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A-level CHEMISTRY > Module 5.1 > Flashcards

Module 5.1 Flashcards

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1
Q

Meaning of rate of reaction

A

it is the change in concentration of reactant or a product per unit time .

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2
Q

what is the order

A

the order , with respect to a reactant is the power to which the concentration of the reactant is raised in the rate equation .

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3
Q

what is the rate constant k

A

is the constant that links the rate of equation with concentration of the reactants raised to power of their orders in the rae equation .

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4
Q

the rate equation

A

the rate equation for a reaction A+B–. C is given by ; rate =k[a[m[B]N , where m is the order of reaction with respect A and n is the order of reaction with respect to B .

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5
Q

what is the overall order

A

of a reaction is the sum of the individual orders m + n .

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6
Q

We learnt about collision theory in book 1 , how is this in relation to the rate equation

A

-reactions only proceed when successful collisions occur - that is particl emust collide with the correct orientaiton and with enough energy to overcoe the activation barrier 9i.e activation energy 0 . HOw frequently these successful collision soccur will determine the rate of a reaction .

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7
Q

what is the rate of reaction equaiton

A

rate of reaction = change in conc of reactant or product / time

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8
Q

How are rates measured

A

rates measured in moldm-3s-1 , (mol per dm3 per s 0 , but other units may sometimes , be more appropriate . If a reaction is very slow , a larger time scale (such as min 0 may be used . In this case , the units would e mol dm-3 min-1 . I

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9
Q

-If it is difficult to measure the concentration , you may use another measurement which allows you to monitor the amt of a product or reactant

A

. For example , if a gas is porduced in a reeaction , you could measure hte volume of gs porduced over a time period .

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10
Q

what is the order of reaction used for

A

-If more than one reactant is invovled in a reaction , each reactant can affect the rate of the reaction differently . The effect of the individul is described by stting an order with respect to each reactant . COnsider a reactant A , it is concentration affects the rate of a reaction . This can be expressed mathemticlly check page 10 .

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11
Q

learning tip about square brackets

A

square brackets are used to show a concentration
-the x denotes any power to which the concentration of a is raised .
fish sign means is proportional to .

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12
Q

meaning of zero order

A

zero order is if the order of 0 with respect to reactant A then rae is propotional to A to power of zero .

-THe rate is unaffected by changing the concentration of A .
-Note that any number to the power 0 is equal to 1 ..

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13
Q

meaning of first order

A

-If the order is 1 with respect to a reactant B then rate is proportional to [B]1 .

-If [B] , increased by 2 times , the rate also increases by 2 times .
-if [B] is increased by 3 times , the rate also increases by 3 times .

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14
Q

meaning of second order

A

If the order is 2 with respect to a reactant C , then rate is [c]2 . The change in rate will be equal to the change in concentration squared .
-If [C] increased by 2 times , the rate increased by 2squared = 4 times
-If [C] increases by 3 times , the rate increases by 32 squared , 9 times .

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15
Q

Rate reactions and overall orders

A

Chemist use rate equations to mathematically express the influence each reactant has on a reaction .
-Take the reaction A+B+C—. products .

-If the order for A , B and C were x y and z perspectively you could wirte the following expression for the rat .

rate is proportionla [A][B][C]

-The sign for proportional can be removed if a constant is addded into the equation . Chemists use the rate constant , k , he rate constant links the concentrations and orders of reactants to the rate . The expressiono then becomes a rate equation .

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16
Q

if any reactant is zero order , willit appear in the rate equation

A

no , that is because it does not affect the rate . For instance , in our example above , where the order are 0 , 1 , and 2 , with respect to A , B and C , the rate equation would be
rate=k[A]0[B]1[C]2

however , BECAUSE ANY NUMBER raised to the power of zero is one ,t he reactant can be remove d, powers of 2 can also e omited from outisde the brackets . but do not remove the reactant fromt he rate wquaiton .

rate ewution will become
RATE=K[B][C]2

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17
Q

what will the overall order of this reaction be

A

the overall order of a reaction is the sum of the individual orders .

-In this exampe abobe rate =k[B][C]2 and the overall order is +=
-the rate equation cn be determined only from experimental results . NOte that the orders are not the same as the numbers used to balance an equaiotn .

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18
Q

check calculating the value and units for rate constants

A

zero
first
second
third

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19
Q

do worked example 1 - page 12 - 13

A

…

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20
Q

what is the half life

A

the half life of a reactant is the time taken for the concentration of the reactant to reduce by half

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21
Q

why do chemists use concentration time graphs

A

chemists can use concentration time graphs to fi dout about the rate and order of reactions . Theshapes of conentration time graphs depend on the order of the reaction .

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22
Q

finding rates from concentration - time graphs (1) - what can we measure for reactions involving cids or bases

A

pH changes by carrying out titrations .
pH changes using a pH meter

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23
Q

for reactions that produce gases , what can we measure

A

the change in volume or pressure .
-the loss in mass of reacants .

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24
Q

for reactions that produce visual changes what can we measure

A

the formation of a precipitiate
-colour change

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25
Q

how are visual changes monitoed

A

using a colorimeter , as the intensity of colour is directly related to the concentration of a coloured subtance .

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26
Q

example used - sulfur dichlordie dioxide decomposes t produce sulfur dioxide and chlorine

A

-was monitotired conc every 500 seconds , and hen plotted .
RATE OF REACTION found by taking a gradeint at the purple line , As the dat produces curve ,t angent

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27
Q

where does the tangent on page 1 shows lines at

A

t=0 and t=3000s

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28
Q

how is the rate calculated on page 14 (1)

A

after t=0 , (shown by the red , steeper tangent line ) .

Initial rate = change in concentration SO2CL2/ time for the change to take place .
0.500-0.00/3300-0

=1.5*104moldm-3s-1

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29
Q

how is the rate calculated on page 14 (2)

A

after t=3000s (shown by the green shallower tangent line )
rate = change in concentration of SO2CL2/ time for the change to take place

(0.38-0.14)/(4000-0)
=6.0*10-5moldm-3s-1

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30
Q

check the symbol for half life

what do the graphs look for zero order and first order reactions . in figure 2 and figue 3

A
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31
Q

how to workout the worked examples 1

A

…

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32
Q

how do u use half lives to find the rate constant

A

because first order reactions have constant half-lives , the valule of the half-life , can be used to determine the value of the rate constant .
-The half-life and the rate constant are relaed bby the following equation .

K=In2/tand a half this equation ONL APPLIES TO FIRST ODER REACTIONS . other order reactions have no half lives .

