NCEA Level 3 Chemistry (91392) 2014 Page 5 of 5

NCEA Level 3 Chemistry (91392) 2014 — page 1 of 6

Assessment Schedule – 2014 Chemistry: Demonstrate understanding of equilibrium principles in aqueous systems (91392) Evidence Statement

Achievement with Q Evidence Achievement Achievement with Merit Excellence

ONE  Equation correct.

(a)(i) – + OR HOCl + H2O ⇌ OCl + H3O FOUR species correctly (ii)  ALL species and order  ALL species and identified. correct AND partial order correct AND explanation to support complete  Recognises HOCl partially + – – + – – the order of the species. justification. HOCl > H3O > OCl > OH or HOCl > H3O = OCl > OH dissociates. OR HOCl partially dissociates, and so the equilibrium lies to the LHS/favours the One correct justification. reactants; therefore HOCl is present in the greatest amounts. + – H3O and OCl are produced in equal amounts / there is a small contribution to H O+ from water therefore H O+ > OCl–  Any two correct  Complete comparison. 3 3 relationships. + – Because there is a relatively high [H3O ], the [OH ] is very low (or links to Kw).

(b) Hydrofluoric acid is a stronger acid/more acidic/dissociates more because it has a smaller pKa (larger Ka) than hypochlorous acid. + + So HF will therefore have a higher [H3O ]. As [H3O ] increases, the pH decreases, so HF will have a lower pH than HOCl. (pH HF = 2.09, HOCl = 4.27)

(c) - + [F ][H3O ] Ka =  Correct method but error  Correct answer [HF]  Writes correct Ka or pH expression. in calculation / units with units. [F - ]´10-4.02 10-3.17 = OR missing / unit incorrect. 0.0500 + Calculates Ka or [H3O ]. [F- ] = 0.354 mol L-1 n(NaF) = 0.354 mol L-1 ´ 0.150 L = 0.0531 mol  Correct ‘n’ and ‘m’ step with m(NaF) = 0.0531 mol ´ 42.0 g mol-1 = 2.23 g ¯ incorrect [F ]. NCEA Level 3 Chemistry (91392) 2014 — page 2 of 6

NØ N1 N2 A3 A4 M5 M6 E7 E8

No response or no 1a 2a 3a 4a 2m 3m 2e with minor 2e relevant evidence. error / omission.

Q Evidence Achievement Achievement with Merit Achievement with Excellence

TWO (a)(i)  Both (i) and (ii) correct. 2+ – (ii) PbCl2(s) ⇌ Pb (aq) + 2Cl (aq)

(iii) 2+ – 2 Ks = [Pb ][Cl ]  Method correct, for  Correct answer for solubility calculation of solubility. and both [Pb2+] and [Cl–]. [Pb2+] = x [Cl–] = 2x 3 Ks = 4x K x = 3 s 4 1.70 ´10-5 = 3 4 = 1.62 ´10-2 mol L-1

[Pb2+] = 1.62  10–2 mol L–1  One calculation step correct.  One calculation error AND  Answer correct with [Cl–] = 3.24  10–2 mol L–1 Compares Q and Ks to make a supporting calculation and valid conclusion. correct conclusion. (b) 2.00 g n(Pb(NO3)2) = 331 g mol-1  Compares incorrect Q and = 6.04 × 10–3 mol Ks to make a valid conclusion. [Pb2+] = 6.04 × 10–3 mol / 0.500L = 1.21 × 10–2 mol L–1

Q = (1.21 × 10–2) x (0.440)2 = 2.34 × 10–3

As Q > Ks, a precipitate will form. NCEA Level 3 Chemistry (91392) 2014 — page 3 of 6

(c)  Writes the equilibrium  Partial explanation for BOTH equation. changes in pH, not fully 2+ – Zn(OH)2(s) ⇌ Zn (aq) + 2OH (aq) related to the effect on the  Recognises solubility equilibrium. increases at pH of less than When pH is less than 4 / low, [OH–] is decreased due to the 4 (acidic conditions) due to OR + – reaction with H3O to form water, removal of OH . One change in pH fully  Complete explanation for + – H3O + OH  H2O OR explained. BOTH changes in pH. so equilibrium shifts to the right to produce more [OH–], Recognises the solubility therefore more Zn(OH)2 will dissolve. increases at a pH greater than 10 due to formation When pH is greater than 10 / high, then more OH– is available of a complex ion. and the complex ion (zincate ion) will form.

