A2 - Physical Flashcards

Question

in electrochem how to make full equation of cell using reduction half eqs

Answer 1

- use half eqs - combine wo H+ , H2O, e- - remember reducing agent eq is reversed - combine, MULTIPLYING by e-, then add H+ etc

Answer 2

different ligands

Answer 3

lithium reacts with electrolyte/water

Answer 4

due to not taking place exactly/strictly in standard conditions

Answer 5

to allow passage of ions

Answer 6

- metal is reused so there are no landfill problems

Answer 7

It releases CO2 upon combustion and fermentation So that CO2 can then be reuptaken by photosynthesis

Answer 8

To remove oxide layer (of eg.Aluminium) and remove grease

Answer 9

- show emf difference - so *RHS* will oxidise *LHS* to…

Answer 10

To provide surface for e- transfer

Answer 11

1. Equilibrium moves left (of half eq) 2. Releaseing more e- so moving Zn2+/Zn electrode potential 3. More negative (bc more e-)

Answer 12

The enthalpy of atomisation of an element is - the enthalpy change when 1 mole of gaseous atoms is formed - from the element in its standard state Na (s) > Na(g) [ΔatH = POSITIVE kJ mol-1] or ½ O2 (g) > O (g) [ΔatH = POSITIVE kJ mol-1]

Answer 13

The first ionisation enthalpy is the enthalpy change - required to remove 1 mole of electrons - from 1 mole of gaseous atoms to form - 1 mole of gaseous ions with a +1 charge Mg (g) → Mg+ (g) + e [ΔIE 1H]

Answer 14

The bond dissociation enthalpy is standard molar enthalpy change - when one mole of a covalent bond is - broken into two gaseous atoms (or free radicals) Cl2 (g) > 2Cl (g) [ΔdissH = POSITIVE kJ mol-1] .. For diatomic molecules the ΔdissH of the molcule is - the same as 2x ΔatH of the element

Answer 15

The second ionisation enthalpy is the enthalpy change to - remove 1 mole of electrons from - one mole of gaseous 1+ ions - to produces one mole of gaseous 2+ ions Mg+ (g) → Mg2+(g) +e- (ΔIE 2H)

Answer 16

The first electron affinity is the enthalpy change that occurs - when 1 mole of gaseous atoms gain - 1 mole of electrons to form - 1 mole of gaseous ions with a -1 charge O(g) +e → O- (g) [ΔEA 1H] = NEGATIVE kj mol-1]

Answer 17

The second electron affinity is the enthalpy change when - one mole of gaseous 1- ions gains - one electron per ion to produce - gaseous 2- ions. O- (g) +e- → O2 (g) [ΔEA 2H= POSITIVE kJ mol-1]

Answer 18

The enthalpy of lattice formation is the - standard enthalpy change when - 1 mole of an ionic crystal lattice is formed - from its constituent ions in gaseous form. Na+ (g) + Cl- (g) → NaCl (s) [ΔLattH = NEGATIVE kJ mol-1)

Answer 19

The enthalpy of lattice dissociation is - the standard enthalpy change when 1 mole - of an ionic crystal lattice form is separated - into its constituent ions in gaseous form. NaCl (s) → Na+ (g) + Cl- (g) [ΔLattH = POSITIVE kJ mol-1]

Answer 20

Enthalpy change when - one mole of gaseous ions become aqueous ions. X+/-(g) + aq → X+/-(aq) Eg. Li+ ΔhydH = NEGATIVE kj mol-1) Eg. F- ΔhydH = NEGATIVE kj mol-1

Answer 21

The enthalpy of solution is the - standard enthalpy change When one mole - or an lonic solid dissolves in a - large enough amount of water to ensure that - dissolved ions are well separated and dont interact with one another. NaCl (s) + aq → Na+ (aq) + Cl- (aq)

Answer 22

1. ΔfH: v MgCl2 2. ΔatH (Mg): Mg(s) ^ Mg(g) 3. Δie 1H (Mg): Mg(g) ^ Mg1+ + e 4. Δie 2H (Mg) 5. ΔatH (Cl): Cl2 ^ 2Cl(g) . 6. 2 x ΔeaH: 2e- + 2Cl v 2Cl- 7. ΔLattH: v MgCl2

Answer 23

1. The sizes of the ions - The larger the ions, the less negative the enthalpies of lattice formation - (i.e. a weaker lattice). As the ions are larger, charges become further apart - and so have a weaker attractive force between them. - - The lattice enthalpies become less negative down any group. e.g. LiCi, NaCl, KCI, etc

Answer 24

2. The charges on ion - The bigger the charge of ion, the greater attraction between ions so - the stronger lattice enthalpy (more negative values).