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33
Q

check worked example 2 p.g 15

A

…

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34
Q

initial rates (1)

A

The rate at the very start of a reaction given the shorthand , t=0 , (which stands fo r time =0) , is known as the initial rate . Ifa tangent line is drawn at t=0 , on a concentration-itme graph, the gradient of this line is equal to the initial rate .

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35
Q

how to determine the order with respect , to each reactant , you carry out he reaction several time .

A

-to determine the order with respect of each reactant , to carry out the reaction sseveral times . Each time , you vary the concentration of one of the reactants . You can obtain the initial rte for eaach of these different concentrations and plot , the result , on a graph to produce a rate-concentration graph for each reactant ..

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36
Q

clock reactions (1)

A

Some reactionsNwill produce visible changes such as the formation of a ppt , or a dramatic colour change . Measuring the time taken for such changes to occur canbe used o find out how concenration affects the initial rate .

-Only the initial rate is consididered as the visile changes that occur are assumed to last occur as the reaction first happens , the rate is GENERALLY FASTEST .

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37
Q

-The time taken , for the visible event to occur is inversely proportional to the initial rate ;

A

the shorter the time taken for the change the faster the reaction must be . this cn be expressed mathematically proprotional 1/t

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38
Q

this can be expresed mathemastically as rate proportional 1/t

A

this means that when a graph is plotted 1/t is taken as a good approximation of the initial rate .

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39
Q

example of this rection between sodium thisosulfate , hcl

A

a cross is drawn on a piece of paper and placed under a beaker . The reactants are then added together in the beaker . The products , form a cloudy o solution as a ppt is ormed , so a stopwatch is used to time how long takes for the cross under the beaker to be obscured y solutoin.

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40
Q

prt 2 - odiumthiosulfate

A

the experiment , is then carried out using various concentrations of both sodium thiosulfate and hydrochloric acid to determine how thier concnerations affect the rate . graph 1/t , against concentation is plotted .

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41
Q

check f-1 , nd the iodine clock reaction(pagee 16)

A
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42
Q

What happens rate-concentration graphs - zero order

A

If the order is 0 , with respect to a given reactant , A , the rate-concentration graph will appear as in .

THE figure 2 , on 17 shows
-This graph , shows the rate proportional to A .
-Changes in the concentration of this rectant have no effect on the rate .

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43
Q
A
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43
Q

what happens in a first order reaction

A

-if the first order is 1 with respect to given reactant , B , the rate concentration graph will apper as in .

-This graph shows the rate proportional B .
-If the concentration of reactaaant B , is doubled , the rate will double .
-If the concentration of reactant B , is tripled the rate will triple .
-If the concenration of reactant B , is increased 100 fold the rate will increase 100 fold .

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44
Q

Determining , the rate constant from first order rate-concentration graphs

A

If a rate-concnetration , is first order , with respect , to a given reactant , it is possible to use the graph to determine the ate cosntant , k .

For the first order reactions ;
-Rate proportional [reactant]
-this means the rate=k[reactant]
If we arrange this to find k ;
k=rate/concentration

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45
Q

-Rate is ploted on he yaxi s, ont he rate-concentration graph ,

A

and conentration is plotted on the xacis so calculating , the rate is the samr as calcualting the gradient , as gradient =change in y divided by change in x .
-tje imot pf l , will depend on the raye equation for hthe givenr eaction.

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46
Q

what is a second order reaction

A

-if the order is 2 , with respect to a to a given reactant C , the rate cocnentration graph will apper as figure 4 .

–This graph shows the rate proportional [C]2
-if the concentraiton of reactant C is doubled the raye will incfrase by 2 squard .
-if the concenration b is tripled the rate will incrase by 3 squared .
if the concnetratio of rectant c is increased 100 fold , the rate will increase by 100 squared .

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47
Q

a reaction mechanism is what

A

a series of steps that together make up the overall reaction .

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48
Q

what is the rate determinising step

A

it is the slowest step is the slowest step in the reaction mechnism of a multistep reaction .

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49
Q

what is an inermediate

A

it is a species formed in one step of a multi-step reaction that is used , up in a subsequent step annd is not seen as either a reactant or a porduct or the overall eequation .

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50
Q

why is the rate determinig step useful for

A

reactions can occur in one step or in many steps Experimental resutls can be used to predict , how many steps will occur during reaction . The series of steps that occur during a reaction is called a reaction mechnaisms .

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51
Q

what will the slowest step in the reaction dictating

A

how quicklyt he reaction wil proceed . This slowest step is called the rate -d etermining step .

-when you measure the rate of any reaction , that has a multi-step reaction mechanism , you are effectively measuring the rate of this determinig step .

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52
Q

predicitng reaction mechanims from rate equaiton

A

-As you have learnt in previous topics , some reactants hvae no influence on the rate of reaction .
-Thee reactant have an order of zzero and do not appear in the rate equaiton .
-I fthey have no influence on the rate , they cannot be involved in the rate-determinig step .

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53
Q

check page 19 for hdyrolysis of the bromoethane

A

-What does this rate equation show .
-THIS SHOWS that only the concentraiton of CH3BR influences the rate . Therefore , only this moleucle is involed int he rate-determingi step .

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54
Q

step 2 - bromoethane

A

this suggests that the rate dereminig step must involve the c-BR bond breaking , breaking the OH- can take its place .
S1-CH3BR–CH3++BR-
STEP2;CH3+ +OH—-CH3OH

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55
Q

step 3

A

in this example , CH3+ must be involved in the second step , in the order to be used up as it does not apper in the overall equation .

-The sepecies invovled in the two steps , cancel one aonther to leav the overall equation .

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56
Q

Orders and the rate determini step -what doe the order show for the bromoethane hdyrolysis

-check the rate equaiton ofr htis

A

[CH3BR] is raised to the power of 1 , ( rememebr ]a 1 is not hown in the rate equaiton )
-This means tht the rate-determinig step invovles just one moelcule CH3br

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57
Q

consider the hydrolysisi of bromoethane - what if it was order is 2

A

-If the order is 2 with respect go 2e.g rate =k[x]2 , there will be wo moleucles of this reactant invovled in the rate-deerminign step . Doubling the concentration of each of the two
moelucle will double the rate so verall the rate is quadurpled .

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58
Q

what should be in place to be able to work out the reaction mechanism

A

a rate - determining step is given that involves the number of moleucles shown by the order for that reactant .

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59
Q

what should be in place to be able to work out the reaction mechanism (2)

A

subsequen steps are hown that eventually geenrate the products shown in the balanced equation

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60
Q

what should be in place to be able to work out the reaction mechanism (3)

A

any intermediate generated is not present int he overall equation (i.e it is used up within subsequent steps ).

