Assessment Schedule – 2013 Chemistry: Demonstrate understanding of equilibrium principles in aqueous systems (91392) Evidence Statement




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NCEA Level 3 Chemistry (91392) 2013 – page of

Assessment Schedule – 2013

Chemistry: Demonstrate understanding of equilibrium principles in aqueous systems (91392)

Evidence Statement

Q

Evidence

Achievement

Achievement with Merit

Achievement with Excellence

ONE

(a)



HCl < CH3NH3Cl < CH3NH2

HCl, a strong acid, reacts completely with water to form 1 mol L–1 H3O+ and hence a low pH.



HCl + H2O  H3O+ + Cl

CH3NH3Cl dissociates completely in water to form CH3NH3+ and Cl. CH3NH3+, a weak acid, partially reacts with water to form less than 1 mol L–1 H3O+ and hence a higher pH than HCl.



CH3NH3Cl  CH3NH3+ + Cl

CH3NH3+ + H2O  CH3NH2 + H3O+

CH3NH2, a weak base, partially reacts with water to form OH ions. So there are more OH ions than H3O+ ions and the pH is thus high.



CH3NH2 + H2O  CH3NH3+ + OH

  • Correct order.

  • TWO equations correct.

  • Recognises that HCl dissociates completely in water.

    OR

    Recognises that CH3NH3+OR CH3NH2 only partially react with water.



  • THREE correct equations.

  • Recognises that HCl dissociate completely in water.

    AND


    Recognises that CH3NH3+ or CH3NH2 only partially react with water.



  • Discusses all the reactions correctly including concentrations of OH and H3O+ ions.




(b)

HCl = CH3NH3Cl > CH3NH2

CH3NH3Cl and HCl will dissociate completely in water to produce 2 mol L–1 ions.

CH3NH2 will only partially react with water to produce less than 1 mol L–1 of ions.


  • CH3NH2 written last.




  • Links concentration of ions to degree of conductivity.




  • CH3NH2 written last and discusses HCl / CH3NH3Cl AND CH3NH2.
    Links concentration of ions to degree of conductivity.

  • Correct order with valid discussion. Links concentration of ions to degree of conductivity.




(c)(i)


Candidates should not be penalised for using ratio of volume and getting correct answer.

  • Correct Ka expression.

    OR


  • pH = pKa + log

    OR

    Correct concentrations or number of moles.



  • Correct process with minor error.




  • Correct answer.




(ii)

When a small amount of acid (H3O+) ions are added, they will react with the CH3NH2(aq) molecules to form CH3NH3+(aq) ions.

CH3NH2(aq) + H3O+(aq)  CH3NH3+(aq) + H2O()

The added acid (H3O+), is mostly consumed, and the pH of the solution changes very little.


  • Correct equation.

    OR


    Shows understanding that CH3NH2(aq) reacts with added acid.

    OR

    Discusses minor reaction of OH + H3O+.



  • Correct equation.

    AND


    Shows understanding that CH3NH2(aq) reacts with added acid.

Correct equation and correct discussion of reaction.





Not Achieved



No response or no relevant evidence

N1

2a

N2

3a

Achievement

A3

4a

A4

5a

Merit

M5

3m

M6

4m

Excellence

E7

3e with minor error / omission / additional information.

E8

4e



Q

Evidence

Achievement

Achievement with Merit

Achievement with Excellence

TWO

(a)


    Ks = [Ag+]2[CrO42–]

  • Correct Ks expression.







(b)(i)

(ii)




  • Correct process

    OR

    Correct answer with limited working





  • Correct ratio of [Ag+] : [CrO42–]



  • Uses 4s3 with incorrect answer.

  • Correct concentration of silver chromate calculated.



  • Correct solubility concentration values for each ion and Ks value.



(c)

Dissolving 0.0100g of silver chromate in 50 mL water will result in solid being present, as the required amount to make a saturated solution is 1.44  10–3 g in 50 mL, so any more than this will form a solid.

If the same mass is added to 50 mL of ammonia, more will dissolve and less solid will be present due to the formation of a complex ion.

The Ag2CrO4 will dissociate completely and form an equilibrium.

Ag2CrO4  2Ag+ + CrO42–

Ag+ + 2NH3  [Ag(NH3)2]+

The silver ion will then react further with NH3, removing it from the above equilibrium. Thus, more Ag2CrO4 will dissolve to re-establish equilibrium.



  • Recognises that more dissolves in B.

  • Recognises that a complex ion forms.

  • Recognises that more dissolves in beaker B with link to an equation.

  • Recognises that in ammonia a silver complex ion will form.



  • Links equilibrium of silver chromate with silver & ammonium complex ion removal and hence more dissolves.

  • Recognises
    0.0100 g > 1.44  10–3, therefore solid Ag2CrO4 is present.

  • Correct equation of formation of complex ion.




Not Achieved



No response; no relevant evidence.

N1

1a

N2

2a

Achievement

A3

3a

A4

4a

Merit

M5

2m

M6

3m

Excellence

E7

2e

E8

3e


Q

Evidence

Achievement

Achievement with Merit

Achievement with Excellence

THREE

(a)





  • Correct process.

  • Correct pH.



    (a) & (b) correct.





(b)

Halfway to equivalence point, half of the ethanoic acid has been used up. There are now equimolar quantities of ethanoic acid and sodium ethanoate.

According to the equation when [CH3COOH] = [CH3COO]


then Ka = [H3O+]

So pKa = pH.



  • Recognises that there are equimolar quantities of ethanoic acid and sodium ethanoate.

  • Relates equation correctly to explanation.



(c)(i)

NaOH(aq) + CH3COOH(aq)  NaCH3COO(aq) + H2O(l) (1)

[CH3COO] increases as it is formed in reaction (1).

[Na+] increases as NaOH is added (1).

[CH3COOH] decreases as it reacts with NaOH (1).

[H3O+] decreases because [CH3COO] / [CH3COOH] increases and Ka is a constant.

[OH] increases because [H3O+] decreases and [H3O+] [OH] is constant.



  • Correct equation minor error.



  • Correct statement relating to change in concentration of 1 species.

    Correct equation and correctly describes the change in concentration of 2 species.



  • Correct equation.

    AND


    Correctly describes the change in concentration of the 4 species.



(c)(ii)

n(CH3COOH at start) = 0.0896  20 10–3

= 1.79  10–3 mol



n(NaOH added) = 0.1  5  10–3

= 5  10–4 mol

After 5 mL NaOH added:

n(CH3COOH) = 1.29  10–3 mol

n(CH3COO) = 5  10–4 mol

[CH3COOH] = 0.0516 mol L–1

[CH3COO] = 0.0200 mol L–1

[H3O+] = 4.48  10–5 mol L–1

pH = 4.35

Candidates will not be penalised for not calculating concentrations.



  • Correct n for CH3COOH OR NaOH at the start.

  • Correct process to identify either of the species after 5 mL has been added (mol or mol L–1).

  • Correct answer.

Not Achieved



No response; no relevant evidence.

N1

1a

N2

2a

Achievement

A3

3a

A4

4a

Merit

M5

2m

M6

3m

Excellence

E7

2e

E8

3e with one minor error



Judgement Statement




Not Achieved

Achievement

Achievement with Merit

Achievement with Excellence

Score range


0 – 6

7 – 12

13 – 18

19 – 24


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