Answer 25

There is a tendency towards covalent character in ionic substances when • the positive ion is small •the positive ion has multiple charges •the negative ion is large •the negative ion has multiple negative charges. - - - When a compound has some covalent character- it tends towards giant covalent - so lattice is stronger than if it was 100% ionic. So the - Born-Haber value would be larger than the theoretical value. - - - When the negative ion becomes distorted and more covalent we say it becomes polarised. - The small metal cation is called polarising if it polarises the larger negative ion.

Answer 26

A spontaneous process (e.g. diffusion) will proceed on its own without any external influence. - - - A reaction that is more exothermic will result in products that are more - thermodynamically stable than the reactants. - - This is a driving force behind many reactions and causes them to be - spontaneous (occur without any external influence). > some are still endothermic (entropy)

Answer 27

- The Born Haber lattice enthalpy is the real experimental value. - When a compound shows covalent character, the theoretical and the born Haber lattice enthalpies differ. - The more the covalent character the bigger the difference between the values.

Answer 28

>> a description of number of ways atoms can share quanta of energy. - If number of ways of arranging the energy (W) is high, - then the system is disordered and entropy (S) is high. .. Elements, Simpler compounds, pure substances ..tend to have lower entropies than… Compounds, Complex compounds, Mixtures - - When a solid increases in temp its entropy increases as particles vibrate more. - here is a bigger jump in entropy w boiling than that with melting. - gases have large entropies as they are much more disordered 0K substances = zero entropy. - no disorder as particles are stationary.

Answer 29

- In general, a significant increase in the entropy will occur if: -there is a change of state from solid or liquid to gas - theres a significant increase in number of molecules between products and reactants. .. NH4CI (s) → HCI (g) + NH3 (g) AS = +ve •change from solid reactant to gaseous products •increase in number of molecules both will increase disorder • Na (s) + ½ Cl2 (g) → NaCl (s) AS = -ve •change from gaseous and solid reactant to solid •decrease in number of molecules both will decrease disorder

Answer 30

- not possible for a substance to have a standard entropy of below 0 The unit of entropy is J K-1 mol-1 ΔS°= ΣS products - ΣS° reactants .. - Elements in their standard states do not have zero entropy. - Only perfect crystals at absolute zero (0 K) will have zero entropy.

Answer 31

- The balance between entropy and enthalpy determines feasibility - of a reaction; whether it can actually take place - given by the relationship : ΔG = ΔH - TΔS (/1000 to convert units) - Gibbs free energy combines effect of enthalpy and entropy into one number. - For any spontaneous change, ΔG will be negative. - A reaction w increasing entropy (+ve ΔS) and is exothermic (-ve ΔH) - will make ΔG be negative and will always be feasible. .. - if ΔG is negative theres still a possibility reaction wont occur or - will occur so slowly that effectively it doesn't happen. - If reaction has a high activation energy, reaction will not occur.

Answer 32

ΔG : Kj mol-1 ΔH : Kj mol-1 TΔS: J K-1 mol-1 (but always /1000 as needs to convert to kj) >> from C to K (+273)

Answer 33

1. Reaction will be feasible if G<= 0 so is 0 in equation 2. Know change in H and S and insert to eq 3. Find T in K > the statement should say... > T must be above (ans) to be feasible

Answer 34

- As physical phase changes like melting and boiling are equilibria, - the ΔG for such changes is zero.

