TOPIC 12 HW MS

1.	(a)     Any two from:
Weigh by difference or rinse weighing bottle and add to beaker
Rinse beaker and add washings to graduated flask
Invert flask several times to ensure uniform solution
Use a funnel to transfer to the flask and rinse the funnel
Use a stirrer to prepare the solution and rinse the stirrer
If more than two answers apply the list rule.
Max 2
(b)     Ka = [H+]2 / [HA]
Allow any correct expression relating Ka, [H+] and [HA]
1
[HA] = (10–2.50)2 / 1.07 × 10–3
M2 also scores M1
1
= 9.35 × 10–3 (mol dm–3)
Do not allow 9.4 (answer is 9.346).
Correct answer only scores 1 mark.
Do not penalise precision but must be to at least two significant figures.
1
(c)     (b) × 138.0 / 4
1
= 0.322
Using 8.50 × 10–3 gives 0.293
Correct answer scores M1 and M2.
Do not penalise precision but must be to at least two significant figures.
1
(d)     (c) × 100 / 0.500 = 64.5%
Using 0.293 from (c) gives 58.7%
Using 0.347 gives 69.4%
Do not penalise precision.
1
[8]

2.	
(a)     (i)      –log[H+]
Penalise missing [ ] here and not elsewhere
1
(ii)     [H+][OH–]
1
(b)     (i)      [H+] = 2.34 × 10–7
1
pH = 6.63
Penalise fewer than 3 sig figs but allow more than 2 dp
1
(ii)     [H+] = [OH–]
1
(iii)     M1    [H+] = Kw/[OH–]
if upside down or CE, allow M3 only for correct use of their [H+]
1
M2    (= 5.48 × 10–14/0.140) = 3.91 × 10–13
1
M3    pH = 12.4(1)
not 12.40 (AE from 12.407)
1
Penalise fewer than 3 sig figs but allow more than 3 sfs
For values above 10, allow 3sfs - do not insist on 2 dp.
For values below 1, allow 2dp – do not insist on 3 sig figs
Not allow pH = 14 – pOH but can award M3 only for pH = 13.1(46)
Can award all three marks if pKw = 13.26 is used

(c)     M1    mol NaOH = mol OH– = (30 × 10–3) × 0.20 = 6.0 × 10–3
mark for answer
1
M2    mol H2SO4 = (25 × 10–3) × 0.15 = 3.75 × 10–3
mark for answer
1
M3    mol H+ = (25 × 10–3) × 0.15 × 2 = 7.5 ×10–3
OR XS mol H2SO4 = 0.75 × 10–3
if factor of 2 missed or used wrongly, CE - lose M3 and next mark gained. In this case they must then use Kw to score any more.
see examples below
1
M4    XS mol H+ = 1.5 × 10–3
1
M5    [H+] = (1.5 × 10–3) × (1000/55) = 0.0273
if no use or wrong use of volume, lose M5 and M6 except if 1000 missed
AE –1 (pH = 4.56)
1
M6    pH = 1.56
Penalise fewer than 3 sig figs but allow more than 3 sfs
For values above 10, allow 3sfs - do not insist on 2 dp.
For values below 1, allow 2dp – do not insist on 3 sig figs
1
[14]

3.	(a)     Proton donor or H+ donor
Allow donator
1
(b)     (i)      B B
Both need to be correct to score the mark
1
(ii)     A A
Both need to be correct to score the mark
1
(iii)     B A
Both need to be correct to score the mark
1

(c)     M1    [H+] = 10–1.25 OR 0.05623
1
M2    mol HCl = (25 × 10–3) × 0.0850 (= 2.125 × 10–3)
Mark for Working
1
M3    vol   = 0.0378 dm3 or 37.8 cm3
allow 0.0375 – 0.038 dm3 or 37.5 – 38 cm3
Units and answer tied
Lose M3 if total given as (25 + 37.8) = 62.8 cm3
Ignore “vol added = 12.8 cm3 ” after correct answer
1
(d)     (i)      4.52
Must be 2dp
1
(ii)     Ka =        ignore =  but this may score M1 in (d)(iii)
Must have all brackets but allow ( ) Allow HA etc
NO mark for 10–pKa
1
(iii)     M1    Ka =  or with numbers
Allow [H+] = √(Ka × [HA]) for M1
1
M2    [H+] = (√(3.01 ×10–5 × 0.174) = √(5.24 ×10–6) )
= 2.29 ×10–3 - 2.3 ×10–3
Mark for answer
1
M3    pH = 2.64             (allow more than 2dp but not fewer)
Allow 1 for correct pH from their wrong [H+]
If square root forgotten, pH = 5.28 scores 2 for M1 and M3
1

(e)     M1    mol OH– = (10.0 × 10-3) × 0.125 = 1.25 × 10–3
Mark for answer
1
M2    orig mol HX = (15.0 × 10–3) × 0.174 = 2.61 × 10×3
Mark for answer
1
M3    mol HX in buffer = orig mol HX – mol OH–
Mark for answer
= 2.61 × 10–3 – 1.25 × 10–3 = 1.36 × 10–3
Allow conseq on their (M2 – M1)
([HX] = 1.36 × 10–3/25 × 10–3 = 0.0544)
If no subtraction, max 3 for M1, M2 & M4 (pH = 4.20)
If [H+] = [X–] & √used, max 3 for M1, M2 & M3 (pH = 2.89)
1
M4    mol X– in buffer = mol OH– = 1.25 × 10–3
([X–] = 1.25 × 10–3/25 × 10–3 = 0.05)
May be scored in M5 expression
1
M5    [H+]   
If use Ka =  no further marks
=    OR 
(= 3.27 × 10–5)
If either value of HX or X– used wrongly or expression upside down, no further marks
1
M6    pH = 4.48 or 4.49   (allow more than 2dp but not fewer)
Do not allow M6 for correct calculation of pH using their [H+] - this only applies in (d)(iii) - apart from earlier AE
1
[18]

