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

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1 Assessment Schedule 2013 NCEA Level 3 Chemistry (91392) 2013 page 1 of ass-2013.doc Chemistry: Demonstrate understanding of equilibrium principles in aqueous systems (91392) Evidence Statement Q Evidence Achievement Achievement with Merit Achievement with ONE (a) HCl < CH 3 NH 3 Cl < CH 3 NH 2 HCl, a strong acid, reacts completely with water to form 1 mol L 1 H 3 O + and hence a low ph. HCl + H 2 O H 3 O + + Cl CH 3 NH 3 Cl dissociates completely in water to form CH 3 NH 3 + and Cl. CH 3 NH 3 +, a weak acid, partially reacts with water to form less than 1 mol L 1 H 3 O + and hence a higher ph than HCl. CH 3 NH 3 Cl CH 3 NH Cl CH 3 NH H 2 O CH 3 NH 2 + H 3 O + CH 3 NH 2, a weak base, partially reacts with water to form OH ions. So there are more OH ions than H 3 O + ions and the ph is thus high. CH 3 NH 2 + H 2 O CH 3 NH OH Correct order. TWO equations correct. Recognises that HCl dissociates completely in water. Recognises that CH 3 NH 3 + CH 3 NH 2 only partially react with water. THREE correct equations. Recognises that HCl dissociate completely in water. AND + Recognises that CH 3 NH 3 or CH 3 NH 2 only partially react with water. Discusses all the reactions correctly including concentrations of OH and H 3 O + ions. (b) HCl = CH 3 NH 3 Cl > CH 3 NH 2 CH 3 NH 3 Cl and HCl will dissociate completely in water to produce 2 mol L 1 ions. CH 3 NH 2 will only partially react with water to produce less than 1 mol L 1 of ions. CH 3 NH 2 written last. Links concentration of ions to degree of conductivity. CH 3 NH 2 written last and discusses HCl / CH 3 NH 3 Cl AND CH 3 NH 2. Links concentration of ions to degree of conductivity. Correct order with valid discussion. Links concentration of ions to degree of conductivity.

2 NCEA Level 3 Chemistry (91392) 2013 page 2 of ass-2013.doc (c)(i) K a = NH 2 ][H 3 O+ ] NH 3 + ] [H 3 O + ] = K a NH + 3 ] NH 2 ] NH 2 ] = = mol L 1 NH 3 + ] = = mol L 1 [H 3 O + ] = mol L 1 Correct K a expression. ph = pk a + log [base] [acid] Correct concentrations or number of moles. Correct process with minor error. Correct answer. ph = 10.8 Candidates should not be penalised for using ratio of volume and getting correct answer. (ii) When a small amount of acid (H 3 O + ) ions are added, they will react with the CH 3 NH 2 (aq) molecules to form CH 3 NH 3 + (aq) ions. CH 3 NH 2 (aq) + H 3 O + (aq) CH 3 NH 3 + (aq) + H 2 O( ) The added acid (H 3 O + ), is mostly consumed, and the ph of the solution changes very little. Correct equation. Shows understanding that CH 3 NH 2 (aq) reacts with added acid. Discusses minor reaction of OH + H 3 O +. Correct equation. AND Shows understanding that CH 3 NH 2 (aq) reacts with added acid. Correct equation and correct discussion of reaction.

3 NCEA Level 3 Chemistry (91392) 2013 page 3 of ass-2013.doc Not Achieved Achievement Merit NØ N1 N2 A3 A4 M5 M6 E7 E8 No response or no relevant evidence 2a 3a 4a 5a 3m 4m 3e with minor error / omission / additional information. 4e

4 NCEA Level 3 Chemistry (91392) 2013 page 4 of ass-2013.doc Q Evidence Achievement Achievement with Merit Achievement with TWO (a) K s = [Ag + ] 2 [CrO 4 2 ] Correct K s expression. (b)(i) n(ag 2 CrO 4 ) = 332 = mol in 50 ml [Ag 2 CrO 4 ] = = mol L 1 [Ag + ] = = mol L 1 [CrO 4 2 ] = mol L 1 Correct process Correct answer with limited working Correct ratio of [Ag + ] : [CrO 4 2 ] Correct concentration of silver chromate calculated. Correct solubility concentration values for each ion and Ks value. (ii) K s = ( ) 2 ( ) = Uses 4s 3 with incorrect answer. (c) Dissolving g of silver chromate in 50 ml water will result in solid being present, as the required amount to make a saturated solution is 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 Ag 2 CrO 4 will dissociate completely and form an equilibrium. Ag 2 CrO 4 2Ag + + CrO 4 2 Ag + + 2NH 3 [Ag(NH 3 ) 2 ] + The silver ion will then react further with NH 3, removing it from the above equilibrium. Thus, more Ag 2 CrO 4 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 g > , therefore solid Ag 2 CrO 4 is present. Correct equation of formation of complex ion.

5 NCEA Level 3 Chemistry (91392) 2013 page 5 of ass-2013.doc Not Achieved Achievement Merit NØ N1 N2 A3 A4 M5 M6 E7 E8 No response; no relevant evidence. 1a 2a 3a 4a 2m 3m 2e 3e

6 NCEA Level 3 Chemistry (91392) 2013 page 6 of ass-2013.doc Q Evidence Achievement Achievement with Merit Achievement with THREE (a) K a = [H 3 O+ ] COO ] COOH] ph = pk a + log [base] [acid] Correct process. Correct ph. [H 3 O + ] = mol L 1 = mol L 1 ph = log[h 3 O + ] = 2.90 (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. Recognises that there are equimolar quantities of ethanoic acid and sodium ethanoate. Relates equation correctly to explanation. As K a = [H 3 O+ ] COO ] COOH] According to the equation when COOH] = COO ] then K a = [H 3 O + ] So pk a = ph. (c)(i) NaOH(aq) + CH 3 COOH(aq) NaCH 3 COO(aq) + H 2 O(l) (1) COO ] increases as it is formed in reaction (1). [Na + ] increases as NaOH is added (1). COOH] decreases as it reacts with NaOH (1). 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. [H 3 O + ] decreases because COO ] / COOH] increases and K a is a constant. [OH ] increases because [H 3 O + ] decreases and [H 3 O + ] [OH ] is constant.

7 NCEA Level 3 Chemistry (91392) 2013 page 7 of ass-2013.doc (c)(ii) n(ch 3 COOH at start) = = mol n(naoh added) = = mol After 5 ml NaOH added: n(ch 3 COOH) = mol n(ch 3 COO ) = mol COOH] = mol L 1 COO ] = mol L 1 [H 3 O + ] = mol L 1 ph = 4.35 Candidates will not be penalised for not calculating concentrations. Correct n for CH 3 COOH 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 Achievement Merit NØ N1 N2 A3 A4 M5 M6 E7 E8 No response; no relevant evidence. 1a 2a 3a 4a 2m 3m 2e 3e with one minor error

8 Judgement Statement NCEA Level 3 Chemistry (91392) 2013 page 8 of ass-2013.doc Not Achieved Achievement Achievement with Merit Achievement with Score range