Answer 35

Changing temp will change value of -TΔS in the equation - If reaction involves ΔS as +ve then - increasing temp will make more likely ΔG is negative so more likely the reaction occurs BUT - OPPOSITE when decrease in entropy .. - the reaction has a ΔS close to zero: temp will not have a large effect on - feasibility of the reaction as -TΔS will be small; ΔG will not change much e.g. N2(g) +O2(g) → 2NO (g)

Answer 36

y= mx+c to the ΔG = ΔH - TΔS equation. - - The positive gradient means ΔS is negative which corresponds - to the equation above showing increasing order. - When ΔG <0 then the reaction is spontaneous. .. - slope of the line would change below melting point - would be a liquid and the entropy change would be different. Y is ΔG X is T С is ΔH m is ΔS

Answer 37

- When an ionic lattice dissolves in water it involves breaking up - the bonds in the lattice and forming new bonds between metal ions and water molecules. - - When an ionic substance dissolves lattice is broken up. - The enthalpy of lattice dissociation is equal to energy needed to break up lattice (to gaseous ions). - This step is endothermic. .. - The size of the lattice enthalpy depends on the size and charge on - the ion. The smaller the ion and the higher its charge, the stronger the lattice. For MgCl, 1. ΔHsolution: v aqueous ions 2. ΔHL dissociation: ^ MgCl2(s) + aq → Mg2+ (aq) + 2Cl (aq) 3. v is ΔhydH for each gaseous ion (x by how many ions)

Answer 38

ΔH solution = ΔL dissociation + ΣΔhydH ΔH Solution = -ΔL formation + ΣΔhydH Remember difference between the two

Answer 39

- Generally H solution is not very exo or endothermic so hydration enthalpy - is about the same as lattice enthalpy. - - In general substance is more likely to be soluble if ΔH solution is exothermic. - If a substance is insoluble it is often because the lattice enthalpy is much - - larger than the hydration enthalpy and it is not energetically favourable - to break up the lattice, making ΔH solution endothermic

Answer 40

- We must consider entropy too, to give us full picture about solubility. - When a solid dissolves into ions the entropy increases as there is more disorder - as solid changes to solution and number of particles increases. - - This positive ΔS can make AG negative even if ΔH solution is - endothermic, especially at higher temperatures.

Answer 41

• exothermic - salt will always dissolve at all temps ΔG (always -ve) = ΔH (-ve) -TΔS (S is +ve bc increased disorder as more particles)

Answer 42

• endothermic - salt may dissolve depending on if -TΔS is more negative than ΔH is positive ΔG = will dissolve if negative ΔH = positive TΔS = ΔS is +ve bc of increased disorder as more particles Increasing temp will make it more likely ΔG will be negative > so reactions feasible and salt will dissolve

Answer 43

In a mixture of gases, its the pressure that the gas would have if it alone occupied the volume that the whole mixture does P= p1 + p2 + p3

Answer 44

Number of moles of gas/total number of moles of all gases

Answer 45

Kp = p(products)^moles / p(reactants)^moles > p is partial pressure of that gas; only include gases in expression and leave out solids or liquids

Answer 46

- larger kp is, greater the amount of products - if kp is small we say equilibrium favours reactants . - kp only changes w temp; pressure conc or catalysts have no effect on value

Answer 47

In this equilibrium which is exothermic 1. If temp increases reaction shifts to oppose change 2. And move in backwards direction and equilibrium position shifts left 3. Kp value gets smaller as fewer products

Answer 48

1. If pressure increases reaction shifts to oppose change 2. And move in forward direction to side w fewer moles of gas 3. Equilibrium position shifts right and value of kp stays same . 4. Mole fractions of products increase and of reactants, decreases 5. Top of kp expression so increases and bottom decreases until original kp value is restored

Answer 49

-When connected together (zinc anode and copper cathode) the zinc half cell has more tendency to oxidise - Zn2+ and release electrons than copper half cell - - more electrons so build up on zinc electrode than copper, so potential difference is created between - two electrodes. Zinc strip is negative terminal and copper, positive .. Potential diff measured w high resistance voltmeter (E)

Answer 50

- stops current flowing in circuit so is possible to measure max possible potential difference > so reactions wont be occurring as current stops flowing

Answer 51

because the metal wire would set up its own electrode system w solutions > salt should also be unreactive w electrodes and solutions

Answer 52

- if voltmeter is removed (and eg replaced w bulb) a current flows > reactions will then occur separately at each electrode so voltage falls to 0 as reactants are used up

Answer 53

- single vertical line represents boundary between phases (like solid | solution) - double line represents salt bridge between two half cells

Answer 54

- provides conducting surface for electron transfer > used when no metal is present for use as an electrode > its unreactive and can conduct electricity

Answer 55

- not possible to measure absolute potential of a half electrode on its own - only pot difference between two electrodes - has to be connected to another half cell of known potential, and potential diff calculated between electrodes

