Acid-base Equilibria Mark Scheme

8/11/2019 Acid-base Equilibria Mark Scheme

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ANSWERS TO ACID BASE EQUILIBRIA EXAM QUESTIONS

1. (a) (i) A proton donor (1)

(ii) Fully ionised or fully dissociated (1)

(iii) 10 1014

(1)

mol2dm6 (1) 4

(b) (i) 150 10325 1000 (1)

! 006 mol dm3 (1)

12 (1)

(ii) "ol #$added ! 50 01501000 ! %5 103

(1)

"ol $

&

used ! 15 10

3

(1)

"ol #$e'cess ! 60 103

(1)

#$! 60103%51000 ! *0 102 (1)

$& ! 1014*0102 ! 125 1013 (1)

p$ ! 12+ (1)*

(c) (i) 03 mol dm3(1)

(ii) $& at p$ ! 0% is 02 mol dm3(1)

m1,1! m2,203 25 ! 02 , (1)

$ence , ! 3%5 (1)

-ater added ! 3%5 25 ! 125 (1)5

[17]

2. (a) (i) Ka!$AA$ +

(1)

Ka!$A

$ 2+

(1)

$& ! 102.*2! 1.514 103mol dm3 (1)

Ka !

( )15.0

10514.123

! 1.53 105 (1) mol dm 3(1)2

(ii) /ecreases (1)

uilibrium sifts to rit (endotermic process ) (1)


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$& as (1)

p$ ets smaller *


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(b) (i) 3

2

n (1)

(ii)

nn3

1

3

2

(1)! 2 (1)

(iii) Ka ! 2$

$$

++

= (1)

! 2.1 104mol dm3 (1) 5

(c) -ea acidstron base p$ at eui,alence >% (1)

metyl orane as colour cane at p$


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4. (a) Definition of a base ?roton acceptor (1)

Essential feature ransfer of protons (1)

Equation $&& #$ $2##@

$&& 7 7$&(1)3

(b) only partially dissociated in solution (1) 1

(c) Ka!##$(a)$$

(a)##$(a)$$

23

23

+

(1)

mol dm3(1) 2

(d) (i) resists cane in p$ (1)on addition of small amounts of stron acid or base (1)

(ii) correct 8ea acidcoBbaseor correct 8ea basecoBacid (1)

(iii) any suitable use (1) 4[10]

5. (a) (i) 0.12 11.* ! " 25 (1)molarity ! 0.05% (1)

(ii) C a!$A

A$ +

(1)

(iii) Volume of NaOH(aq) added 11.*2 ! 5.+ cm3(1)

pH 4.3 to 4.35 (1)

(i,) As $A ! A (1) Ca! $& (1)

p$ ! lo10$& (1) ence Ca! 10

4.3

! 5.0 105(1)Dote: "ar Caconseuentially to p$ in a(iii) +

(b) (i) e added #$reacts 8it $A or $&(1)

e euilibrium= $A $&& A= displacement to rit

or $A ionises (1)

(ii) e added $&reacts 8it A(1)

e euilibrium= $A $&& A= displaced to left (1) 4[13]

. (a) # 2

3 & $&$#3 or Da2#3& $l Da$#3& Dal (1)

$#3 & $&$2# & #2or Da$#3& $l Dal & #2& $2# (1)or $2#3

2

(b) 15 cm3 1


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! 15 0.34 103! 5.10 103 (1)

Enitial moles $A ! 25 0.45 103! 0.01125 (1)

"llow #$#%%# & #$#%%'

"oles Da#$ added ! 0.00510

"oles $A remainin ! 6.15 103

(1)

"llow ($# & *$+#) ,%#&'

-ar. conseq

(ii) 5.10103moles Ain (15 & 25) cm3

$ence A ! 5.10 103 100040 ! 0.12%5 (1)

"llow#$%+/ $%+0 and #$%'

6.15 103moles $A in 40 cm3

$ence $A ! 6.15 103 100040 ! 0.153* (1)

"llow #$%1/ & #$% and #$%

Allo8 mars in (ii) conse to ans8ers in (i)

