources OCR A Level Chemistry A Your notes Acid-base & Redox Reactions Contents Acids Acid-base Titrations Redox...

Head to www.savemyexams.com for more awesome resources OCR A Level Chemistry A Your notes Acid-base & Redox Reactions Contents Acids Acid-base Titrations Redox Page 1 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Acids Your notes Acids, Bases & Dissociation Strong acids dissociating Strong acids will fully dissociate in solution For example, when hydrogen chloride (HCl) is dissolved in water 100% of the product will be ions. HCl (aq) → H+ (aq) + Cl- (aq) In this case hydrogen ions are released, H+ (aq) The same applies with strong bases Strong bases dissociating Strong bases will fully dissociate in solution NaOH (aq) → Na+ (aq) + OH- (aq) In this case hydroxide ions are released, OH- (aq) Weak acids dissociating Weak acids only partially dissociate in solution, only a small percentage of the products will be ions In an equilibrium reaction, the products are formed at the same rate as the reactants are used This means that at equilibrium, both reactants and products are present in the solution For example, ethanoic acid (CH3COOH) is a weak acid that partially dissociates in solution CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq) The same applies with weak bases Weak bases dissociating Weak bases only partially dissociate in solution, only a small percentage of the products will be ions An equilibrium is established containing reactants and products For example, ethylamine (CH3CH2NH2) is a weak base and will partially dissociate in solution and produce hydroxide ions CH3CH2NH2 (aq) + H2O (l) ⇌ CH3CH2NH3+ (aq) + OH- (aq) Examples of acids and bases Page 2 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Neutralisation A neutralisation reaction is one in which an acid (pH 7) react together to form water (pH = 7) and a salt The proton of the acid reacts with the hydroxide of the base to form water The spectator ions which are not involved in the formation of water, form the salt Page 3 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The diagram shows a neutralisation reaction of HCl and NaOH and the two individual reactions that take place to form the water and salt The name of the salt produced can be predicted from the acid that has reacted Acid Reacted & Salt Table Metals and acids The typical reaction of a metal and an acid can be summarized as acid + metal → salt + hydrogen For example: Page 4 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 2HCl (aq) + Zn (s) → ZnCl2 (aq) + H2 (g) hydrochloric acid + zinc → zinc chloride + hydrogen Your notes Clearly, the extent of reaction depends on the reactivity of the metal and the strength of the acid Very reactive metals would react dangerously with acids and these reactions are not usually carried out Metals low in reactivity do not react at all, for instance copper does not react with dilute acids Stronger acids will react more vigorously with metals than weak acids. What signs of reaction would be expected to be different between the two? Faster reaction, seen as more effervescence the metal dissolves faster more exothermic Metals and oxides The reaction of an acid with a metal oxide forms two products: acid + metal oxide → salt + water For example: 2HCl (aq) + CaO (s) → CaCl2 (aq) + H2O (l) hydrochloric acid + calcium oxide → calcium chloride + water Metals and hydroxides The reaction with a metal hydroxide and an acid follows the same pattern as an oxide: acid + metal hydroxide → salt + water For example H2SO4 (aq) + Mg(OH)2 (s) → MgSO4 (aq) + 2H2O (l) sulfuric acid + magnesium hydroxide → magnesium sulfate + water Metals and carbonates The reaction between a metal carbonate and an acid produces three products: acid + metal carbonate → salt + water + carbon dioxide For example: Page 5 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 2HNO3 (aq) + CuCO3 (s)→ Cu(NO3)2 (aq) + H2O (l) + CO2 (g) nitric acid + copper carbonate → copper nitrate + water + carbon dioxide Your notes Page 6 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Acid-base Titrations Your notes Acid-base Titrations Volumetric Analysis Volumetric analysis is a process that uses the volume and concentration of one chemical reactant (standard solution) to determine the concentration of another unknown solution The technique most commonly used is a titration The volumes are measured using two precise pieces of equipment, a volumetric or graduated pipette and a burette Before the titration can be done, the standard solution must be prepared Specific apparatus must be used both when preparing the standard solution and when completing the titration, to ensure that volumes are measured precisely Some key pieces of apparatus used to prepare a volumetric solution and perform a simple titration 1. Beaker 2. Burette 3. Volumetric Pipette Page 7 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 4. Conical Flask 5. Volumetric Flask Your notes Making a Standard Solution Chemists routinely prepare solutions needed for analysis, whose concentrations are known precisely These solutions are termed volumetric solutions or standard solutions They are made as accurately and precisely as possible using three decimal place balances and volumetric flasks to reduce the impact of measurement uncertainties The steps are: Page 8 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Volumes & concentrations of solutions The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of solution The solute is the substance that dissolves in a solvent to form a solution The solvent is often water A concentrated solution is a solution that has a high concentration of solute A dilute solution is a solution with a low concentration of solute Concentration is usually expressed in one of three ways: moles per unit volume mass per unit volume