# Physical Sciences Grade 12 P2 Gauteng June 2025 Memo (Afr)

> Grade 12 Physical Sciences Paper 2 memorandum from the Gauteng June Exam 2025 (Afrikaans), 16 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/physical-sciences-p2-english-and-afrikaans-june-2025-possible-answers/6895aedd-a963-4061-adc3-2b61d18c58d7
- Updated: 2026-09-27
- Subject: Physical Sciences
- Grade: Grade 12
- Language: Afrikaans
- Document type: Memorandum
- Year: 2025
- Exam period: Gauteng June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 16

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## About this document

Physical Sciences · Grade 12 · Gauteng June Exam · 2025 · Afrikaans. Memorandum, 16 pages. Read online or download the PDF.

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### Page 1

JUNE EXAMINATION
                    JUNIE EKSAMEN

                   GRADE/GRAAD 12

                           2025
               MARKING GUIDELINES/
                 NASIENRIGLYNE

                     PHYSICAL SCIENCES/
                    FISIESE WETENSKAPPE

                     (PAPER/VRAESTEL 2)

14 pages/bladsye

### Page 2

PHYSICAL SCIENCES
                   MARKING GUIDELINES
                                        FISIESE WETENSKAPPE
                   NASIENRIGLYNE
                                        (PAPER/VRAESTEL 2)    GR12 0625

QUESTION/VRAAG 1

1.1   C                                                        (2)
1.2   C                                                        (2)
1.3   D                                                        (2)
1.4   D                                                        (2)
1.5   B                                                        (2)
1.6   A                                                        (2)
1.7   C                                                        (2)
1.8   D                                                        (2)
1.9   B                                                        (2)
1.10 A                                                         (2)
                                                                [20]

                                  2

### Page 3

PHYSICAL SCIENCES
                            MARKING GUIDELINES
                                                        FISIESE WETENSKAPPE
                            NASIENRIGLYNE
                                                        (PAPER/VRAESTEL 2)            GR12 0625

QUESTION/VRAAG 2
2.1     Secondary (alcohol) 
        The carbon bonded to the hydroxyl group/functional group is bonded to
        two other carbon atoms.  (2 or 0)

        Sekondêre (alkohol)
        Die koolstof wat aan die hidroksielgroep (funksionele groep) verbind is, is
        aan twee ander koolstofatome verbind. (2 of 0)                                   (3)
2.2
        2.2.1   G                                                                       (1)
        2.2.2   A                                                                       (1)
        2.2.3   B                                                                       (1)

2.3.1       H       H   H    H
        H   C   C C     C    C   H
            H   H       H    H
         Marking criteria/Nasienkriteria:
         •  Functional group between C2 and C3/Funksionele groep tussen
            C2 en C3 
         •  Whole structure correct/Hele struktuur korrek                               (2)

2.3.2

                       (2 or 0)
        Carbon must have 4 bonds/koolstof moet vier bindings hê                          (2)

2.3.3               H

                H   C   H

            H   H       H    H       O

        H   C   C   C   C    C   C

            H   H   H   H    H       H
                                                                                         (2)
         Marking criteria/Nasienkriteria:
         • Functional group/Funksionele groep 
         • Whole structure correct/Hele struktuur korrek                                (2)
2.4
2.4.1   pent-2-ene/pent-2-een        Accept 2-pentene

         Marking criteria/Nasienkriteria:
         • Correct stem/Korrekte stam 
         • IUPAC name completely correct/IUPAC-naam heeltemal korrek 

         If pentene only 1/2/indien penteen slegs 1/2                                    (2)

                                               3

### Page 4

PHYSICAL SCIENCES
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               4

### Page 5

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2.4.2   4-methylhexan-3-one/4-metielheksan-3-oon

         Marking criteria/Nasienkriteria:
         • Correct stem (hex)/Korrekte stam (heks) 
         • All substituents correctly identified/Alle substituente is korrek 
         • IUPAC name completely correct/IUPAC-naam heeltemal korrek 
                                                                                         (3)