– 2– Zn(OH)2(s)+ 2OH  [Zn(OH)4]

2+ – 2– OR Zn + 4OH  [Zn(OH)4] This decrease in [Zn2+] causes the position of equilibrium to

shift further to the right, therefore more Zn(OH)2 dissolves.

NØ N1 N2 A3 A4 M5 M6 E7 E8

No response or no 1a 2a 3a 4a 2m 3m 2e with minor 2e relevant evidence. error / omission. NCEA Level 3 Chemistry (91392) 2014 — page 4 of 6

Achievement with Q Evidence Achievement Achievement with Merit Excellence

+ THREE At point A, [CH3NH2] ≈ [CH3NH3 ]. So the solution has buffering  Recognises near point A  Correct equation linked to (a) properties in the proximity of point A. When HBr is added, the solution is a buffer / neutralisation / absorption of + + H3O is consumed: + H3O . [CH3NH2] ≈ [CH3NH3 ]. + + H3O + CH3NH2  CH3NH3 + H2O + Since the H3O is removed from the solution (neutralised), the pH +  Identifies H3O or HBr is does not change significantly. neutralised / removed by CH3NH2. + –11.8 -12 (b) [H3O ] = 10 = 1.58 x 10

+ – + [CH NH ][H O ]  Calculates [OH ] / [H3O ] / K = 3 2 3 a [CH NH + ] Kb 3 3 [CH NH ][H O+ ] = 3 2 3  Uses suitable process with [OHK -]  Correct method but an error in  Correct answer with a w more than one error. the calculation. clear method. [CH NH ]´ (10-11.8 )2 OR 2.29  10–11 = 3 2 -14 1´10 Rearranges Kb / Ka (2.29 ´10-11 ) ´ (1´10-14 ) expression so [CH3NH2] is [CH NH ] = the subject. 3 2 (10-11.8 )2

= 0.0912 mol L–1 OR K 10-14 [OH–] = w = [H O+ ] 10-11.8 3 = 6.31  10–3 mol L–1 [OH - ]2 K = b [CH NH ] 3 2 (6.31´10-3 )2 4.37  10–4 = [CH NH ] 3 2 (6.31´10-3 )2 [CH3NH2] = -4 4.37 ´10 –1 [CH3NH2] = 0.0912 mol L

+ – + (c)(i) CH3NH3 , Br , CH3NH2, H3O TWO OF: NCEA Level 3 Chemistry (91392) 2014 — page 5 of 6

At the start, before addition of HBr there is a solution of weak base  ALL species correct.  Full explanation of the electrical  Compares and contrasts (CH3NH2) which only partially reacts with water to produce a conductivity and species present the electrical (ii) relatively low concentration of ions. of either the initial CH3NH2 conductivity of BOTH As a result, the initial CH3NH2 solution will be a poor electrical  Recognises ions are solution or the solution at point the initial CH3NH2 conductor. required for electrical B. solution and the solution conductivity in a solution. at point B, including a OR consideration of the + - CH3NH2 + H2O ⇌ CH3NH3 + OH  One correct equation. for an answer discussing each differing concentrations solution separately: of each solution.

– + + Therefore species present are CH3NH2 > OH ≥ CH3NH3 > H3O TWO OF: Species and comparative concentrations within each At point B, there is a solution of the salt CH3NH3Br present which solution for both solutions / two is dissociated completely into ions. Therefore there is a relatively + – of the three equations / high concentration of ions (CH3NH3 and Br ) present in the conductivity of each solution solution, so it will be a good electrical conductor / electrolyte. with reasons. + – CH3NH3Br  CH3NH3 + Br + CH3NH3 reacts with water according to the equation

+ + CH3NH3 + H2O ⇌ CH3NH2 + H3O

– + + – Species present are Br > CH3NH3 > H3O ≥ CH3NH2 > (OH )

NØ N1 N2 A3 A4 M5 M6 E7 E8

No response or no 1a 2a 3a 4a 2m 3m 2e with minor error / 2e relevant evidence. omission.

Cut Scores

Not Achieved Achievement Achievement with Merit Achievement with Excellence NCEA Level 3 Chemistry (91392) 2014 — page 6 of 6

Score range 0 – 7 8 – 13 14 – 18 19 – 24