Answer 56

it is porous and can absorb H2 gas

Answer 57

The position of redox eq will change w conditions > eg. Increase in conc of M^n+ would move equilibrium to right so making potential more positive

Answer 58

SHE is hard to use, so diff standards are used as easier > other standards are themselves calibrated against SHE (primary standard) is known as using secondary standard > eg. Silver, silver chloride

Answer 59

- when an electrode system is connected to hydrogen electrode system - and standard conditions apply the potential difference measured

Answer 60

Electrons are given off and lost , travelling to positive electrode. Electrons arrive at this electrode and absorbed /gained

Answer 61

1. More negative E° will reduce and more positive will oxidise > E° F2/F- > E° O2/H2O, > F2 will oxidise H2O to O2 .. 2. 2F2 + 2H2O > 4F- + O2 + 4H+

Answer 62

- measure of how far from equilibrium the cell reaction lies > more positive Ecell more likely reaction occurs .. If current flows Ecell falls to zero as reactants concs drop when reactants proceeds > follows le chateliers principle: so increase of conc increases Ecell so decreasing vice versa

Answer 63

- most cells are exothermic in spontaneous direction so applying le chateliers principle to a temp rise - results in ecell decrease as equilibrium reactions shift backward - - if ecell is positive reaction might occur, but may not occur happening so slowly it effectively doesnt happen > high activation energy means wont occur - High pressure can help counteract high temps causing ecell to fall due to exothermic reactions

Answer 64

Used as a commercial source of electrical energy - cells can be non rechargeable, rechargeable and fuel cells

Answer 65

Cells are non rechargeable when reactions that occur in them are non reversible

Answer 66

Foreard reaction occurs on discharge giving out charge ; charging causes reverse reaction > reversible cells only work if product stays attached to electrode and doesnt disperse

Answer 67

Reagents in cell are absorbed onto powdered graphite that acts as support medium > this allows ions to react in solvent absence like water

Answer 68

Use energy from reaction of fuel w oxygen to create voltage > maintaining constant voltage over time as are continuously fed w fresh O2 and H2 so > maintaining constant concentration of reactants . > differs from ordinary cells where voltage drops over time as reactant concs drop

Answer 69

+ less pollution and less CO2 as pure H2 only emits water; H2 rich fuels produce only small amts of air pollutants anf CO2 + greater efficiency .. — expensive (produced in water hydrolysis) — storing and transport of H2 in terms of safety, feasibility of pressurised liquid and limited life cycle of solid absorber — limited lifetime requiring regular replacement and disposal, w high production costs —use of toxic chemicals in their production

Answer 70

- as a liquid under pressure - adsorbed on surface of solid material - absorbed within solid material

Answer 71

- compared to hydrogen fuel cells they have advantages - can be made from renewable sources in a carbon neutral way like fermentation - less explosive and easier to store than H2 .. Eq at oxygen electrode 4e- + 4H+ + O2 >2H2O . Eq at ethanol electrode C2H5OH + 3H2O > 2CO2 + 12H+ + 12e- . Overall: C2H5OH + 3O2 > 2CO2 + 3H2O

Answer 72

The rate equation relates mathematically the rate of reaction to the conc of the reactants. - For the following reaction, - aA + bB → products, the generalised rate equation is: -> r = k[A]^m[B]^n - r is rate; units (mol dm-3 s-1) - k is rate constant

Answer 73

1. The units of k depend on the overall order of reaction. >must be worked out from rate equation 2. The value of k is independent of concentration and time. It is constant at a fixed temperature 3. The value of k refers to a specific temperature and increases if we increase temperature

Answer 74

- rearrange r = k[A] to make k subject K = rate/[A] - insert units to cancel - mol dm-3 s-1 / mol dm-3 - = s-1

Answer 75

- When we follow one experiment over time recording change in conc - we call it a continuous rate method. - The gradient represents rate of reaction. - The reaction is fastest at start where gradient is steepest. - The rate drops as the reactants start to get used up; their conc drops. - The graph will eventually become horizontal and the gradient - becomes zero which represents the reaction having stopped.

Answer 76

This works if theres a change in number of moles of gas in reaction. - Using a gas syringe is a common way of following this. - works well for measuring continuous rate but gas syringe only measures 100ml of gas - don't want a reaction to produce more than this

Answer 77

- In reactions with several reactants, if the conc of one of the reactant is kept in a large excess - then reactant wont affect rate and will be pseudo-zero order. - its conc stays virtually constant and does not affect rate.