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61
Q

learning tip on page 20

A

…

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62
Q

worked example page 20

A

…

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63
Q

what is the arnhenius plot

A

is a graph of ln k=lnA -Ea/R
*1/T where ln k is plotted against 1/T

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64
Q

rate constant k

A

reacttion rates depend on both the rate constant and the concentrations of the reactants presents in the rate equatioN

check the rate ewution

-The larger the value of K , the faster the reaction .

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65
Q

the effect of temperature on the rate constant k (1)

A

an increase in temperature , gives more energy to the molecules . This means that collisions are more frequent , and more of the collisions exceed the activation energy of the reaction . EXPLAINED IN BOLTZMANN DISTRIBUTION BOOK 3.2.

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66
Q

the effect of temperature on the rate constant k (2)

A

the key factor affecting the reaction rate is the number of collision that exceed the activtion energy . This mean that rate increases with temperature by much more thn can be expallined soley from any increased frequency of collisons .

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67
Q

what happens in this reaction , if the rate increases with increasing temperature , when the concentrations are the same , then the rate constant must increase with temperature .

A

-raising the temperature speeds up the rate of most reactions by incresin the rate constant k .

-for many reactions , the rate doubles increse in 10 degrees in tmp . This reflect the getaer number of recting particles that exceed the acctivation energy .

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68
Q

proportionality of doubling the rate , what does this mean

A

-typically , doubling the rate will double the value of the rate constant k
-check figure 1

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69
Q

what is the ARRHENIUS equation showing
write equation
-what does the equation mean

A

-arrhenius equation is used to describe , mathematiclly , the exponential relationship between the rate constant and the temperature .
k-rate constant
Ea -activation energy
T - temperature (in kelving )
e=mahematicl cosntant value with value 2.71828 -log
A - preexpoentnial factor
R-gas constant .

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70
Q

what can the arrhenius equation tell us (1)

A

to be able to react , moleucles have to collide with enough energy , to overcome the activation energy . It has been found that at any given emperature , T, rate constant , k is proprotional to e-EA/RT , this is expressues amthetmitcallyas .

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71
Q

what can the arrhenuius euation tell us (2)

A

this can be turned into an equation by adding a constant , the ‘pre-expoenntial facotr’ , A giving k==Ae -Ea/R . This is the Arrhenuius ewqution

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72
Q

What does the Arrhenius equation tell us (3)

A

-Temperature , T, and the rate constant , k , are related exponentially .
-As temperture increases , the rate cosntant increases .

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73
Q

what does adding a catalyst do to the rate constant , k .

A

-adding a catalst provides an alternative reaction path , with a lower activiation enery . If ou follow this through maethematiclaly , a lower value of Ea increases the rate constant K .

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74
Q

Taking logarithims of thr ARRHENIUS EUATION

A

…
check the end of page 22 .
figure 2 , and worked example
im being lazyyy

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75
Q

what does the equillibrium law state

A

state that for the equilibirum look a tpage 23

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76
Q
A
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77
Q

what dos a homogeneous equlibirum show

A

it is an equlibirum in which ALL THE SPECIES , making up the reactants and products , are in the same physical state .

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78
Q

what does a heterogenous equlibrium show

A

it is an equlibirum in which species making up the reactants and products are in different physical states .

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79
Q

dynamic equlibrium , what is it and Kc

A

-not all reactions go to completion . An equiliibirum ma be established , this is when the raes of the forward and reverse reactions are equal .

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80
Q

in a dynamic equlibirum 0 does equilibrium occur straight away

A

no , the reverse reaction cannot take place until the forward reaction has produced a high enough concentration of the product .

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81
Q

example of decompostion of n204 a dynamic equlibrium ,w hat happens (1)

A

-at first the cocnentration of N204 is high and the concentration of N02 , is zero .

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82
Q

example of decompostion of n204 a dynamic equlibrium ,w hat happens (2)

A

as the reaction proceeds , the concentration of N204 , falls and the cocnenration of N02 , rises . The reaction mixture , appers , colourless intiially then a pale brownn colour will appear .

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83
Q

example of decompostion of n204 a dynamic equlibrium ,w hat happens (3)

A

once equliibrium is reached he concentration of each subsance remians unchanged - the reaction has a brown appearnce .

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84
Q

example of decompostion of n204 a dynamic equlibrium ,w hat happens (4)

A

if conditions change , the equilibrium will need to be re-establsihed . A colour change will be observed until the equilibrium is resored .

According to to the equlibrium law - for the reaction .

-equilibirum consant Kc , can be calcualed as follwos .
-once an equilibrium is established , kc remians constant unless the emp of reaction changes .

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85
Q

units of Kc dh3dk page 23

A

…

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86
Q

how to dertermine concentrations at equlilrium - what are the two common ways

A

-in order to determine a value for Kc , cocnentration of reactans and prodcuts at equlibirum need ot be known .
titrtion using a colorimeter

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87
Q

how to determine the concentrations at equilibirum - TITRATION

A

titrating one of the reactants or products agaisnt a subance , with a known concetration will enable us to find out how much of the reactant / product is present .

-

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88
Q

how to determine the concentrations at equilibirum - TITRATION DRAWBACKS

A

-Reactant or product has to be removed from the reaction mxiture , and this will alter the position of the equilibirum .

-Also other subtnces may be present that can affect he results of titrations

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89
Q

how to determine the concentrations at equilibirum - TITRATION DRAWBACKS (2)

A

e,g - if an alakli is used to titrate agaisnt the equlibirum subances in question and an acid catalsyt is also used , both equilibirum subtance and atalyst will react with the alakli .
-The amount o any such sucbances must be carefullyd etermined when the titration results are used to determine equiilbrium cocnetations .

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90
Q

equilibirum investiagion and worked example 1

A

…

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91
Q

calcualting unknown equlibirum concentraitons - for the balanced equation , what are the two things you can find (stochiometric)

A

-the reactaitng quntities neeedto prepare a requried quanitty of a product .
-the quantities of product formed by reching togeher known qunatities of reactnt .

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92
Q

what does kc help us to determine

A

helps us to deterime the equilibrium concentrations of the components in a equilibirum mixtrue .

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93
Q

worked example 2

A

…

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94
Q

learning tip on page 25

A

…

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95
Q

if a heterogenous equilibrium is present then what happens to the molar concentration and volumes

A

-molar concenraions for solids and pure liquids do not change BECAUSE THEIR VOLUME REMAINS CONSNA .