2.5.1   Positional isomers. 

        Posisionele isomere.                                                             (1)

2.5.2   Same molecular formula, but different positions of the side chain,
        substituants or functional group on the parent chain 

        Dieselfde molekulêre formule, maar verskillende posisies van die
        syketting, substituente of funksionele groep op die stamketting                  (2)
2.6
2.6.1   Ester/Ester                                                                     (1)

2.6.2   (Concentrated/Gekonsentreerde) sulphuric acid/swawelsuur 
        If the formula is given then zero
        Indien die formule gegee word, dan nul                                           (1)

2.6.3       H   H    H        O
        H   C   C    C    C       H   H

            H   H    H        O   C   C   H

                                  H   H

         Marking criteria/Nasien kriteria:
           • Functional group/Funksionele groep 
           • Whole structure correct/Hele struktuur korrek                             (3)
                                                                                        [25]
QUESTION/VRAAG 3
3.1     The pressure exerted by a vapour at equilibrium with its liquid in a closed
        system. 

        Die druk uitgeoefen deur ŉ damp in ewewig met sy vloeistof in ŉ geslote
        sisteem.                                                                         (2)

3.2     Alkane/Alkaan                                                                   (1)

                                              5

### Page 6

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3.3     •   Compound B has a longer chain/larger surface area / bigger molar
            mass than compound A. 
        •   Both have the same London forces but the intermolecular forces in B
            are stronger. 
        •   More energy is needed to overcome the intermolecular forces in
            compound B. 

        OR/OF
        • Compound A has a shorter chain/smaller surface area / smaller molar
           mass than compound B. 
        • Both have the same London forces but the intermolecular forces in A
           are weaker. 
        • Less energy is needed to overcome the intermolecular forces in
           compound A. 

        •   Verbinding B het ŉ langer ketting/groter oppervlakte/groter molêre
            massa as verbinding A.
        •   Albei het dieselfde Londonkragte maar die intermolekulêre kragte in
            B is sterker.
        •   Meer energie is nodig om die intermolekulêre kragte in verbinding B
            te oorkom.

        OR/OF
        • Verbinding A het ŉ korter ketting/kleiner oppervlakte/kleiner molêre
           massa as verbinding A.
        • Albei het dieselfde Londonkragte maar die intermolekulêre kragte in
           A is swakker.
        • Minder energie is nodig om die intermolekulêre kragte in verbinding A
           te oorkom.                                                                 (3)
3.4
3.4.1   A chemical formula that indicates the element and numbers of each of the
        atoms in a molecule.  (2 or 0)

        'n Chemiese formule wat die element en getalle van elk van die atome in
        'n molekule aandui. (2 of 0)                                                  (2)

3.4.2   E or propanoic acid/E of propanoësuur                                        (1)

3.4.3   •   Compound E has hydrogen bonds 
        •   Compound C has dipole-dipole forces 
        •   Stronger intermolecular forces in compound E than in C 
        •   More energy required to overcome the intermolecular forces in E 

        •   Verbinding E het waterstofbindings
        •   Verbinding C het dipool-dipool kragte
        •   Sterker intermolekulȇre kragte in verbinding E as in C
        •   Meer energie word benodig om die intermolekulȇre kragte in E te
            oorkom                                                                    (4)
                                            6

### Page 7

PHYSICAL SCIENCES
                           MARKING GUIDELINES
                                                          FISIESE WETENSKAPPE
                           NASIENRIGLYNE
                                                          (PAPER/VRAESTEL 2)               GR12 0625

3.5     Marking criteria/Nasienriglyn:
        • Calculate the % of oxygen in the alcohol. 
                                                             𝑚𝑚
        • Substitute 61; 11,86 and 27,14 in the formula 𝑛𝑛 = 𝑀𝑀 
        • Divide by the smallest mols 1,696 
        • Correct empirical formula 