(iii) C a ! $& A $A ! 2.00 104

$& ! 2.00 104 0.153* 0.12%5 (1)

! 2.41 104 (1)

"llow (+$+ & +$11) , %#&1

p$ ! 3.62 (1)

"llow '$*% & '$*1 and '$*

-ar. conseq to answers in (ii) +[10]

#. (a) only partially ioni


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En $En En increases (1)not9ust euilibrium sifts to rit 2

(iii) eui,alence point end point of titration belo8 p$ % more acidic lo8ertan penolptalein rane is about p$ 4 (1)not9ustte p$ rane is 8ron 1

[10]

1$. (a) (i) p$ ! lo(10) $& DoteH (a) not reuiredH Dot ln $& (1)

(ii) C a! $& $ DoteH (a) not reuired (1)

"llow 2"&3 and 2H"3

/o D# allo8 $&2$

(iii) C a! 4.25 105! $&2$ (1)

$& !51025.445.0 (1)

! 4.3% 103 (1)

not a conseq mar.

p$ ! 2.36 "ar conse to $& abo,e (1)

or p$ ! 21

pCa21

lo10$ ! 21

4.3% & 21

0.346 ! 2.36 (1) (2) (1)

Dote p$ ! 2.4 scores ma' 3 6

(b) (i) $& ! 0.25 0.+5 ! 0.23%5 (1)

"llow #$+'/ & #$+'0 and #$+1

p$ ! 0.62 (1)

"llow #$*+ & #$*'

#nly allo8 p$ mar if $& is correct

(ii) $& ! I ! 0.23%5 (or a ,alue from b(i)) (1)

$I ! 0.05 0.25 ! 0.0125 (1)"llow #$#%+ & #$#%'

Ca! $& I $I

! (0.23%5)2 0.0125 (1)

Ca! 4.51 (1)

"llow 1$' & 1$0

I4nore units

CE if 2H53 is incorrect 6[12]


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11. (a) pC a& lo10 Ca 1

(b) C a ! 1.+0104(1)

Ca ! $&20.52 or $&! (1)

$& ! (1.+01040.52) ! +.+4103(1)

p$ ! lo10$& ! 2.00 (1)

or p$ ! J pC aJ lo $

! 1.*6 (0.142) ! 2.00 4

(c) Ca $& $ (1)

$ ! at alf neutralisation (1)

$ence Ca! $& and pCa ! p$ (1) 3

(d) ere is no rapidsarpsteep cane in p$ durin a 8eaacid B 8ea base titration (1)

Endicator need a sarp p$ rise to cane colour uicly (1) 2[10]

12. (a) (i) C a!##$$

##$$

3

3

+

(1)

(ii) (1) C a! ##$$

$

3

2+

(1)

(2) $& ! 220.010%4.1 5

! 1.+6 103(1)

(3) p$ ! Blo10$& (1)

can score independently

(4) p$ ! 2.%1 (1)2 d.p. essential

If forget can score (1) and (3) for pH = .42

5

(b) (i) moles acid ! 1000

25

0220 (1) ! 5.50 103

!310

x

0.150

x! 25 150.0220.0

o% 5.50 103 150.0

1000

! 36.% (or 3%) cm3(or 36.6) (1)!"# 3$ !"% 37.0&nits '&st 'atc

(ii) Indicator: tymol blue (1)Explanation: 8ea acid stron base (1)eui,alent at p$ >% (1)

or i p$


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5

(c) (1) mol Da#$ added ! 0.40

0.2

! 0.050 (1)If rong *r+ ,- lose 'ars (1) and (2) ten 'ar on

conse/&entially 'a 4

(2) mol $3##$ left ! 0.220 B 0.050 ! 0.1%0 (1)

(3) mol $3##formed ! 0.050 (1)

(4) $& ! Ca saltacid

#@ p$ ! pCa &

loH"

"

etc (1)If epression rong no 'ars for 4 $

can score (1) to (4) in ()

(5) $& ! 1.%4 105 )05.0(

)1%0.0(

#@ p$ ! 4.%6 & lo

1%.0

05.0

(1)