Answer 78

- rate at the start of the reaction, where it is fastest. - calculated from the gradient of a continuous monitoring conc vs time graph at time = zero. - - A measure of initial rate is preferable as we know the concs at the start of the reaction.

Answer 79

The initial rate can be calculated from taking gradient - of a continuous monitoring conc vs time graph at time = zero - - Initial rate can also be calculated from clock reactions where time taken to reach a fixed conc is measured. Initial rate represented as 1/t .. - In clock reactions there are often two successive reactions. - end points achieved when one limited reactant runs out, resulting in sudden colour change. - By repeating experiment several times, varying conc of a reactant e.g. I- (keeping other reactants constant conc) - you can determine order of reaction with respect to that reactant

Answer 80

Hydrogen peroxide reacts w iodide ions to form iodine. - thiosulfate ion then immediately reacts iodine formed, in second reaction as shown below. H202(aq) + 2H+(aq) + 2l-(aq) → I2(aq) + 2H20(l) 2S2032-(aq) + I2(aq) → 2I-(aq) + S4062-(aq) - When the I2 produced has reacted with all of limited amount of thiosulfate ions present, - excess I2 remains in solution. - - Reaction with starch then suddenly forms a dark blue-black colour. - A series of experiments is carried out, in which the conc of iodide ions is varied, - while keeping concs of all of the other reagents the same. - In each experiment, time taken (t) for reaction mixture to blue is measured.

Answer 81

Working order out graphically - in an experiment where conc of one reagent is changed and reaction rate measured, - its possible to find order graphically - Log both sides of equation Log rate = log k + n log [Y] Y= c + mx A graph of log rate vs log [Y] will yield a straight line where the gradient is equal to the order n

Answer 82

To calculate order for a particular reactant, its easiest to - compare two experiments where only that reactant is being changed. - If conc is doubled and rate stays same: order= 0 - If conc is doubled and rate doubles: order= 1 - If conc is doubled and rate quadruples : order= 2 Add orders of all reactants in equation to find equation's order - - if two reactants are changed then the effect of both individual changes - of conc are multiplied to give overall change in rate

Answer 83

Increasing temp increases value of the rate constant k - the relationship is given by Arrhenius equation (k = Ae^-Ea/RT - where A is Arrhenius constant, - R is the gas constant, - and Ea is activation energy .. k vs temp increases same rate as 2nd order graph

Answer 84

k = Ae-Ea/RT The Arrhenius equation is usually rearranged to Ln k= Ln A -Ea/(RT) > shd be able to rearrange and substitute values into both these equations. Units: -Temp uses the unit K -R = 8.31 J mol-1K-1 -Activation energy will need to be in J mol-1 to match the units of R -The unit of Arrhenius constant A will be same as the unit of rate constant **diagram 6 for examples**

Answer 85

Using the rearranged version Ink=In A - Ea/(RT) k is proportional to rate of reaction so ln k can be replaced by ln(rate) From plotting a graph of ln(rate) or ln k against 1/T the activation energy - can be calculated from measuring - gradient of the line .. Eg. - in a graph: y is ln(rate) and x is 1/T - gradient = Ea / R - Ea = gradient x R (rearranged) >> gradient is always negative **diagram 7 for example**

Answer 86

- A mechanism is a series of steps through which reaction progresses, - often forming intermediate compounds. -> If all steps added together theyll add up to overall equation for reaction > - Each step can have a diff rate of reaction. - slowest step will control overall rate of reaction. - The slowest step is **rate-determining step.** - - The molecularity (number of moles of each substance) of molecules - in the slowest step will be same as order of reaction for each substance. -> e.g. 0 moles of A in slow step wd mean A is zero order. 1 mole of A in the slow step would mean A is first order **diagram 8** lots of examples

Answer 87

1. Substance that donates a proton 2. Substance that accepts a proton .. Each acid is linked to a conjugate base (species that remains after acid donates proton) on other side of eq HCl > Cl- (base) H2O > H3O+ (acid)

Answer 88

pH (2dp) = log[H+]

Answer 89

They completely dissociate > conc of H+ in monoprotic strong acid will be same as conc of acid

Answer 90

1 x 10^-pH

Answer 91

All aqueous solutions and pure water the following equilibrium occurs H2O <> H+ + OH- .. Bc H2O conc is much bigger than ion concs we assume its value is constant > equals Kw Kw = [H+][OH-] At 25c Kw value for all aq solutions is 1x10^-14 mol2 dm-6

Answer 92

Pure water/neutral solutions are neutral bc [H+]= [OH-] Kw = [H+]^2 At 25c [H+] = v1x10-14 = 1x10-7 so pH7 . At different temps pH of pure water changes > water dissociation is endothermic so increasing temp pushes eq to right > so bigger H+ conc and lower pH

Answer 93

For bases, given OH- conc > to find pH we need to work out H+ conc with Kw expression

Answer 94

Only slight dissociation upon dissolving in water giving equilibrium mixture - HA + H2O <> H3O+ + A- Simplified : HA <> H+ + A- (Put into expression wch is for Ka) .. Ka for ethanoic acid is 1.7 x 10^-5 Larger ka, stronger the acid

Answer 95

Ka values are quoted as this > pKa = -log Ka > SO Ka = 10^-pKa

Answer 96

Make calculations easier two assumptions made to simplify Ka expression - H+ conc = A- conc as have dissociated in 1:1 ratio - Ka expression simplifies [H+]^2 / [HA] .. - amt of dissociation is small so initial conc of undissociated acid remains constant HA conc at eqm = conc initial

Answer 97

Method differs if acids strong or weak for partially neutralised case - work out moles of original acid/base and so moles H+ and OH- > find wch in excess .. If excess acid, find new conc of excess H+ ions, Or OH- if excess alkali [H+/OH-] = moles excess H+/OH- / tot volume in dm3 > find [H+] with kw expression if excess alkali > Find pH with H+ conc .. if bases or acids are diprotic, multiply H+/OH- moles by 2 when working out

Answer 98

Find moles of orig acid and base and find excess - if excess acid find new conc of excess HA [HA] = initial HA mol - moles OH- / total volume dm3 - then find conc of salt formed [A-] [A-] = moles OH- added / total volume dm3 . Rearrange Ka expression to get H+ conc and find pH

Answer 99

When a weak acid has been reacted w exactly half neutralisation volume of alkali.. - at half neutralisation HA conc = A- conc - so H+ conc = Ka And pH = pKa .. pH = pKa = -log(ka) Use vols and concs to find pH of half neutralisation

Answer 100

pH of strong diluted acid Conc H+ = old H+ conc x old volume/new volume find pH with H+ .. With pH of alkali , find conc of OH- in same way - then H+ conc found by kw/OH- conc

Answer 101

Where pH doesnt change significantly if small amts of acid or alkali are added .. - acidic buffer is made from weak acid and salt of weak acid > from reacting weak acid w strong base (ethanoic acid and sodium ethanoate) .. - basic buffer is from weak base and salt of that weak base > from reacting weak base w strong acid (ammonia and ammonium chloride)

Answer 102

In ethanoic acid buffer It dissociates to produce conjugate base and H+ > in buffer theres a higher conc of the salt CH3CO2- ion than in pure acid > buffer contains reservoir of HA and A- ions

Answer 103

If small amounts of acid is added to the buffer: - Then the above equilibrium will shift to the left - removing nearly all the Ht ions added, . CH3CO2- (aq) + H+ (aq) > CH3CO2H -As there is a large concentration of the salt ion in the buffer, -the ratio [CH3CO2H]/ [CHCO2] stays almost constant, so the pH stays fairly constant.

Answer 104

If small amounts of alkali is added to the buffer, OH ions will react with H+ ions to form water. - The equilibrium will then shift to the right to produce more H+ ions. - CH3CO2H <> CH3CO2- + H+ (aq) - - Some ethanoic acid molecules are changed to ethanoate ions but - as there is a large concentration of the salt ion in the buffer, the ratio - [CH3CO2H/ [CH3CO2-] stays almost constant, so the pH stays fairly constant.

Answer 105

Still using weak acids dissociation expression >But here we assume [A-] conc is due to added salt only > expression rearranged to [H+] = .. - The salt content can be added in several ways: a salt solution could be added - to the acid or some solid salt added. A buffer can also be made by partially neutralising - a weak acid with alkali and therefore producing salt. . - We also assume the initial concentration of the acid has - remained constant, because the amount that has dissociated or reacted is small.