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96
Q
A
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97
Q

how does expression of rkc change with heteroegenous equilibrium

A

-the concentraions of solid substances are not included in the expression .
-the concentrations of pure liquids are not included in hte xpression .

e.g – when phosphorus reacts with oxygen , following hetereogeneous equillibrium is esbalished check p.g 25

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98
Q

example of a heterogenous catalyst - when phosphorous reacts with oxygen , the following equilibrium is established .

A

P4(s)+502P4O1-

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99
Q

What is a mole fraction ?

A

The mole fraction of a substance is a measure of how much of a given substance , is present in a reaction mixture .

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100
Q

What is the partial pressure ?

A

The partial pressure of a substance is the pressure an individual gaseous substance would exert if it occupied a whole reaction vessel on its own.

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101
Q

How is the mole fraction of a substance calculated ?

A

mole fraction Xa = number of moles of substance A / total number of moles of all substances .

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102
Q

How can we calculate the partial pressure if we know the total pressure ?

A

For substance A ; partial pressure , PA = mole fraction * total pressure
The concentration of a substance is proportional to its partial pressure .

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103
Q

round brackets are used for …

A

partial pressures

104
Q

square brackets are used for …

A

concentrations

105
Q

check end of page 26

A

equilibrium expressions can be written using partial pressures instead of concentrations .
Check how to write kp .

106
Q

go through the worked example on page 27

A

…

107
Q

what is the expression for kp when a reaction is heterogenous it will alter as dollows

A

-solids will not be included in the expression .
-pure liquids will not be included in the expression .

108
Q

what does the magnitude of the equilibrium constants kc and kp indicate ?

A

they indicate the extent of a chemical reaction .

109
Q

what would an equilibrium constant with a value of 1 indicate

A

that the position of equilibrium is halfway between reactants

110
Q

when k is greater than 1 what does it suggest in an equilibrium constant

A

the reaction is product-favoured .
-the products on the right hand side predominate at equilibrium .

111
Q

when k is less than 1 what does it suggest in an equilibrium constant

A

the reaction is reactant favoured .
the reactants on the left hand side predominate at equilibrium .

112
Q

how does an increase in temperature shift the position of k .

A

an increase in temperature shifts the position of equilibrium in the endothermic directions .

113
Q

how does a decrease in temperature shift the position k ?

A

a decrease in temperature shifts the position of the equilibrium in the exothermic direction .

114
Q

what does the ^_ H values get affected in an equilibrium

A

they have the same magnitude but have opposite signs .

115
Q

what is the equilibrium constant , controlled by ?

A

rate constant , which changes its vale only with changes in temperature . When any other changes occur to conditions k will remain constant .

116
Q

what will happen to k , if the forward reaction is endothermic .

A

the equilibrium , yield of the products on the RHS , increases .
The equilibrium yield , of the reactants on the LHS , decreases .

117
Q

what will happen to k , if the forward reaction is exothermic

A

-the equilibrium , yield of the products on the RHS , decreases .
-equilibrium yield of the reactants on the LHS , increases .

118
Q

k increases when
k decreases when

A

k increases as the temperature rises if the forward reaction is endothermic .
-k decreases as the temperature rises if the forward reaction is exothermic /

119
Q

changes in concentration - exam Q –> if N204 , is doubled from 0.200mol dm-3 to 0.400moldm-3 (1)

A

-work out the kp expression which now gives the ration x .
-the system is no longer in equilibrium .

120
Q

changes in concentration - exam Q –> if N204 , is doubled from 0.200mol dm-3 to 0.400moldm-3 (2)

A

the equilibrium position must shift to restore th ratio to the Kc value of x . The system must ;
-Increase x - on the top expression for Kc .
-Decrease x on the bottom of the expression of kc .

121
Q

changes in concentration - exam Q –> if N204 , is doubled from 0.200mol dm-3 to 0.400moldm-3 (3)

A

-This will cause a shift in the equilibrium , position from the left to right , the same will be true , for the rate constants Kp , as t his will also remain constant when conc or the equivalent partial pressures are changed .

122
Q

what will changes in pressure do to kc - if the pressure is doubled , what will happen to the concentrations of both gases

A

it will also effectively double , .

123
Q

learning tip for pressure and volume - being inversely proportional

A

if the pressure is increasing , and there are no changes in temperature , or concentration therefore then the volume must be decreasing .

-concentration and volume can be linked by n=cv .

-If the number of moles stay the same , and volume decreases , concentration must increase .
-The molar volume of any gas is proportional to its partial pressure , so increases concentration = increases partial pressure .

124
Q

what if the value of kc is now 25.6

A

the system is no longer in equilibrium so the equilibrium positions must shift , to restore the ratio to the kc so the value of 12.8 , the system must ;
-Decrease of NO2 on the top
Increase in N204 on the bottom .
-Causing shift in the equilibrium position form right to the left , this will also be true for the rate constant Kp.

125
Q

how does the presence of catalsyt affect k

A

it dosent for kc and kp ,
-it affects the chemical reaction but not the position of equilibrium .

-Catalyst speeds up both the forward and reverse reactions in the equilibrium by the same factor . Equilibrium , is reached more quickly , but the equilibrium position and hence the value of the equilibrium constant is unchanged , by the action of a catalyst.

126
Q

What is a Bronsted lowry acid

A

it is a proton H+ donor .

127
Q

what is a Bronsted lowry base

A

it is a portion H+ acceptor

128
Q

what does the Bronsted lowry base model show

A

-the reactions between acids and bases states that ;
-reactions between acids and bases involves the transfer of H+ ions (known as protons ) .
-What is a Bronsted lowry acid - is any substance that can donate a proton .
what is a Bronsted lowry base - it is any substance that can accept a proton .

129
Q

Models for cids and bases over time (1)1770S

A

-In the 170s many chemists were investigating air and gases within it . Joseph Pruestky and Jarl Scheele both reported findings that suggested the existence of the gas we know now as oxygen .
-Another chmist , Antoine - Laurnet , de Lavoiser , found among other things , tht this gas was important int he rusting of metl and in 1778 , he proposed that it was the source of acidity . Lavoiser was wrong , but it was an early step towards , the understadnig of aids .

130
Q

in 1815 What did Humphtry Davy show

A

that some acidicacuduc substances such as HCL did not actuall contain oxygen .

131
Q

what did justus liebig do in 1832

A

he defined an acid as a substance containing hydrogen , that could be repalced by a metal - close to todas definiton .

132
Q

What happened int he alte 1880s ,

A

Swaante Arhenius proposed that acids dissociated in water to form the hydrogen ios H+ . and that bases dissocited in water to form hydroxide ions , OH- this is true , fo rmany acids and bases and they can be referred as Arhenius acicdsa nd bases , However , the model breaks down when the acids and bases are not solution sin water or when bases are not soluble hydroxides .