(6) p$ ! 4.23 (1),orrect anser gets*ar () is for &se of correct al&es of (acid 'oles) and(salt 'oles)if one rong allo pH conse/if ot rong no f&rter 'arse.g. if candidate forgets s&stit&tion in (2)e loses (2) and () &t can score (1) (3) (4) ($) = 'a 4

for pH = 4.12 if salt

acid

&pside don5 anser .26 scores 3

for (1) (2) (3)

6[1$]

13. (a) moles $A !310

25

0.150 ! 3.%5 103(1)

,ol Da#$ ! 20.010%5.3 3

! 1.*%5 102dm3(1)

or 18.7 c'3

2

(b) (i) p$ ! lo10$& (1)

(ii) Kalue abo,e % but belo8 11 (1)


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(iii) penol red tymol blue penolptalein tymolptaleini.e. indicator it 7 pin 11

3


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(c) (i) Only slightly dissociated (1)!"# 9not f&lly dissociated ionised:

(ii) C a!

H"

"H +

(1)

!"#

2

H"

H+

(iii) For 8ea acid alone:

Ca !

2

H"

H+

(1)

$& ! 15.0)10%5.2(5

! 2.03 103(1)p$ ! 2.6+ (1)pH so&ld e gien to 2 deci'al placespenalise anser to 1 d.p. once in /&estion

5

(d) moles #$added ! 1.*%5 103! moles A! moles $A left (1)o% A ! $ACa ! $& o%p$ ! pCa(1)p$ ! 4.56 (1) 3

[13]

14. ?enalise p$ i,en to 1 dp first time it 8ould a,e scored only

(a) (i) C 8! $&

#$

(1)

(ii) p$ ! lo $& (1)or in words or below unless contradiction

(iii) Calculation: $& !14104*.5 (1)

! 2.34 10%

p$ ! 6.63 or 6.64 (1)

Explanation: pure 8ater $& ! #$ (1) 5

(b) (i) #$

! 0.150 $&

! 1014

0.15 ! 6.66 1014

or p#$ ! 0.*2

p$ ! 13.1* (1)or p$! 13.1%

(ii) moles #$! (35 103) 0.150 ! 5.25 103(1)a

moles $&! (40 103) 0.120 ! 4.*(0) 103(1)b

e'cess moles of #$! 4.5 104(1)c

#$ ! (4.5(0) 104) 1000%5d(1)&

$& !3

14

1000.610 ! 1.66 1012or p#$ ! 2.22


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p$ ! 11.%* (1)f

or 11.%% *

(c) (i) C a![ ]$$ +

(1)

(ii) $

&

! 1.*0 10

2

0.150 ! 2.%0 10

3

(1)

Ca!

[ ][ ]

( )150.0

10%0.2

$

$ 232 +

=(1)! 4.*6 105(1)mol dm3(1) 5

or

( )14%3.0

10%0.223

6 1$ , %#&

Not&'

(a) Ef C 8includes $2# allo8 6.63 if seen oter8ise no mars liely

(b) (ii) Ef no ,ol= ma' 4 for a= b= c= f ans8r ! 10.65

Ef 8ron ,olume ma' 5 for a= b= c= e= fEf no substraction ma' 3 for a= b= dEf missin 1000 ma' 5 for a= b= c= d= f ans8er ! *.%*Ef uses e'cess as acid= ma' 4 for a= b= d= f ans8er ! 2.22Ef uses e'cess as acid and no ,olume= ma' 2 for a= b ans8er ! 3.35

(c) Ef 8ron C ain (i) ma' 2 in part (ii) for $& (1)and conse units (1)

but mar. on full7 from minor errors

e4 no 2 3 or c8ar4es missin4

[18]

15. (a) $ydroen bondin (1)bet8een $2# and D$3(1) 2

(b) (i) D$3& $2# D$4&& #$(1)

(ii) Ammonia is 8ea base (1)NO9 partiall7 ionised

uilibrium to left or incomplete reaction (1) 3

(c) A proton donor (1) 1

(d) :uffer solution: A solution 8ic resists cane in p$ (1)8en small amounts of acid or base added or on dilution (1)