        •    Bereken die % suurstof in die alkohol.
                                                            𝑚𝑚
        •    Vervang 61; 11,86 en 27,14 in die formule 𝑛𝑛 = 𝑀𝑀
        •    Deel deur die kleinste mol hoeveelheid 1,696
        •    Korrekte empiriese formule
                                                              m
                                                        n =
             Element    % m = 100g        M                   M               Ratio
                                                    61
                                                         = 5,08       5,08
                C              61        12         12                         = 2,99
                                                                      1,696
                                                11,86
                                                         = 11,86      11,86
                H          11,86          1         1                           = 6,99
                                                                      1,696
                                                27,14
                                                         = 1,696        1,696
                O         27,14         16      16                              = 1
                                                                       1,696

        Correct empirical formula/Korrekte empiriese formule : C3H7O                          (4)

3.6         C3H8    +    5O2    →     3 CO2 + 4 H2O
                                                                balancing/balansering       (3)
                                                                                              [20]

QUESTION/VRAAG 4

4.1      Unsaturated,  there are one or more multiple bonds between two carbon
         atoms in the hydrocarbon chain. 

         Onversadig, daar is een of meer meervoudige bindings tussen C-atome in
         die koolwaterstofketting                                                              (3)
4.2
4.2.1    Addition/Hydrogenation  Addisie/Hidrogenering/Hidrogenasie                           (1)

4.2.2    Substitution/ halogenation     Substitusie/halogenering                              (1)

4.2.3    Elimination/dehydrohalogenation  Eliminasie/dehidrohalogenering                      (1)

                                                7

### Page 8

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4.3                                             H   Cl   H       H
                                          H     C   C    C       C       H

                                                H   H    H       H

        Marking criteria/nasienriglyne
         Chlorine on carbon 2/Chloor op koolstof 2
         Whole structure/Hele struktuur                                                                   (2)
4.4
4.4.1   Hydration  Hidratering/Hidrasie                                                                   (1)

4.4.2   Excess water/H2O 
        Small amount of strong acid as catalyst  / sulfuric acid (H2SO4) or
        phosphoric acid (H3PO4).

        Oormaat water/H2O
        Klein hoeveelheid van ‘n sterk suur as katalis / swawelsuur (H2SO4) of
        fosforsuur (H3PO4).                                                                                (2)

4.4.3   Butan-2-ol  accept: 2-butanol
        Butan-2-ol    aanvaar: 2-butanol                                                                   (2)

4.5     2-methylpropane or methylpropane 2-metielpropaan of metielpropaan                                (2)

4.6
                H   H   H       H                                    H       H   H    H              
                                           
        H       C   C   C       C   H   + HBr                H       C       C   C    C   H   + H2O
                H   H   O       H                                    H       H   Br   H
                                   
                        H

         Marking criteria/Nasienkriteria:
         • Correct structural formula of butan-2-ol  the OH is on the correct
             carbonand HBr 
         • Correct structural formula of 2-bromobutane and H2O 

            • Korrekte struktuurformule vir butan-2-ol die OH is op die regte
              koolstof en HBr
           • Korrekte struktuurformule vir 2-bromobutaan en H2O
        If condensed structural formulae used max 2/5
        Indien gekondenseerde struktuurformules gebruik maks 2/5                                           (5)
                                                                                                          [20]

                                                         8

### Page 9

PHYSICAL SCIENCES
                        MARKING GUIDELINES
                                                    FISIESE WETENSKAPPE
                        NASIENRIGLYNE
                                                    (PAPER/VRAESTEL 2)              GR12 0625

QUESTION/VRAAG 5
5.1   Reaction rate is the change in concentration of reactants or products per
      unit time   (2 or 0)

      Verandering in konsentrasie van reaktante of produkte per eenheid tyd.
      (2 of 0)                                                                         (2)