133
Q

what happened in 1929

A

the Bronsted Llowry definiton discussed , above was proposed by the Danish chemist , Johannes Bronsred , and Brithish chemist thomas lowry .

134
Q

exampleof bronsted lowry of acid in acrion ,
bronsted lowry acid hcl

A

for hydrogen chloridw HCL
HCL — H+ CL-
hcl has donated a H+ ion it i a bronsted lowry svif .

135
Q

exampleof bronsted lowry of acid in acrion ,
bronsted lowry acid h2s04

A

sor sulfuric acid
H2s04– H+—HS04-
-H2s04 has donated an H+ ion , it i a bronsted lowr aicd , h2s04+ can actually go on to donte another H+ ion , (sulfuric acid is dibasic ) this will e covered alter .

136
Q

bases accept protons - give an example o a bronsted lowr base

A

For ammonia , NH3- , NH3+H+— NH4+
-The ammonia has accepted (h+) , IT IS A BRONSTED WORY BASE .

137
Q

Why does a bronsed acid base invovle ionic equations

A

because it involve the transger of H+ ions , as they re usuall epresnted by ionic equations .

138
Q

what do ionic equations show (1)

A

-the H+ ions being donted
-tje ions tht is acepting the H+ ion

139
Q

what do ionic euqations show (2)

A

-the product is formed
-all changes correctly balanec (they cancel out )
-state sumbolds for all substances inclduing ions .

140
Q

what is the ionic equation for the rection between hydrochloric aicd and sodium hydroxide

A

H+ +OH- – H2-

141
Q

what are monobasic aicds

A

HCL IS A MONOBASIC AICD BECAUE EACH MOLEUCLE CAN RELEASE ONE PROTON .
hcl– h+-CL-

142
Q

what are dibasic acids

A

H2S04 , is a dibasic acid becsue each moelcule can release two protons . This is done in two stages
-H2sO4 – H+ (aq) +Hs04
-HSO3 – H+ +SO42-

143
Q

What are tribasic acids

A

Tribasic acids are where each moleucle can release three protons this is done in three syages
H3PO4 – H+ + H2PO4
-H2PO4– H+ +HPO42-
-HPO42- — H+ + PO42-

144
Q

What are conjugate acid bases

A

an acid base pair is a set of two species that transform into each other by gain or loss of a proton .
-F1 - gives an example of conjugate aicd-base pair , for the dissociation of a nitrous acid HN02 . It also shows how the acid and bases in acid - base pair are linked by H+

144
Q

how many acid - base pairs do ewuilibria inovvle ,
-give an example reaction

A

acid - base equilibria invovle two acid - base apirs . The equilibbrium below shows the dissociationf of ntiorus aicd HN02 in water .
HN02 +H20 -_ , H20+ + NO2-

144
Q

What does the frowad reaction show for acid base equilibria

A

-The acid HN02 releases a proton to form its conjugate bases H2- .
-tHE BASES nO2- accepts the proton to form its conjugate acid HN02 -

145
Q

What does the frowad reaction show for acid base equilibria (2)

A

-The acid H30 + reeleases a proton to form its conjugate ase H20 .
HN02 and NO2- accepts the proton , to form its conjugate acid HNO2- .

145
Q

pair one

A

HN02 AND NO2- differ by H+ and make up one aicd - base pair

146
Q

pair two

A

H3-P+ AND H20 differ by H+ , and make up a second acid - base pair .

147
Q

What is an alkali

A

it is a vase that dissolves in water forming OH ions .

148
Q

What is neutrakisation

A

It is a cgenucak reacuib ub which an acud abd a vase react together to produce a salt and water .

148
Q

What is a strong acid

A

is an acid that completely dissociates in solution .

149
Q

what is a weak acid

A

is an acid that partiall dissociation in solution .

150
Q

What is the acid dissociation constant Kp

A

check sheet it is of an acid Ha is defined …

151
Q

what -pKa

A

-log 10 Kp

152
Q

What is Kp

A

10-pkA

153
Q

typical acid base reactions

A

-aqueous takes part in typical acid base reactions with carbonates bases and alkalis .
-In all of these reactions a neutrlisaiton reaction occurs and water is formed as one of the products .

154
Q

Learning tip one - figure 1 - when acids release protons in water

A

-when hcl is added to water for example whaat reaction occurs
HCL+H2O– h20 +CL-
-H+ ABD H20+ can be used interchangeaky when describing aqueous acid solutions and H+ is usually used in ionic equations fro acid -nase reactuibs wgucg us wgt tge a vive reaction can be simpliied

HCL– H+CL-

155
Q

Reactions with carbonates
-In the reactions of acids , that follow ionic equations are used to show the important role of the H+ ions .

A

-The ionic equation show only H+ . from the acid . This H+ , could be provided by hdyrochloric acid HCL , ntrici aicd HNO3 , or sulfuric acid H2S04 .
-ThESE ACIDS have the same ionic equaiton for each type of reaction becasue they ll release H+ ions , whent hey disovl in wter .

156
Q

Reactions with carbonate part two - auqesous acids can react with solid carbonaed to form - a salt , carbon dioxide and water .

A

2HCL – aCO –> Cacl2 + Co2 ,, +H20 .
all ions ;
-2H+2CL- check the PAGEEE UHH H FOR CANCELATIONS

THen the ionic euation for that
2H+– CACI2 –CA3+ +CO2+H20

157
Q

what should be the sum of charges on either side of an equation

A

-the sum of the charges on either side of an ionic equaiton need notbe zero , but must be the same on the two sides . In this xample , the two H+ ions are balanced by one Ca+ ion - there is a charge of ++ one ach side of the ewuation .

158
Q
A
159
Q

If the carbnate is in solution , what does the final ionic euaion simplife too when the carbonate is dissociated

A

check to see the full euation an then wehn all the ions cancel out , the ionic equation .

160
Q

should there be state symbols in ionic ewuations

A

yes , this is especially important when an acid reacts with a solid as solids will not be dissociated into ions .
-Notice , in the ions , euaation the species which remian unchanegd are being called out , there are spectator ions .

161
Q

aqueous cids reacting with bases - forming salt and waer
-dheck the full equation
-and then the ionic equaiton

A

-full equaiton .
-ionid equation .

162
Q

aqueous acids recting with alkali forming a salt and water

A

-Ful Equation .
-Ionic equationnn check page

163
Q

the reaction of a metal with an acid is what kinda rection

A

-it is redox resction , so does not fit in with acid-base model . Howebeer , theya re a coommon , acid reaction .

-The generl equation ;
acid+metal– salt+hydrogen

164
Q

aqueous acids recting with metlas to form show in the equation
ful equatin and ionic equation , but hwat is the probelm

A

although most acids react with metals in this way , you have o be carefuls .
-Some acids such as sulfuric acids and nitric acids are powerful , oxidising agents , and othe reactions ma also take place especiall when the acids are cconcentrated .

-CLEAR AWAY FROM H2S04 AND HNO4 when giving examples of cid reactions iwth emtals .

165
Q

strong acids n weak acids

A

-check page 34
-in aqueous solution , acid dissociate and an equilibrium , is set up . The equilibrim shows the dissociation of an acid HA in water .

-the strength of an acid HA is the extent of its disscoaition into H+ and A- ions .

166
Q

what is the difference between strength and concentration - LEARNING TIP

A

-Make sure youa re clar about the differences between strength and concentration .
-STRENGTH is the extent , an acid dissociate into H+ and A- .
-Concentration , of an acid is how many moles of the acid are present in a given vlume .

-It is perfectly possible to hae a dilute soltuion of strong acid and a concentrated soltuion of a weak acid .

167
Q

SA1

A

-HCL - hdrochloric acid

168
Q

SA 2

A

HN03 nitric acid

169
Q

SA 4

A

H2S04 - SULFURIC AICDS
HI HYDROIC ACID
HCLO4 - CHLROIC (VII) PECHLORIC ACID

170
Q

how do weak acids dissociate

A

a weaj acud, only partially dissociates in an aqueous solutions , man naturally occueing cids are weak .

171
Q

what is the equilibrium et up when ehtnaoic aicd reacts with water

A

CH3COOH ____ H+ CH3COO-
-The equilibrium position lies well over to the elft .

-There are onlys mll concentrations of dissocaited ions , H+ and CH3COO- compared witht he cocnentration of undissociated CHCOOH .
-We can actually say that CHCOO- is a very good base , ti i er good at accepting the dissociated H+ back .

172
Q

what is the acid dissociation contant Kka

A

-the actul extent of aicd dissociatio is measured by an equilibrium constant called the acid dissociation constnat Ka .

  • weak acid Ha , has the following equilibriumin aquesous soltuoin ,

chck pahe for more .

173
Q

what dies a large ka value indiciate

A

a large extent of acid dissociationthe acid is stong .

174
Q
A
175
Q

what does a small ka value indiicate

A

a small extent of dissociation , the acid is wek .

176
Q

pka is ore manageable than ka , a ka is vsats so what are the two values of ka

A

pka=-logK
-kA =-pKa

177
Q

-whatd oes a low value of Ka mean

A

high vaue for pKa

178
Q

what does a high value of Ka match with

A

a low value fo rpka

179
Q

the smaller the pKa value …

A

the stronger the acid .
-check table 2 .

180
Q

how to work out pH with hdrogen ions

A

-log (H+)

181
Q

how o work out ph with hdyroge ions

A

-Pj

182
Q

skim page 36 and worked example and 2 .

A

…

183
Q

How to calculate the Ph of strong acids

A

-A strong monobasic aicd HA , has virtually complete dissociaiton inw ater .

Meaning hydrogen of h+ of a strong acid is approimatelye wwul to the cocnentration of the acid HA
-hdyrogen ion = HA .

U CALCUATE PH JUST BY -LOG check worked example 4

184
Q

calculating the Ph of weak acids is slightly mroe difficult why

A

as a weak monobasic aicd only partialll dissociaites thereofre , WHEN SETTING UP THE EQUILIBRIUM .
-We can no longer assume hdyrogen+ is ewqual tot hte concentration of the aicd , we needto work out the vle of K for the reaction - to find the extent of disscoiationt hat has occured to dictate the value of H+ .

Ka= (h+)(A-)/HA

184
Q

things to know about calculaitng weak acids (1)

A

for weak acids (H+) is much less than HA because the extent of dissociation is so SMALL .
-When HA moleucles dissociate , H+ and A- ions are formed in equal quantities and it is iassumed that any water present will l hae dissocited to such a negligible maount htat it will not affect the cocnentration of H+

184
Q

thigns to note about calcualting Ph of weak acid (2)

A

thereofre h+ and a - can e consdiered equal . -This means the expression H+ and A- in the Ka , cslculation becomes the H+)2/

185
Q

as the ha moelcules have dissocited (HA) WILL HAEVE REDUCES SLIGHT SO WHAT DOES THE EQUILIBRIUM COCNENTRATION BECOME

A

CHECK PAHGE 37 FO RHIGY

and hen worked exampel 4.

186
Q

check

A
186
Q

the leanign tip on pGW

A

38

186
Q
A
187
Q

What is the limiition of the scientifc approximation

A

-when calclaitng the Ph ofor a weak acid , it is assumed that so ie , of the originl cids , has dissocitedtha tthe cocnenration of the acid a the equilibrium is ffectiely the ame as the cocnentration of the oiginal amount o aciid .

-THIS CAN BE EXPRESSED AS HA EQUILBRIUM = HA undissociated

188
Q

What is the limiition of the scientifc approximation (2)

A

it is actually the case that there will be some acid , whidh does dissociate even if only for a smll amount . If this is the amoutn less than 5% of the total for Hha IS UNDISSOCIATED it , then it deemd for safe ,t ouse the abov approximation this iwll be the case for very weka acids iwll very low Ka vlaues .

188
Q

what isntuemnts do you have to use to calcuualt ek for a weaka cid

A

use a ph meter - also need the cocnentraiton o the weak acid - see worked example 5 .

189
Q

What is the ionic product of water

A

Kw is defined as Kw = H+OH- at 25 degrees = 1.00*10-14

190
Q

is water an acid or a base

A

water cn act as a n acid by donating a proton
water can act as a base bya ccepting a proton .
-This back and fourth transfer of porotns happens conitnously in water . In fact ,w ater exists t euilibrium ,,, known as ionisation of water .

-EUWILIBRIUM LIES well to the elft , and onlya n extremel tinya mt of wate ris dissociated at aigven time .

chekc forkc expression .

191
Q

as the amount of dissociaiton , ionisaiton of water is so small what does thi mean

A

it is consdiered a contant like kc , we rearrange the equation as follows to combine hte two cosntants

Kc*(H20) = (H+)(OH-)
-This can be furhter simplified by reerring to the expressio kc *H20 as a new cosntant KW .

192
Q

Check sheet fo rhte implfiyin gof kw , what is kw

A

the ionic product of eate r.
-the units of kw are always that ebcuae

-25d egress the measured ph of water is 7 and hydrogen =10-7mol .