parts per million Performing the Titration The key piece of equipment used in the titration is the burette Page 9 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Burettes are usually marked to a precision of 0.10 cm3 Since they are analogue instruments, the uncertainty is recorded to half the smallest marking, in Your notes other words to ±0.05 cm3 The end point or equivalence point occurs when the two solutions have reacted completely and is shown with the use of an indicator The steps in a titration A white tile is placed under the conical flask while the titration is performed, to make it easier to see the colour change Page 10 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The steps in a titration The steps in a titration are: Measuring a known volume (usually 20 or 25 cm3) of one of the solutions with a volumetric pipette and placing it into a conical flask The other solution is placed in the burette To start with, the burette will usually be filled to 0.00 cm3 A few drops of the indicator are added to the solution in the conical flask The tap on the burette is carefully opened and the solution added, portion by portion, to the conical flask until the indicator starts to change colour As you start getting near to the end point, the flow of the burette should be slowed right down so that the solution is added dropwise You should be able to close the tap on the burette after one drop has caused the colour change Multiple runs are carried out until concordant results are obtained Page 11 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Concordant results are within 0.1 cm3 of each other Recording and processing titration results Your notes Both the initial and final burette readings should be recorded and shown to a precision of ±0.05 cm3, the same as the uncertainty A typical layout and set of titration results The volume delivered (titre) is calculated and recorded to an uncertainty of ±0.10 cm3 The uncertainty is doubled, because two burette readings are made to obtain the titre (V final – V initial), following the rules for propagation of uncertainties Concordant results are then averaged, and non-concordant results are discarded The appropriate calculations are then done Percentage Uncertainties Percentage uncertainties are a way to compare the significance of an absolute uncertainty on a measurement Page 12 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources This is not to be confused with percentage error, which is a comparison of a result to a literature value The formula for calculating percentage uncertainty is as follows: Your notes Adding or subtracting measurements When you are adding or subtracting two measurements then you add together the absolute measurement uncertainties For example, Using a balance to measure the initial and final mass of a container Using a thermometer for the measurement of the temperature at the start and the end Using a burette to find the initial reading and final reading In all these example you have to read the instrument twice to obtain the quantity If each you time you read the instrument the measurement is ‘out’ by the stated uncertainty, then your final quantity is potentially ‘out’ by twice the uncertainty Acid-base Titration Calculations Volumes & concentrations of solutions The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of solution The solute is the substance that dissolves in a solvent to form a solution The solvent is often water conc entration mol dm − 3 = number of moles of s olute 3 mol ( ) volume of s olution dm ( ) ( ) A concentrated solution is a solution that has a high concentration of solute A dilute solution is a solution with a low concentration of solute When carrying out calculations involve concentrations in mol dm-3 the following points need to be considered: Change mass in grams to moles Page 13 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Change cm3 to dm3 To calculate the mass of a substance present in solution of known concentration and volume: Your notes Rearrange the concentration equation number of moles (mol) = concentration (mol dm-3) x volume (dm3) Multiply the moles of solute by its molar mass mass of solute (g) = number of moles (mol) x molar mass (g mol-1) Worked Example Neutralisation calculation 25.0 cm3 of 0.050 dm-3 sodium carbonate was completely neutralised by 20.00 cm3 of dilute hydrochloric acid. Calculate the concentration in mol dm-3 of the hydrochloric acid. Answer Step 1: Write the balanced symbol equation Na2CO3 + 2HCl → 2NaCl + H2O + CO2 Step 2: Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing the volume by 1000 to convert cm3 to dm3 amount (Na2CO3) = 0.025 dm3 x 0.050 mol dm-3 = 0.00125 mol Step 3: Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry 1 mol of Na2CO3 reacts with 2 mol of HCl, so the molar ratio is 1 : 2 Therefore 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl Step 4: Calculate the concentration, in mol dm-3, of hydrochloric acid Page 14 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources concentration (HCl) (mol dm-3) = 0.125 mol dm-3 Your notes Page 15 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Redox Your notes Oxidation Number Oxidation Number Rules A few simple rules help guide you through the process of determining the oxidation number of any element Remember, you are determining the oxidation state of a single atom Oxidation Numbers The oxidation state of an atom is the charge that would exist on an individual atom if the bonding were completely ionic It is like the electronic ‘status’ of an element Oxidation numbers are used to Tell if oxidation or reduction has taken place Work out what has been oxidised and/or reduced Construct half equations and balance redox equations Oxidation Numbers of Simple Ions Oxidation Rules Table Page 16 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Molecules