5.2   •   Number of particles per unit volume decrease 
      •   Fewer particles with enough energy available for collision 
      •   Fewer effective collisions per unit time,/lower frequency of effective
          collisions
      •   Reaction rate decreases/Lower reaction rate/Reaction slows down

      •   Hoeveelheid deeltjies per eenheid volume verlaag.
      •   Minder deeltjies het genoeg energie beskikbaar vir botsings
      •   Minder effektiewe botsings per eenheid tyd/laer frekwensie van
          effektiewe botsings
          Reaksietempo verlaag/Laer reaksietempo/Reaksie word stadiger                 (4)

5.3                    ∆V(CO2 )
      Rate/Tempo =
                           ∆t
                       0,3 − 0
                   =
                       20 − 0                                                         (3)
                   = 0,015  (dm3.s-1)

                                             9

### Page 10

PHYSICAL SCIENCES
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                                  NASIENRIGLYNE
                                                        (PAPER/VRAESTEL 2)         GR12 0625

5.4     Marking criteria/Nasienkriteria:
                                            𝑉𝑉
         (a) Substitute/ Vervang 24 in n = 𝑉𝑉 correctly with 0,5/korrek met 0,5
                                              𝑚𝑚
         (b) USE/GEBRUIK mole ratio/mol verhouding 1:1
             MgCO3 : CO2 
         (c) Molar mass of MgCO3 (84) in the correct formula 
             Molêre massa van MgCO3 (84) in die korrekte formule
         (d) Final answer range 1,68 g – 1,75 g 
             Finale antwoordreeks 1,68 g – 1,75 g

        V(CO2) = 500 cm3 /1000 = 0,5 dm3
        Vm = 24 dm3
               V
        n=
             Vm
             0,5
         = 24 (𝒂𝒂)

         = 0,02 mol
        MgCO3 : CO2
                   1:1
             0,02: 0,02  (b)
                          m
        n(MgCO3) =
                          M
                   m
        0,02 =            (𝐜𝐜)
                   84
        m = 1,75 g  (d)                                                              (4)
5.5
5.5.1   Graph/Grafiek B                                                              (1)

5.5.2   • A peak/maximum at a higher kinetic energy/peak shifted to the right. 
        • More molecules have Ek > Ea 

        • ‘n Piek/maksimum by ‘n hoër kinetiese energie/piek skuif na regs.
        • Meer molekule het Ek > Ea                                                   (2)

5.5.3   LEFT/LINKS                                                                   (1)

                                                   10

### Page 11

PHYSICAL SCIENCES
                           MARKING GUIDELINES
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5.6
5.6.1   Exothermic/Eksotermies                                                           (1)

5.6.2   B  or D                                                                          (1)

5.6.3   10 (kJ)                                                                          (1)
                                                                                         [20]

QUESTION/VRAAG 6
6.1     Marking criteria/Nasienkriteria

        If any one of the underlined key phrases in the correct context is omitted,
        deduct 1 mark.

        Indien enige een van die onderstreepte sleutelfrases in die korrekte konteks
        weggelaat word, trek 1 punt af.
        When the equilibrium in a closed system is disturbed, the system will re-
        instate a new equilibrium by favouring the reaction that will oppose the
        disturbance.  (do not accept isolated system)

        Wanneer die ewewig in ŉ geslote sisteem versteur word,stel die sisteem ŉ
        nuwe ewewig in deur die reaksie wat die versteuring teenwerk te
        bevoordeel. (moet nie geïsolleerde sisteem aanvaar nie)                           (2)