!ea4ent: D$4l (1) 3"llow a correct stron4 acid

(e) (i) C a! $& A $A (1)

! $& 0.125 4 (1) 1.00$& ! 1.%0 105 0.125 4 ! 3.40 105(1)

p$ ! lo10$& ! 4.4% (1)

"llow pH conseq to 2H;3 if + place decimals 4i


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1. (a) C a!$A

A$ +

1("ll t8ree sets of square brac.ets needed= penalise missin4

brac.ets or missin4 c8ar4e once in t8e question)

(Don>t penalise extra 2H;3+?2H"3)

(b) C a !$A

$ 2+

or $& ! A1

$& ! 25.0)10(1.45 B4

! 6.02 103

1p$ ! 2.22

1(must be to +dp)

(allow 1t8 mar. consequential on t8eir 2H;3)

%

(c) (i) p$ (almost) uncaned 1(-ust be correct to score explanation)

$&remo,ed by Aformin $Aor acid reacts 8it saltor more $A formed 1

(ii) $& ! 103.5+! 2.5% 104or 2.6 104

1

A ! $

$ACa+

1

!4

4

105%.2

25.01045.1(

1! 0.141 (mol dm 3)

1("llow #$%' to #$%1% and allow #$%1)

(If not used '$= to find 2H;3 can onl7 score -+ for wor.in4)

(If '$ used but 2H;3 is wron4= can score -+ for correct

met8od and conseq -1)

If wron4 met8od and wron4 expression= can onl7 score -%)

(ii) "lternati


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17. (a) (i) pH = log [H+] (1)

(ii) Expression for Ka: Ca![ ]$

$ +

(1)

Calculation: p$ ! 2.56 $& ! 2.%5 103(1)

Ca![ ][ ]$$ 2

+

!( )

12.010%5.2 23

! 6.32 105(1) (mol dm3)

or $& ! (1) 5dependin4 on approximate made=


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(b) (i) lo$&H 1

(ii) $& ! 1.25+ 1012(or 1.26 or 1.3) O! p#$! 14 p$H 1

12

14

1025*.1

10#$

=

O!

! 2.10H1

! %.+(4) 103H 1(if 2H

;3 is wron4 allow % for 2OH3 6 KB?2H

;3 or as numbers)

(c) (i) C a! $&2$3$2##$

O!

$&2$A

O!

$& ! A etcH 1

$& ! l.35l05 0.11% or e'pression 8itout numbersH 1

! 1.25% l03

p$ ! 2.+0H 1

(iii) C a! $&

O!

pCa ! p$H 1

p$ ! 4.*%H 1(penalise %dp once)

[13]

1#. (a) lo $& 14.5% 103 1

allow 1$* , %#&'

ecf if 2 3 wron4 and alread7 penalisedi4nore units


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(b) (i) Ka= $$ B+

allow HA etc

not 2H@3

32H ++

but mar. on

Ef e'pression 8ron allo8 conse units in (ii) but no oter mars in (ii) 1

(ii) $$ 2+

! 1500)105%4(

23

1If use 1$* , %#

&'

Ka6 %$1(%) , %#&1

and pKa 6 '$0 1

! 1.3+ 104

allow %$' & %$1% , %#&1

mol dm3

(iii) pKa! 3.*6 1enalise dp of final answer or + in pH once in paper

(c) (i) 1000

30

0.4*0 ! 0.0144 or 1.4(4) 102 1-ar. is for answer (-%)

(ii) 1000

1*

0.350 ! 0.0063 or 6.3 103 1-ar. is for answer (-+)

(iii) 0.0144 2(0.0063) ! 1.*0 103 1-' is for (i) & +(ii)

If x + missed= CE i$e$ lose -' and t8e next mar. 4ained

(i,) 1.*0 103 4*

1000

! 0.03%5 (0.03*) 1-1 is for answer

Ef ,ol is not 4* 103(unless A) lose "4 and ne't mar ainedEf multiply by 4* B tis is A B i.e. lose only "4Ef multiply by 4* 103tis is A B i.e. lose only "4

(,) 1014 0.03%5 (1014 0.03*)1

- for K8?2OH3

(! 2.66 1013) (! 2.63 1013) 1or pOH

or p#$ ! 1.426 (or p#$ ! 1.420)If no attempt to use Kwor pOH lose bot8 - and -*

p$ ! 12.5% (12.5*) "6 1"llow -* conseq on "E in - if met8od OK

[13]


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2$. (a) (i) $$ 1

lo $& 1

(ii) $& ! #$ 1

(iii) (2.0 103) 0.5 ! 1.0 103 1

(i,) $& !3

14

100.1

1002.4

( ! 4.02 1011) 1

p$ ! 10.40 1

(b) (i) Ca ! $&$3$2## 1

$3$2##$

! $&1

$3$2##$

$& ! L(1.35 105) 0.125 ( ! 1.30 103)1

p$ ! 2.*+ 1

(c) (i) (50.0 103) 0.125 ! 6.25 103 1

(ii) (6.25 103) (1.0 103) ! 5.25 103 1

(iii) mol salt formed ! 1.0 103 1

($&) ! Ca $3$2##$ 1

$3$

2##)

! (1.35 105) K)100.1(

K)1025.5(3

3

( ! %.0** 105)1

p$ ! 4.15 1[1$]

21. (a) oncentration of acid: m1,1! m2,2ence 25 m1! 1*.2 0.150

#@

moles Da#$ ! 2.%3 103H 1

m1! 1*.2 0.15025! 0.10+H 1

(b) (i) Ka!$&A$A not

Ka! $&2 $AH

1

(ii) pC a! loCaH 1

(iii) A

! $AH 1enceKa! $

& A $A ! $&


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and loCa ! lo$&H 1

(c) ratio A : $A remains constantH1

ence as $& 6 Ka$A AH $& remains constantH

1

(d) (i) p$ of 0.250 mol dmB3$l ! 0.60

and p$ of 0.150 mol dmB3$l ! 0.*2H1

p$ cane ! 0.22H1

(ii) moles $l ! 30 0.250 103! , 0.150 103! %.50 103

#@

, !30 0.250 103 0.150 103!50H 1

8ater added ! 50 30 ! 20 cm3H 1[12]

22. (a) proton or $&donor (1) 1

(b) (i) partially ionised or dissociated (1)

not full7

(ii) D$3 (1)

not NH1OH & but allow in equation

not H+O & but allow in equation if bot8 wea. acid and base stated

D$3& $2#+4D$ & #$ (1) 3

(c) (i) $##$(a) & $2#(l) $##(a) & $3#

&(a) (1)

allow H+O(aq)

(ii) $2# or 8ater (1)

$##or metanoate ion (1)

(iii) $##$$##$ +

=aK (1)

allo8 #$3& 4

(d) (i) addition of small amounts of acid (1)addition of small amounts of base (1)

allow


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allo8 salt of metanoic (or tis) acid not 9ust an ion (metanoate)

(iii) #$added (1)

or base

$&reacts 8it #$ (1)

or formin4 water"ore $##$ dissociates to restore euilibrium (1)

allow equilibrium mo


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If no , + for :a(OH)+ t8en H;is in @F -"@ 1 ex *

If no


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m,1000 ! 0.*00 10.51000 ! *.4 103(1)

(ii) "ol $ remainin ! oriinal mol $ mol Da#$ added (1)(25 0.+21000) *.4 103! 0.0146 (1)

(iii) Concentration of @& *.4 103 1000(25 & 10.5)! 0.23% (1)

Concentration of H@ 0.0146 100035.5! 0.411 (1)

pH of solution Ca! $& $ (1)

$& ! 5.25 105 0.4110.23% (1)p$ ! 4.04 (1)

"ared conseuentially to b(iii)+

(c) C8an4e in pH Kery small fall or slit cane (1)

Explanation $&& $ (1)

uilibrium restored $ $&& #@3

[16]

2. (a) -onoprotic acid An acid 8ic i,es only one proton(1)Example $l etc(1)

2

(b) (i) lo10$& or in 8ords(1)

(ii) 1.5* " (allo8 1.6)(1) 2

(c) "ol $&! 25 0.15 103(1)! 3.%5 103

"ol #$! 35 0.12 103(1)! 4.20 103

'cess #$ ! 4.5 104(1)

#$ ! 4.5 104 100060(1)! %.5 103

$& ! 1014%.5 103(1)!1.33 1012(1)p$ ! lo101.33 10

12

! 11.+(1)

D7 onseuential marin if #$ not calculated to ma'imum of 5 %[11]

2!. (a) Equation for HCl(4) $l() $&(a) & l(a) (1)

Equation for KOH(s) C#$(s) C&(a) $(a) (1)2

(b) K8! $& #$ 1

(c) stron base= fully dissociated (1)or #$ ! 0.016 " (1)

$

&

!

#$ 8K

(1)

!016.0

10 14

! 6.25 10

13

"(1)

(2)

p$ ! lo10$& (1)


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p$ ! 12.2 (1)

neutral solution= $& ! #$ (1)

p$ ! % (1) %

(d) (i) %55 cm3 of 0.012 " acid contain

1000%55012.0

mol $&! +.06 103moles (1)

moles #$used for neutralisation ! +.06 103(1)

(ii) p$ ! 11.6 $& ! 1011.6! 2.5 1012" (1)

#$ ! ! $8

+

K

(1)#$ ! 3.+* 103" (1) (2)

in %55 cm

3

tere are1000

%5510+*.3 3

! 3.0 103 mol (1)

(iii) otal moles ! (+.06 & 3.0) 103! 0.012 mol (1)

(i,) -r! 3+ & 16 & 1 ! 56

m ! 56 0.012 ! 0.6* (1) *[18]

2". (a) $F & $2# $3#&& F(1)

$l & $2# $3#& & l(1) 2

for $F= must a,e re,ersible arro8allo8 (a) in $l euation

(b) (i) p$ ! lo10$& o%eui,alent 8ord definition (1) 1

allo8 Blo$3#& o%Blo$&(a)

(ii) $&! 0.050 mol dm3

p$ ! 1.3(0) (1) 1

if correct definition demonstrated in (ii)= but 8ord definition in (i)8ron= allo8 mar transfer from (ii) to (i)

(c) (i) Ka!$F

F$ ++

(1) 1

do note'pression allo8 8it $2#H allo8 $3#&

allo8 conseuential mar from 8ron euation in (b) pro,idin$& present

(ii) Ka!$F

$ 2+

or

$& ! $FaK (1)

$& ! 050.0106.54


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! 0.0053 (1)

p$ ! 2.3 2.2* (1) 3

allo8 mar for correct p$ from 8ron $&

(d) ydroen fluoride o%$F (1)

donates a proton (to te nitric acid) (1)

con9uate base F (tis mar dependent on correct identification of acid) (1) 3[11]

2#. (a) (i) proton donor (1) 1

(ii) partially dissociated (into ions) (not 8ealy dissociated) (1) 1

(b) (i) Ka!##$$

$##$

3

3

+

(1) 1

allo8 eiter $3#&

or $&

in e'pressionmust include caresif $2# included ten no mar

(ii) pKa! lo(l.% 105) 4.%% 4.*4.*0 (1) 1

(c) (i) indicator: penolptalein (1)e'planation: 8ea acidBstron base p$ cane abo,e %.0 (1)

lin bet8een p$ cane and indicator rane (1) 3

(ii) colourless to red pin purple (1) 1

if metyl orane named as indicator B 8ron but allo8 second e'planationmar and colour cane mar in (ii) (ie pin red yello8 orane)

(iii) Da#$ & $3##$ $3##Da & $2# (1)

if cares so8n tey must be correctaccept $&& #$$2# 1

(d) e'cess acid & Da#$ acid & Da#$ ! Dat (1)

in 2:1 ratio of ,olumes 2:1 mol 1:1 acid: salt (1) 2

(e) (i) #$ !# bonds are polar (8ords or on diaram)

due to different electroneati,ities of # and $ (or # and ) (1)

lone pairs of electrons on # atoms (1)

attraction of & $ atom in #$ for # atom lone pair in !#bet8een different molecules (1) "a' 3

[14]

3$. (a) Bea. acid An acid 8ic only partially ionises(1)Example tanoic= carbonic etc(1)


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2

(b) Expression Ka !