or Compounds In molecules or compounds, the sum of the oxidation number on the atoms is zero Oxidation Number in Molecules or Compounds Page 17 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Because CO2 is a neutral molecule, the sum of the oxidation number must be zero For this, one element must have a positive oxidation number and the other must be negative How do you determine which is the positive one? The more electronegative species will have the negative value Electronegativity increases across a period and decreases down a group O is further to the right than C in the periodic table so it has the negative value How do you determine the value of an element’s oxidation number? From its position in the periodic table and/or The other element(s) present in the formula The oxidation states of all other atoms in their compounds can vary By following the oxidation number rules, the oxidation state of any atom in a compound or ion can be deduced The position of an element in the periodic table can act as a guide to the oxidation number Oxidation Numbers & the Periodic Table Page 18 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Test your understanding on the following examples: Worked Example Deducing oxidation numbers Give the oxidation number of the elements in bold in these compounds or ions: a. P2O5 b. SO42- c. H2S d. Al2Cl6 e. NH3 f. ClO2- Answers Page 19 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Are oxidation numbers always whole numbers? The answer is yes and no When you try and work out the oxidation numbers of sulfur in the tetrathionate ion S4O62- you get an interesting result! Page 20 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The oxidation number of sulfur in S4O62- is a fraction The fact that the oxidation number comes out to +2.5 does not mean it is possible to get half an oxidation number This is only a mathematical consequence of four sulfur atoms sharing +10 oxidation number Single atoms can only have an integer oxidation number, because you cannot have half an electron! Roman numerals Roman numerals are used to show the oxidation states of transition metals which can have more than one oxidation state Iron can be both +2 and +3 so Roman numerals are used to distinguish between them Fe2+ in FeO is written as iron(II) oxide Fe3+ in Fe2O3 is written as iron(III) oxide Redox Reactions & Equations Metals can react with acid to form a salt and hydrogen Metal + acid → salt + hydrogen During this reaction, there are changes in oxidation number This means that the reaction can be classified as a redox reaction Worked Example Page 21 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Explain why each of the following reactions is a redox reaction: 1. Zinc + hydrochloric acid → zinc chloride + hydrogen Your notes 2. Magnesium + sulfuric acid → magnesium sulfate + hydrogen Answer 1 Step 1: Write the balanced symbol equation Zn + 2HCl → ZnCl2 + H2 Step 2: Deduce the changes in oxidation number Zinc - starts at 0, changes to +2 Hydrogen - starts at +1, changes to 0 Chlorine - remains at -1 throughout Step 3: Explain which species is reduced / oxidised Zinc is oxidised as its oxidation number increases from 0 to +2 Hydrogen is reduced as its oxidation number decreases from +1 to 0 Answer 2 Step 1: Write the balanced symbol equation Mg + H2SO4 → MgSO4 + H2 Step 2: Deduce the changes in oxidation number Magnesium - starts at 0, changes to +2 Hydrogen - starts at +1, changes to 0 Sulfate ion - remains at -2 throughout Step 3: Explain which species is reduced / oxidised Magnesium is oxidised as its oxidation number increases from 0 to +2 Hydrogen is reduced as its oxidation number decreases from +1 to 0 Examiner Tips and Tricks Remember that oxidation number increases in oxidation reactions and decreases in reductions reactions If you are asked to explain why a reaction is a redox reaction, you should always talk about one of the following: Gain / loss of oxygen Gain / loss of hydrogen Page 22 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Gain / loss of electrons Changes in oxidation numbers Simply saying that a reaction is a redox reaction because "reduction and oxidation happen at the Your notes same time" is describing, not explaining Interpreting Redox We can identify the oxidation and reducing agents in a reaction by using the oxidation state. For example Zn (s) + H2SO4 (aq) → ZnSO4 (aq) + H2 (g) If we look at zinc, Zn, in the reaction above we can see that it increases from 0 to +2 in zinc sulfate, ZnSO4 An increase in oxidation number indicates oxidation has occured Therefore zinc is the reducing agent If we look at sulfuric acid, H2SO4, the oxidation state of hydrogen has decreased from +1 to 0 in H2 A decrease in oxidation number indicates reduction has occurred Therefore sulfuric acid is the oxidising agent Worked Example Identify the oxidising agent and reducing agent in the following reaction: 2NH3 + NaClO → N2H4 + NaCl + H2O Answer Step 1: Deduce the oxidation numbers of nitrogen and chlorine in the equation (hydrogen = +1, oxygen = -2, sodium = +1 N in NH3 is -3 Cl in NaClO is +1 N in N2H4 is -2 Cl in NaCl is -1 Step 2 Identify which species has been oxidised and which has been reduced by looking at the oxidation numbers Nitrogen is increasing in oxidation number, therefore has been oxidised Chlorine is decreasing in oxidation number, therefore has been reduced Page 23 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Step 3 Your notes Identify the oxidising and reducing agent NH3 is the reducing agent (it has been oxidised itself) NaClO is the oxidising agent (it has been reduced itself) Remember, the whole species is the reducing agent, not just the element (e.g. NH3 is the reducing agent, not N on its own) Page 24 of 24 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers