NCEA Level 3 Chemistry (90700) 2011 — page 1 of 4Assessment Schedule – 2011Chemistry: Describe properties of aqueous systems (90700)Evidence Statement Question Evidence Achievement Merit Excellence ONE NH3 weak base TWO of: THREE from part (a) correct. (a) NaCl neutral NH4Cl weak acid • THREE from part (a) correct. AND HF weak acid (b)(i) NH3 + H2O NH4+ + OH– ONE explanation to Merit level. Equilibrium is to the left, so the greatest concentration of a species is NH3. For each NH3 that reacts equal AND amounts of NH4+ and OH– are formed and are greater than the OH– and H3O+ formed by the dissociation of water. ONE full explanation. (all 4 • Correct equation AND correct NH3 > OH– ≥ NH4+ > H3O+ • Correct equation. species). order of species for BOTH OR (b)(i) and (b)(ii). (ii) HF + H2O F– + H3O+ • Correct rank for b(i) or b(ii) OR Equilibrium is to the left, so the greatest concentration • Correct equation, order of AND of a species is HF. For each HF that reacts equal species AND full explanation amounts of F– and H3O+ are formed and are greater than (all 4 species) for EITHER the OH– and H3O+ formed by the dissociation of water. (b)(i) or (b)(ii). Correct answer with units, and HF > H3O+ ≥ F– > OH– appropriate number of sig. fig. EITHER (c) Ka = 6.76 × 10–4 HF + H2O H3O+ + F– • Ka correct. AND Assume [H3O+] = [F–] OR • Correct [H3O+]. • Correct answer with minor [H 3O + ]2 [H 3O + ]2 error (incorrect sig. fig. or Ka = ! [HF] = [HF] Ka units). [H 3O + ] = 4.57 " 10#3 mol L#1 [HF] = 0.0309 mol L#1
NCEA Level 3 Chemistry (90700) 2011 — page 2 of 4TWO Zn(OH)2(s) Zn2+(aq) + 2OH–(aq) TWO of:(a)(i) • Part (a) correct. (ii) Ks = [Zn2+][OH–]2 (b) Let s be solubility Solubility calculated correctly, Solubility calculated correctly, Ks = 4s 3 (incorrect sig. fig.). 3 sig. fig. and s is defined. 3 • Method correct, but error in Ks calculation. (Allow s2 follow on s= from part (a) or 2s3error but if so, AND AND 4 must have calculated s value s = 1.96 ! 10 –6 mol L"1 correctly according to the candidates ONE of: follow on.) • Recognises that a complex ion Complex ion forms, precipitate (c) Raising the pH will increase the concentration of OH– will form and links this to re-dissolves, as equilibrium ions. either less solid remaining or shifts in the forwards direction / This will initially cause additional precipitate to form. equilibrium shifting to the to RHS. This shift to the right • Recognises that [OH–] has increased. Once the pH has been increased sufficiently (enough right. will occur so more Zn2+ and OH– OH- has been added) the formation of a complex ion • Identifies equilibrium shifting will dissolve into solution so that with Zn2+ will occur, lowering OH– ion concentration in • Recognises equilibrium will shift to to the left due to additional the solution becomes saturated solution. the left. OH–. again. Thus the precipitate will redissolve as a complex ion • Explains equilibrium shifting to and less precipitate will be at the bottom of the test tube. the left in terms of the I.P. now exceeding Ks.
NCEA Level 3 Chemistry (90700) 2011 — page 3 of 4THREE TWO of: (a) HG + H2O G– + H3O+ OR • Part (a) and (b) correct. HOCH2COOH + H2O HOCH2COO – + H3O+ (b) [G ! ][H 3O + ] Ka = (must have equilibrium arrow) [HG] • EITHER Correct value for Ka (c) Correct answer with minor error. Correct answer with [H 3O + ] = Ka ! [HG] OR appropriate number of sig. fig. Ka = 1.50 ! 10"4 Correct rearrangement of Ka expression to make [H30+] subject. [H 3O + ] = 9.99 ! 10"3 mol L"1 AND pH = 2.00 AND (d) [H 3O + ] = 1.00 ! 10"4 mol L"1 Correct [G–]. Ka ! [HG] • EITHER [G " ] = + = 1.48 mol L"1 OR [H 3O ] Correct [H3O+]. Thus in 200 mL = 0.2 ! 1.48 = 0.296 mol OR Correct method for [G–] and n(G–) Alternative method Ka expression rearranged for [G–] or calculation but incorrect answer. Correct n(G–) to 3 sig. fig. other appropriate method for [G–] [weak base] pH = pK a + log10 stated and rearranged for [G–]. [weak acid] [base] 4.00 = 3.83+ log10 [acid] log10 [base] = 0.17 [base] = 1.48 mol L–1 Thus, in 200 mL = 00.2 ! 1.48 = 0.296 mol
NCEA Level 3 Chemistry (90700) 2011 — page 4 of 4 FOUR A ONE of: (a) At point A, there is an equi-molar mixture of HEt and • Recognises that at point A there is a Et–. On addition of OH– ions, the acid part of the buffer buffer solution. Describes how a buffer works Shows recognition of equimolar neutralises the OH– ions, by donating a proton. The acid (for when both acid AND base HEt and Et– thus pKa = pH reacts with the base: • States that equimolar amounts of are added) by: and discusses how the buffer HEt + OH– → Et– + H2O acid / base conjugate are present at EITHER solution works and links to A. • Giving equations for the equations. On addition of H3O+, the ethanoate will accept a proton specific buffer from the hydronium ion: OR • States that pH will not change when Et– + H3O+ → HEt + H2O small amounts of acid or base are • Writing about how a buffer added. works in general terms Candidate may discuss equilibrium shift. pKa = pH = 4.76 (accept 4.5 – 4.9) OR • Correct pKa / Ka • Links that due to equimolar (b) B HEt and Et– thus pKa = pH At the equivalence point all the HEt has been AND neutralised by NaOH. AND AND ONE of: HEt + NaOH → EtNa + H2O • Recognises that all the HEt has been • Recognises that none of the used up at B. original HEt remains as ithas The Et– reacts further to a small extent with water. all reacted with NaOH • That the pH of equivalence point is OR Et– + H2O HEt + OH– greater than 7. (must have clearly Uses two equations to explain • That the pH of equivalence why the pH is above 7. (One indicated that point B is the point is greater than 7 with a equation may be implied in the Thus the pH of the equivalence point is above 7 due to equivalence point) valid reason. candidate’s written answer.) presence of OH–.Judgement Statement Achievement Achievement with Merit Achievement with Excellence 3A 2M+1A 2E+1A