[ ] [ ][ ]$A

A$ +

(1)nits mol dm3(or mol l1)(1)

2

(c) (i) e dissociation of 8ater is an endotermic process(1)Ness dissociation on coolin (or euilibrium mo,es to 8ater or

K8 decreases)(1)less $&(or$& lo8er)(1)

(ii) 7ecause $& ! #$B(1) 4

(d) @esists cane in p$(1)on addition of small uantities of acid or base(1) 2[10]


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31. (a) (i) Strong = flly dissociated )

!ea" = partially dissociated ) in a#eos soltion both needed (1) $

(ii) Ka 0.16)(10

$

$ 22.%42 =+

! 2.0% 105 (1) mol dm3(not ") (1) 2

'planation: $ barely dissociates #@ $

&

,ery small(1)so $em! $ oriinal (1)

and $& ! in an aueous solution of 8ea acid(1) 3

(b) (i) moles $&in 1* cm3 ! 1* 0.16 103! 2.** 103 (1)moles #$in Ve,cm

3! 2 (1) (Ve, 0.12 103)

Ve,! 212.016.01*

! 12.0 cm3 (1)

;tart $alf eui,alence

ui,alence /oubleeui,alence

Kolumecm37a(#$)2solution added

0.0 6.0 12.0 24.0

p$ for titration of S 0.*0 1.22 % 12.84

p$ for titration of W 2.%4 4.68 *.5 12.84

3

(ii) St%on a*+d S W&a, a*+d W

Halfequi


27/29

1 4

1 3

1 2

1 1

1 0

+

*

%

6

5

4

3

2

1

0

p

$

K o l u m e 7 a ( # $ ) c m23

0 5 1 0 1 5 2 0 2 5

5

(c) (i) :uffer roperties resists cane in p$ (1)on addin small amounts of acid or base (1)

Was a :uffer ?lenty of present to mop up $& (1)

?lenty of $ present to mop up #$ (1)#@ euations so8in same

e.. $&

&

$= #$

& $

& $2#

(ii) acid buffersact at lo8 p$= basic buffersact at i p$ (1)alfneutralised Wis an acid buffer (1)basic buffer: mi' 8ea base 8it te salt of its coacid (1)#@ correct specific e'ample 6

[30]


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32. (a) p$ ! lo$& 1

A$or##$$

$C

3

2

a == +

+

1

$&! 1.%4 105 0.15 (or 1.62 103) 1

p$ ! 2.%+ (penalise 1 dp or more tan 2dp once in te u) 1

(b) (i) ;olution 8ic resists cane in p$ maintains p$ 1despite te addition of (small amounts of) acidbase (or dilution) 1

(ii) $3## & $&$3##$ 1

must so8 an euation full or ionic in 8ic etanoate ions arecon,erted to etanoic acid

(c) (i) ##$

##$$C$

3

3a=+

if rearranement incorrect= no furter mars1

10.0

15.010%4.1 5 =

1

(! 2.61 105)

p$ ! 4.5* 1

(ii) "l moles $&added ! 10 103 1.0! 0.011

"2 moles etanoic acid after addition! 0.15& 0.01 ! 0.16 1"3 moles etanoate ions after addition! 0.10B0.01 ! 0.0+ 1

"40.0+K0.16K

101.%4##$

##$$C$ 5

3

3a ==+

1

( ! 3.0+ 105)

"5 p$ ! 4.511

e essential part of tis calculation is additionsubtraction of 0.01 moles to ain mars

"2 and "3. Ef bot of tese are missin= only mar "l is a,ailable. ereafter treat eacmar independently= e'cept if te e'pression in "4 is 8ron= in 8ic case bot"4 and "5 are lost.

[1]

alternati


29/29

0+.0

16.0lo%6.4=pH

1

Documents

Acid-base Equilibria Mark Scheme