Answer 106

- If a small amount of alkali is added to a buffer then the - moles of the buffer acid would reduce by the number of moles of alkali added - and the moles of salt would increase by same amount so new calculation of pH can be done wnew values. CH3CO2H +OH- → CH3CO2- + H2O . - If a small amount of acid is added to a buffer then the - moles of the buffer salt would reduce by the number of - moles of acid added and the moles of buffer acid would - increase by the same amount so a new calc of pH can be done w new values. CH3CO2- (aq) + H+ → CH3CO2H (aq)

Answer 107

Diluting a buffer solution with water will not change its pH - This is because in buffer equation below the ratio of [HA]/]A] will stay constant - as both concentrations of salt and acid would be diluted by the same proportion.

Answer 108

Constructing a pH curve 1. Transfer 25cm3 of acid to a conical flask with a volumetric pipette 2. Measure initial pH of the acid with a pH meter 3. Add alkali in small amounts noting the volume added .. 4.Stir mixture to equalise the pH; Measure and record the pH to 1 d.p. 5. Repeat steps 3-5 but when approaching endpoint add in smaller volumes of alkali 6. Add until alkali in excess

Answer 109

- Calibrate meter first by measuring known pH of a buffer solution. This is necessary - because pH meters can lose accuracy on storage. - - Most pH probes are calibrated by putting probe in a set buffer (often pH 4) - and pressing a calibration button/setting for that pH. Sometimes this is - repeated with a second buffer at a different pH .. Can also improve accuracy by maintaining constant temperature

Answer 110

There are four mains curve types: - strong acid strong base - weak acid strong base - strong acid and weak base - weak acid and base .. Y axis is pH and x axis is cm3 of base - long steep area from pH 3-9 - pH at equivalence point =7 (lies at mid point of extrapolated vertical portion of curve) - neutralisation vol worked from titration data (standard titration calcs)

Answer 111

1. Plot initiall and final pH 2. Volume at neutralisation 3. General shape (pH at neutralisation)

Answer 112

-pH starts at 3 - at start ph rises quick then levels off; flattened part is buffer region as buffer solution is made - steep part is 7-9 and equivalence point is >7 > Phenolphthalein

Answer 113

For weak acids Ka expression; at 1/2 neutralisation vol HA conc is A- conc > Ka = H+ conc and pKa = pH .. - we know ka so can find pH at 1/2 V or vice versa - if pH curves plottrd then pH of weak acid at 1/2 V is equal to pKa ..

Answer 114

- steep part is 4-7 - equivalence point is <7 > methyl orange

Answer 115

- theres no steep part of curve > pH probe

Answer 116

- Indicators can be considered as weak acids. The acid must have a different colour to its conjugate base - An indicator changes colour from HIn to In: over a narrow range. - Different indicators change colours over different ranges. - - The end-point of a titration is defined as the point when the colour of the indicator changes colour - The end-point of a titration is reached when [Hin] = [In]. To choose a correct indicator for a titration - - one should pick an indicator whose end-point coincides with the equivalence point for the titration.

Answer 117

HIn <> In- + H+ > in acid solution H+ ions present push eq to reactants so colour A is acidic > in alkaline OH- react and remove H+ so eq shifts to products So B is alkaline .. Indicator works if pH range lies on steep part of titration curve > indicator will change colour rapidly and colour change will correspond to neutralisation point

Answer 118

initial moles OH- = excess OH- moles + neutralised H+ moles

Answer 119

1. When using Ka expression, using start conc of HA and end conc of H+ 2. To find end HA conc, after dissociates 3. When you find ka and need to ise further, the conc of acid in question is eq. >> add change from products to en conc . . 1. When doing partial neutralisation calcs with acid + strong base buffer

Answer 120

• for both ΔH and ΔS = product — reactants INCLUDING MOLS • divide answer for S by 1000 • rearrange for temp

Answer 121

OH- react with H+ HA <> H+ + A- equilibrium Shifts right TO MAINTAIN H+

Answer 122

- find initial pH from moles initially worked out > dont need to convert by dividing w vol becuase cancel out - - find new pH with [HA]<>[H+][A-] - if H+ is added, add its moles to HA mol, and take away from right side - vice versa with OH-

Answer 123

- if an equilibrium , forms an intermediate . (Dont inc in rate eq) - must match rate equation - is the slowesr step

A2 - Physical Flashcards

(148 cards)