<p> NCEA Level 3 Chemistry (91392) 2014 — page 1 of 6</p><p>Assessment Schedule – 2014 Chemistry: Demonstrate understanding of equilibrium principles in aqueous systems (91392) Evidence Statement</p><p>Achievement with Q Evidence Achievement Achievement with Merit Excellence</p><p>ONE  Equation correct.</p><p>(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).</p><p>(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)</p><p>(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</p><p>NØ N1 N2 A3 A4 M5 M6 E7 E8</p><p>No response or no 1a 2a 3a 4a 2m 3m 2e with minor 2e relevant evidence. error / omission.</p><p>Q Evidence Achievement Achievement with Merit Achievement with Excellence</p><p>TWO (a)(i)  Both (i) and (ii) correct. 2+ – (ii) PbCl2(s) ⇌ Pb (aq) + 2Cl (aq)</p><p>(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</p><p>[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</p><p>Q = (1.21 × 10–2) x (0.440)2 = 2.34 × 10–3</p><p>As Q > Ks, a precipitate will form. NCEA Level 3 Chemistry (91392) 2014 — page 3 of 6</p><p>(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.</p><p>– 2– Zn(OH)2(s)+ 2OH  [Zn(OH)4] </p><p>2+ – 2– OR Zn + 4OH  [Zn(OH)4] This decrease in [Zn2+] causes the position of equilibrium to </p><p> shift further to the right, therefore more Zn(OH)2 dissolves. </p><p>NØ N1 N2 A3 A4 M5 M6 E7 E8</p><p>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</p><p>Achievement with Q Evidence Achievement Achievement with Merit Excellence</p><p>+ 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</p><p>+ – + [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</p><p>= 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</p><p>+ – + (c)(i) CH3NH3 , Br , CH3NH2, H3O TWO OF: NCEA Level 3 Chemistry (91392) 2014 — page 5 of 6</p><p>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. </p><p>– + + 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</p><p>+ + CH3NH3 + H2O ⇌ CH3NH2 + H3O</p><p>– + + – Species present are Br > CH3NH3 > H3O ≥ CH3NH2 > (OH )</p><p>NØ N1 N2 A3 A4 M5 M6 E7 E8</p><p>No response or no 1a 2a 3a 4a 2m 3m 2e with minor error / 2e relevant evidence. omission.</p><p>Cut Scores</p><p>Not Achieved Achievement Achievement with Merit Achievement with Excellence NCEA Level 3 Chemistry (91392) 2014 — page 6 of 6</p><p>Score range 0 – 7 8 – 13 14 – 18 19 – 24</p>