4.          (a)     before any KOH added: Ka =  or  (1)
 Ka =  (1)
 [H+] =  = 1.67 × 10–3 (1)
 pH = 2.78 (1)
4
(b)     at 8 cm3 KOH:
Moles KOH added = (8 × 10–3) × 0.210 = 1.68 × 10–3 (1)
 moles of CH3COO– formed = 1.68 × 10–3 (1)
Original moles of CH3COOH = (25 × 10–3) × 0.160 = 4.0 × 10–3 (1)
 moles of CH3COOH left = (4.0 × 10–3) – (1.68 × 10–3)
                                 = 2.32 × 10–3 (1)
          [H+] = Ka ×  (1)
          = 1.74 × 10–5 ×  = 2.40 × 10–5 (1)
 pH = 4.62 (1)
It forget subtraction : max 5
If Ka expression not used max 5
if moles of CH3COOH wrong but substitution used max 5
7
(c)     at 40 cm3 of KOH:
Total moles of KOH = (40 × 10–3) × 0.21 = 8.4 × 10–3 (1)
 excess moles of KOH = (8.4 × 10–3) - (4.0 × 10–3)
                                    = 4.4 × 10–3 (1)
in total volume = 40 + 25 = 65 cm3 (1)
 [OH–] = 4.4 × 10–3 ×  = 0.0677 (1)
 [H+] = 
OR pOH =1.17
= 1.477 × 10–13 (1)
 pH = 12.83 (1)
If volume missed : max 4
If moles of acid wrong but method includes subtraction : max 5
If no subtraction : max 4
6
[16]
 5.	(a)     [H+] =  or = 1.74 × 10 −5 × 
Allow ( )
M1
1
=     3.08 × 10−5
If [HX] / [X-] or  upside down, or any addition or subtraction lose M1 & M2.
M2
1
pH = 4.51                 (correct answer scores 3)
Can score M3 for correct pH conseq to their [H+], so pH = 5.01 scores one
Must be to 2 dp
M3
1
Alternative using Henderson−Hasselbach Equation
pH = pKa − log[HX] / X−] = −log(1.74 × 10−5) − log()
Allow ( )
M1
pKa = 4.76 − 0.248
If [HX] / [X−] or  upside down, can only score 1
M2
pH = = 4.51
so pH = 5.01
Must be to 2 dp
M3

(b)     mol HX after addition (= 0.251 + 0.015) = 0.266
For HX, if no addition or error in addition (other than AE) (or subsequent extra add or sub) MAX 3
M1
1
mol X- after subtraction (= 0.140 − 0.015) = 0.125
For X− if no subtraction or error in subtraction (other than AE) (or subsequent extra add or sub) MAX 3
M2
1
[H+] = (  ) = 
 
If errors above in both addition AND subtraction can only score M3 for insertion of their numbers in rearranged expression. One exception, if addition and subtraction reversed then pH =4.58 scores 2
M3
1
[H+] = 3.703 × 10−5 (mol dm−3)
If [HX] / [X−] upside down, lose M3 & M4 (or next two marks) but can score M5 for correct pH conseq to their [H+], so if M1 & M2 correct, pH = 5.09 scores 3.
M4
1
pH = 4.43
Correct use of HX and X− values from (d) gives
pH= 4.41 and scores 4
If wrong method, e.g. √ or no use of rearranged Ka expression, may score M1 & M2 but no more.
Allow more but not fewer than 2dp here.
M5
1

Alternative using Henderson−Hasselbach Equation
mol acid after addition = 0.251 + 0.015 = 0.266
For HX, if no addition or error in addition (other than AE) (or subsequent extra add or sub) MAX 3
M1
mol salt after addition = 0.140 − 0.015 = 0.125
For X− if no subtraction or error in subtraction (other than AE) (or subsequent extra add or sub) MAX 3
M2
pH = (pKa − log[HX] / [X−]) = −log(1.74 × 10−5) − log(0.266 / 0.125)
If errors above in both addition AND subtraction can only score M3 for insertion of their numbers − except if addition and subtraction reversed then pH = 4.58 scores 2
M3
pH = 4.76 − 0.328
M4
pH = = 4.43
If [HX] / [X−] upside down, lose M3 & M4 (or next two marks) but can score M5 for correct pH conseq to their working, so if M1 & M2 correct, pH = 5.09 scores 3.
Allow more but not fewer than 2dp here.
M5
[8]

6.	(a)     (i)      G
1
(ii)     F
1
(iii)    H
1
(b)     (i)      cresol purple
1
(ii)     yellow to red
both colours needed and must be in this order
1
(iii)    yellow or pale yellow
Not allow any other colour with yellow
1
[6]

7.	(a)     NH4+  →  NH3  +  H+
Accept multiples.
Accept NH4+  +  H2O  →  NH3  +  H3O+
Ignore state symbols, even if incorrect.
1
(b)     Test             indicator / conc HCl
Do not accept ‘smell’.
Do not accept precipitation reactions of aqueous ammonia.
1
Observation  colour for an alkali / white fumes
If wrong test then lose second mark.
1
[3]

8.	(Calibrate) meter with solution(s) of known pH/buffer(s)
Do not accept ‘repeat reading’
1
Adjust meter/plot calibration curve
1
[2]

9.	C
[1]
10.	D
[1]
11.	B
[1]
12.	D
[1]
13.	A
[1]