6.2     Marking criteria/Nasienkriteria:
        (a) USING ratio/GEBRUIK verhouding
            n(CS2): n(Cℓ2): n(CCℓ4): n(S2Cℓ2) = 1:3:1:1. 
        (b) nequilibrium/ewewig = ninitial/aanvanklik –/+ nchange/verandering 
        (c) Divide nequilibrium by the volume 2 dm3 
            Deel newewig deur die volume 2 dm3
        (d) Correct Kc expression (formulae in square brackets). 
            Korrekte Kc uitdrukking (formule in blokhakkies).
        (e) Substitute Kc value 0,36 
            Vervang die Kc waarde 0,36
        (f) Substitution of equilibrium concentrations into Kc expression.
            Vervang ewewigkonsentrasies in Kc uitdrukking
        (g) Solve for 𝑥𝑥 /Los op vir 𝑥𝑥 
        (h) Final answer/Finale antwoord: 1,21 mol (1,20459 mol) 

                                                11

### Page 12

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OPTION 1/OPSIE 1                    CS2(g)       Cℓ2(g)    CCℓ4(g)    S2Cℓ2(g)
Initial amount                             𝑥𝑥    5         0          0
moles/Aanvanklike mol
hoeveelheid
Change in amount                    -0,8         -2,4      +0,8       +0,8 (a)
(moles)/Verandering in
hoeveelheid(mol)
Equilibrium amount (moles)/         𝑥𝑥 – 0,8     2,6       0,8        0,8 (b)
/Ewewigshoeveelheid
Equilibrium concentration/            𝑥𝑥 – 0,8   2,6       0,8        0,8
                                                       =         =       = 0,4
Ewewigskonsentrasie                       2       2         2         2
(mol.dm-3)                                       1,3       0,4        (c)
      [ CCℓ4 ][S2 Cℓ2 ]
Kc =                     (d)
        [CS2 ][ Cℓ2 ]3              No Kc expression, correct substitution
               (0,4)(0,4)
 0,36 (e)=                 (f)    Max 7/8
              [C𝑆𝑆2 ](1,3)3
[CS2 ] = 0,202…                     Wrong Kc expression Max 4/8
x – 0,8
         (𝐠𝐠)= 0,202…              Geen Kc uitdrukking, korrekte vervanging
   2
𝑥𝑥 = 1,20 mol (h) (1,20459         Maks 7/8
mol)                                Verkeerde Kc uitdrukking Maks 4/8
OPTION 2/OPSIE 2
                                    CS2(g)       Cℓ2(g)    CCℓ4(g)    S2Cℓ2(g)
Initial concentration/Aanvanklike        𝑥𝑥      2,5       0          0
konsentrasie                             2       (c)
Change in concentration             -0,4         -1,2      +0,4       +0,4 (a)
/Verandering in konsentrasie
Equilibrium concentration/            𝑥𝑥 – 0,8
Ewewigskonsentrasie                       2      1,3       0,4        0,4 (b)
(mol.dm-3)
      [ CCℓ4 ][S2 Cℓ2 ]
Kc =                     (d)
        [CS2 ][ Cℓ2 ]3              No Kc expression, correct substitution
               (0,4)(0,4)
 0,36 (e)=                 (f)    Max 7/8
              [C𝑆𝑆2 ](1,3)3
[CS2 ] = 0,202…                     Wrong Kc expression Max 4/8
x – 0,8
         (𝐠𝐠)= 0,202…              Geen Kc uitdrukking, korrekte vervanging
   2
𝑥𝑥 = 1,20 mol (h) (1,20459         Maks 7/8
mol)                                Verkeerde Kc uitdrukking Maks 4/8
                                                                                   (8)

                                           12

### Page 13

PHYSICAL SCIENCES
                         MARKING GUIDELINES
                                                     FISIESE WETENSKAPPE
                         NASIENRIGLYNE
                                                     (PAPER/VRAESTEL 2)               GR12 0625

6.3.1 Increase 
      Removing the carbon tetrachloride will decrease the concentration of the
      product the system will try to increase the products by favouring the forward
      reaction concentration of reactants will decrease and products will
      increase.

       Vermeeder
       Die verwydering van die koolstoftetrachloried sal die konsentrasie van die
       produk verlaag, die stelsel sal probeer om die produkte te verhoog deur die
       voorwaartse reaksie te bevoordeel, konsentrasie van reaktante sal afneem
       en produkte sal toeneem                                                          (2)

6.3.2 Decrease 
      Volume is increased then pressure is decreased. The system will counteract
      the change by favouring the reaction that produces the greater amount of
      moles. Therefore, the reverse reaction is favoured the concentration of
      reactants will increase and products will decrease.

       Verminder
       Volume word verhoog en druk verminder. Die sisteem sal die verandering
       teëwerk deur die reaksie te bevoordeel wat die meeste mol produseer.
       Daarom word die terugwaartse reaksie bevoordeel, die konsentrasie van
       reaktante sal toeneem en produkte sal afneem.                                    (2)

6.4.1 Decreased/Verlaag                                                                (1)

6.4.2 •    Temperature INCREASES 
      •    Increase in temperature favours endothermic reaction 
      •    Reverse reaction is favoured 
      •    Reverse reaction is endothermic 

       •   Temperatuur VERHOOG
       •   Toename in temperatuur bevoordeel endotermiese reaksie
       •   Terugwaartse reaksie word bevoordeel
       •   Terugwaartse reaksie is endotermies                                          (4)

6.4.3 22 (minutes/minute)                                                              (1)
                                                                                       [20]

QUESTION/VRAAG 7

7.1    Weak acids ionise incompletely in water  to form a low concentration of
       H3O+ ions.

       Swak sure ioniseer onvolledig in water om 'n lae konsentrasie H3O+-ione
       te vorm.                                                                         (2)

7.2    H2C2O4 ; 𝐻𝐻𝐶𝐶2 O-4  OR/OF    H3O+; H2O                                         (2)

                                             13

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PHYSICAL SCIENCES
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                                                                          FISIESE WETENSKAPPE
                                      NASIENRIGLYNE
                                                                          (PAPER/VRAESTEL 2)                  GR12 0625

7.3     Oxalic acid can donate two protons (H⁺) during its ionisation in an
        aqueous solutions/It ionises to form 2 protons.

        Oksaalsuur kan twee protone (H⁺) skenk tydens die ionisasie daarvan in
        ŉ waterige oplossings/Dit ioniseer om 2 protone te vorm.                                                (1)

7.4     A substance that can act as either an acid or a base.
        Accept: a substance that can donate or accept a proton

        'n Stof wat as beide 'n suur of 'n basis kan reageer.
        Aanvaar: ‘n stof wat ‘n proton kan skenk of ontvang                                                     (1)

7.5.1   OPTION 1/OPSIE 1                    OPTION 2/OPSIE 2
              m                                     m                                n
        c=                                 n=                                 c=        
              MV                                    M                                V

          =
                  2,25
                                               =
                                                    2,25                       (for both formulae/vir beide
              90(0,25)                              90
                                                                               formules)
              = 0,1 mol.dm-3                   = 0,025 mol
                                                                                     0,025
                                                                                 =            
                                                                                     0,25

                                                                               = 0,1 mol.dm-3                  (3)
7.5.2   Positive marking from 7.5.1/positiewe nasien vanaf 7.5.1
        Marking criteria/Nasienkriteria:
                                                                n          ca Va         na
        (a)        Any formula/Enige formule               c=       OR               =
                                                                V          cb Vb         nb
        (b)        Substitution/Vervanging
        (c)        USE mole ratio/GEBRUIK mol verhouding H2C2O4 : NaOH as 1 : 2
        (d)        Final answer/Finale antwoord: 0,18 mol.dm-3
                                (accept/aanvaar 0,175 mol.dm-3)

        OPTION/OPSIE 1                                        OPTION/OPSIE 2
                                 𝑛𝑛                           ca Va       na
        c(H2C2O4) = 𝑉𝑉 (a)                                           =         (a)
                                                              cb Vb       nb
        n = (0,1)(0,025) (b)                                 (0,1)(0,025)                    1
                                                                                  (b) = (c)
          = 0,025 mol                                         cb (0,0286)                     2
                             n
        c(NaOH) =                                             cb = 0,18 mol.dm-3 (d)
                             V
                             2(0,0025)
                         =               (c)
                              0,0286

                         = 0,18 mol.dm-3 (d)                                                                   (4)

                                                              14

### Page 15

PHYSICAL SCIENCES
                          MARKING GUIDELINES
                                                      FISIESE WETENSKAPPE
                          NASIENRIGLYNE
                                                      (PAPER/VRAESTEL 2)               GR12 0625

7.5.3   Phenolphthalein is a suitable indicator for the titration of oxalic acid and
        sodium hydroxide because the reaction involves a strong base (NaOH) and
        a weak acid (H2C2O4). The endpoint of the titration occurs when the pH
        is greater than 7 
        OR
        Because the salt of the titration will undergo hydrolysis and form a basic
        salt solution.

        Fenolftaleïen is 'n geskikte indikator vir die titrasie van oksaalsuur en
        natriumhidroksied omdat die reaksie tussen 'n sterk basis (NaOH) en 'n
        swak suur (H2C2O4) is. Die eindpunt van die titrasie word bereik wanneer
        die pH groter as 7 is.
        OF
        Omdat die sout van die titrasie hidrolise sal ondergaan en 'n basiese
        soutoplossing sal vorm.                                                          (2)

7.6     Marking criteria/Nasienkriteria:
        (a) Substitution/Vervanging (Mg(OH)₂) : (0,115)(0,05) 
        (b) Substitution/Vervanging (HNO₃) : (0,095)(0,025) 
        (c) USE ratio/GEBRUIK verhouding: 2n(HNO₃): = n(Mg(OH)₂) 
        (d) n(Mg(OH)₂)excess/oormaat = ninitial/aanvanklik – nreacted/gereageer
                              = 0,00575 - 0,0011875
        (e) Use the ratio 2 OH- : Mg(OH)₂
        (f) Substitute/Vervang 0,075 dm3 
        (g) Use/gebruik Kw = [H3O+] [OH-] 
        (h) Substitute/Vervang [OH-] = 0,12166 
        (i) Formula/Formula pH = - log [H3O+] 
        (j) Range for final answer/Gebied vir finale antwoord: (13,08 - 13,12) 

                                               15

### Page 16

PHYSICAL SCIENCES
                        MARKING GUIDELINES
                                                         FISIESE WETENSKAPPE
                        NASIENRIGLYNE
                                                         (PAPER/VRAESTEL 2)            GR12 0625

                   n                                          n
c(Mg(OH)₂) =                                   c(HNO₃) =
                   V                                          V

n = (0,115)(0,05)  (a)                        n = (0,095)(0,025) (b)

  = 0,00575 mol                                   = 0,002375 mol

                                                     HNO₃ : Mg(OH)₂
                                                           2:1
                                                 0,002375 : 0,0011875 (c)
n(Mg(OH)₂ )excess/oormaat = ninitial/aanvanklik – nreacted/gereageer
                               =0,00575 – 0,0011875
                               = 0,0045625 mol (d)
             OH-        :   Mg(OH)₂
                    2:1
               0,009125 : 0,0045625 (e)
               n
    c[OH-]=
               V
               0,009125
           =                (𝐟𝐟)
                0,075

           = 0,12166 mol.dm-3
       Kw = [H3O+] [OH-] (g)                  pH = - log [H3O+]  (i)
                                                  = - log (8,219 × 10-14)
1 × 10-14 = [H3O+] [0,12166] (h)
                                                  = 13,082(j)
 [H3O+] = 8,219 × 10-14
                                               OR/OF
                                               pH + pOH = 14
                                               pH = 14 – (– log [OH-])
                                                 = 14 – (- log(0,12166))
                                                 = 13,085                               (10)
                                                                                        [25]
                                                                       TOTAL/TOTAAL:    150

                                                16

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