-when water ionises the samennmber of h and oh ions are relaed

193
Q

meanong of that the cocnentraio of OH- ions are also equalt o 10-7 mol

A

check sheet for equation .

the values o h+ and oh - are only eqal if the water is pure and neutral .

194
Q

how can the balance beteee nH and OH be afected

A

byt he addition of extr H+ or Oh- ions that cause the equilibrium to shift . Remember that as with all euqilibrium follow le chateiller the equilibrium wills hift to restore value of kw , and changes only if TEMP IS ALTERED .

KW CONTROLS BALNCE BETTWEEN H+ AND OH- IN AQUEOUS SOLTUIOSN .

195
Q

at 25 degrees wgt is the ph value of 7 in terms of h and oh-

A

it iss the neutral point at which h+ and oh- cocnwntrations are the same equal 10-7 , moldm-3 applying to water and also to neturals oltuions .
relaativve cocnetration sof h+ and oh- are determined by kw .

196
Q

at 25 degrees is rain wwater neutral

A

-at 25 dehreees rainater is actually acidic , carbon dioxide dissovles inw ater to forma weak acid H2C03 , which is able to dissociate and release H+ ions into the water .

197
Q

in rain water when H+ has icnreas OH ions will fall until

A

H+ and OH- = 1.00x10-14 .
-The overall efect is that Ph decreases ,r ainwter ph is abt 56 ,

mienral water fo rexmaple contains dissovled ions suhc as carbonate co32- , these ions reat with acid and thus remoe H+ ions at a lower (h+) .
(h+) , has creases (oh-) , must increaseuntil

(h+)(oh-( =1.00*10-14mol-14mol2dm6 at (25 degrees )
-The overalle fecct is that Ph increase . tHE ph OF MINTERL wate ris 7-8 abd tgat nebs nuebrak water us akjakube ,

198
Q

check table 1 on page 40

A

…

199
Q

strong bases tend to do what give example
weak bases tend to what give an example

A

strong ases tend to be hdyroxies of the emtals in group1 and 2
ammonia nh3 is a weak base , in aqueous soltuion n equilibriu is set up and the equilirium postion lies well to the LHS .

200
Q

How do you use Kw , to calculate the Ph of strong abses

A

-to workout the pH of stong bses we need to know H+ and this depnds on
-the cocnentriton of the base
-the ionic product of water Kw = 1.00*10-14 (because th is links with (h+) AND (Oh-)

201
Q

a strong monobasic alkali - give an example of this

A

NAOH is comeptlelu dissocited in aqueous solution . This means that (oh-) of a strong base is equal to the cocnentratio of the basw
NAOH —NA+ +OH-
SO (Oh-)=(NAOH)
we can find (H+) from kw and (Oh-)
kw=(H+) and (oh-).
Kw =(h+)(0h-)
(h+) =kw/(oh-)
the pH canbe calcualted using pH=-log10(H+)
see worked exmaple of 41 ., also check learning tip .

202
Q
A
202
Q

What is a buffer solution

A

it is a mixture that minimised pH changes on addition of small amounts of acid or base . The word minimises is essential to this defnition .

202
Q
A
203
Q

a buffer solution cannot prevent the pH from changing slightly but …

A

pPH changes are minimised , at least for as long as the buffer solution reamins .

204
Q

give an example of how a buffer can be made

A

-it can be made from a weak acid and a salt of the weak acid , for example ehtnaoic acid (ch3cooh) and sodium ehtanoate (ch3coona)

204
Q

what is a buffer solution a mixture of

A

a weak acid HA
and its conjugate base A-

205
Q

whta sissociates fully and whagt dissociaes partially in he ch3cooh and ch3coona uffer sytem

A

check page 42 .

-the equilibrium mixture formed contains a high concenraiton of the undissociaed weak aci CH3COOH , and its conjugate abse CH3COO- . The high concenration of the conjugate base pushes the equilibrium ot he elt , so the concentraion of H+ ion sis ver small .

The RESULTING buffer soltuion contains alrge resvoirs of the weak acid and its conjugate base . seen in figure 1 .

206
Q

can buffers be made from a weak acid and a strong alkali

A

-In this situation , a solution containnig a mixture of the salt and excess of weak acid is formed . For exampe , a weka acid sucha s methanoic aic d, HCOOH ,can be partially neutralised by an aqueous alkali such as NaOH .

207
Q

How does a bufffer act to cotnrol pH

A

In a buffer solution ,t he weak acid HA ,a nd the conjugate base , A- , are both responsible for controlling pH . The buffer solution minimises pH changes b ussing the equilibrium

HA —- H+ + A-

OVERALL PRINCIPLE BEING

-the weak acid , HA removes added alkali
-the conjuage bas , A- removes added acid .

208
Q

In a buffer , what does the addition of an acid H+ do to a buffer solution

A

(H+) IS INCREASES
THE CONJUAGTE BASE a- REACTS WITH H+ ions .
-the equilibrium shifts to the elft removing msot of the added h+ ions .

tthe euiqlibirum shifts to the elft HA— H+ + A-

-On addition of an alakali , OH- to buffer mixture
-OH- is increases small cocnentraion of H+ ions react witht he OH- ions .

209
Q

Ha dissociates shifting the equilibrium tot hre right to resotre m ost of the H+ ions that have reacted

A

equilibriu shifts to the right
HA— H+ + A- .

These actions of a buffer are summarised in figure 2
HA — H+ +A-

left added acid right added alakli

210
Q

true orfalse , a buffer cannot remove all of anya cid or alkali that is added

A

TRUE IT JUS MINIMISES THE pH changes .

210
Q

why do we need to check the concentrations of HA and A -

A

as only a very small proprotion o HA dissocites , so , as disucssed in topic 5.1.11 onn calcualitng hte pHo f weka acids we can ssume that Ha equilibrium = HAA undissociated .

210
Q

lculaitng the phH OF BUFFER SOLUTIONS

A

THE pH OF A BUFFER SOLTUIOND EPEDNS ON
-tHE ACID DISSOCIATION CONSTNAT kA , OF THE BUFFER SYTEM .
-THE COCNENTRATION RATIO FO THE WEak acid and its conjugate base .
-gor a buffer consiting of a weak aci HA and its conjuagete base A-

Ka =H+ A-/HA

menaing h+=ka* HA/A

211
Q

what does the salt of the weak acid do

A

it is ionic , and sissociates completel in aqueous solution show example pn page 43 .

211
Q

check learning tip and workd example on page 43

A

…

212
Q

what is the equivalence point

A

it is the point ina titration at which the volume of one solution has reached exxlt the point of volume in the second solution .
This matches the stochiometry of the reaction taking pace .

213
Q

what is the end point in a titration

A

the end point in a titration at which there are equal concentrations of the weak acid and conjugatebase forms of the indicagotr . The colour at the end point is midway beween the colours of the acid and the conjugate base forms .

214
Q

when you carry out a titration , what are you determining

A

you are determining the volume of one solution that reacts exactly with a known volume of another solutoin knowna s the equivalnece point of the titration . At this point ,t he solution in the concical flask has exactly rected with the solution int he burette .

215
Q

what waycan titratios be carrried out aicd to base terms

A

acid to base or base to acid - same pricniples apply to both routes .

216
Q

check figure 1 section 1 on page 45 what is it showing

A

A slight increase in pH occurs as base is added , because the acid is in such excess . It is importaant to note that the section of the curve is not horizonral , pH is rising slightly .

217
Q

check figure 1 section 1 on page 45 what is it showing

A
217
Q

check figure 1 section 2 on page 45 what is it showing

A

the sharp rise in PH occurs the acid is no longe rin excess so any base added has a large impact on the pH .

218
Q

check figure 1 section3 on page 45 what is it showing check figure 1 section 1 on page 45 what is it showing

A

a slight increase in pH occurs as a further base i added . The increase is only slightly because the base is in excess , now and extra base has had little impact on the pH .
-It is importanat to note however that this section is not horizontl - pH is rising slightly .

218
Q

check figure 1 section equivalence point on page 45 what is it showing

A

the equivalence point is the centre of the vertical section of the titration curve

218
Q

ttitration curces

A

pH meters or data loggers can be used to mesure the pH of the reactionmixture , as solution from the burette is added over time .
-The vsarying pH values can then be plotted as grpah , nown as an acid-base pH titration curve . THe cure mabe plotted by hand or computer pH data logger .

219
Q
A
220
Q

Typical cid -base titration

A

pH curve is shown in figure 1 for the reaction between a strong acid and strong base .

CURVE HAS THREE DISTINCT AREAS .
1-A slight increase in pH
2- a sharp increase in pH
3 - a slight increase in pH .

221
Q

wh at i an acid base indicator

A

an acid base indicator is a weak acid often represented as Hln . An indivaotrhas one colour in it acid formHln and a different colour in its conjugate base form IN-

222
Q

Investigation - purpose of a pH meter

A

-an indicator such as universal indicator can only give a general indication of a pH solution where as Ph meter can be accurate two 2 d.p

223
Q

pH meter stage one

A

calibrate so the results are accurate .

224
Q

pH meter stage two

A

-the probe is removed from its storage solution and rinsed with deionised water .

225
Q

Ph meter stage three

A

-The probe is blotted dry and then placed into a solution of a known pH such solutions are usually manufactured professionally .
Ph is allowed to settle before checking that the pH has registered .
-process is repeated for other known Ph’s .

226
Q

what colour is Hln

A

H+ and In-

227
Q

what happens when there are equal amounts of weak acid and conjugate base

A

the indicator is at its end point .

228
Q

how do you know that a titration has reached its end point

A

end points are accompanied by a visual colour change that allows for a titration to be monitored easily . For example ,t he colour of the end point for methyl orange is orange , at which point HIN (red) and (In-) are eqqual .

229
Q

most indicators change colour over what range

A

of about two pH units .
-The pH ranges for the indicators shown in figue 2 on page 46 .

230
Q

How to use the correct indicator for a titration ?

A

-Choose a pH value close to the end point as possible to the pH value of the titrations equivalence point .
-In practice , a suitable indicator , changes colour within the pH range in the vertical section of the titration curve .

231
Q

For the following curves , the ranges of the indicators are not actually on the equivalence point , so how can you show that the reaction has reached this point ?

A

As the equivalence point , occurs halfway through an incredibly sharp rise or fall in pH , the moment at which the indicator passes through its end point is effectively the same as the moment that the equivalence point takes place .

232
Q

STRONG ACID STRONG BASE TITRATION - check figure 3 on page 46

-What points can we notice on the curve

A

-The vertical section of the graph covers a large change in pH , starting around pH 3 , and ending around pH 11 , with an equivalence point at pH 7 .
-Both the indicators , methyl orange and phenolphthalein have end points that fall within this pH range .
-Either indicator would be suitable to use in a titration between strong acid and a strong base .

233
Q

STRONG ACID - WEAK BASE TITRATION - figure 4 on page 47
-What points can we notice on the curve .

A

-The vertical section of the graph , covers a small change in Ph starting around pH 3 , and ending around pH 7.5 , with the equivalence point occurring at more acidic value , i.e. at a Ph lower than 7 .
-Methyl orange has an end point that galls within this Ph range .

234
Q

STRONG ACID - WEAK BASE TITRATION - figure 4 on page 47
-What points can we notice on the curve . (2)

A

-Phenolphthalein does not have an end point that falls within this pH range .
-One methyl orange would be a suitable indicator for a titration between strong acid and a weak base .

235
Q

WEAK ACID - STRONG BASE TITRATION - figure 5 on page 47 (1)

A

-The vertical section of the graph covers a smaller change in pH , and occurs further up toward the higher pH values , starting around pH 6.5 , and ending around pH 11.5 .

-The equivalence point occurs at a more basic Ph i,e pH value above 7 .

236
Q

WEAK ACID - STRONG BASE TITRATION - figure 5 on page 47 (2)

A

-Methyl orange has an end point that falls outside this pH range .
-Phenolphthalein has an end point that falls within this pH range .
-Only phenolphthalein would be a suitable indicator for titration between a strong acid and a weak base .

237
Q

WEAK ACIDS - WEAK BASE TITRATIONS (1) FIGURE 6 47

A

-There is no real vertical section .
-Neither indicator has an end point near the equivalence point .
-Neither indicator suitable .

238
Q

WEAK ACIDS - WEAK BASE TITRATIONS (2) FIGURE 6 47

A

-An indicator would change colour gradually , over a few cm3 , of base added .
-No indicators , are really suitable for a weak acid - weak base titration .

239
Q

Investigation for titrations summary (1)

A

-Set up titrations between acids and alkalis of different strengths . You should select some weak acids and bases and some strong acids and bases .

240
Q

Investigations for titrations summary (2)

A

Carry out titrations using various combinations of weak and strong acids and bases . Use a pH meter to record the pH throughout the titrations . Use your results to plot acid-base titrations curve . Identify the equivalence points for each titration .

A-level CHEMISTRY (11 decks)

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