NZIC 2010

Assessment Schedule for Chemistry 3.7 AS 90700

Chemistry: Describe properties of aqueous systems

While the writers of this assessment have worked to compile a resource that meets NCEA requirements, it has no official status and teachers may wish to adjust questions and the assessment schedule as they see fit.

Q / Evidence / Achievement / Achievement with Merit / Achievement with Excellence
ONE
(a) / NH4+ + H2O NH3 + H3O+
HNO3 + H2O → NO3 + H3O+
CH3CH2NH2 + H2O CH3CH2NH3+ + OH / Three out of five:
Two out of the threeequations correct / Achieved plus: / Merit plus:
(b) (i)
(ii) / HNO3, NH4Cl, CH3CH2NH2
HNO3 – strong acid, fully dissociates to produce a high [H3O+]
NH4Cl – dissociates in solution to form NH4+ and Cl. The NH4+ ions are weakly acidic and partially dissociate in water to give a small increase in [H3O+].
CH3CH2NH2 – weak base, partially dissociates to give a small increase in [OH] / Three solutions listed in the correct order of pH
Solutions identified as acidic or basic using at least two of their equations
HNO3 identified as a strong acid and NH4+ (NH4Cl) as a weak acidORCH3CH2NH2identified as a weak base / Strength of HNO3 and NH4Cl solutions linked to [H3O+] OR strength of CH3CH2NH2 solution linked to[OH]
(c) / Since CH3CH2NH2 is a weak base and therefore only partially reacts with water, there will only be a few CH3CH2NH3+and OHions in the solution, making it a poor electrical conductor/weak electrolyte. However, HNO3 is a strong acid and therefore completely dissociates to produce a higher concentration of ions, specifically H3O+ and NO3, in solution. ThereforeHNO3 will be a better electrical conductor than CH3CH2NH2, since it is the sum of all ions in solution that contributes to a solution’s electrical conductivity. / Solutions conduct due to the presence of ions / Conductivity of HNO3ORCH3CH2NH2 solutions linked to concentration of ions in solution / Conductivity of HNO3ANDCH3CH2NH2 solutions linked to concentration of ions in solution. Must indicate that both ions produced from dissociation contribute to conductivity.
Q / Evidence / Achievement / Achievement with Merit / Achievement with Excellence
TWO
(a) (i) / CH3CH2COOH + H2O CH3CH2COO + H3O+ / Four out of six:
Correct equation / Achieved plustwo of: / Merit plus:
(ii) / Ka = [CH3CH2COO] [H3O+] / [CH3CH2COOH] / Correct Ka expression
(iii) / 1.35 × 105 = [H3O+]2 / 0.0250
[H3O+]2 = 1.35 × 105× 0.0250
[H3O+] = 5.81 × 104 mol L1
pH = log 5.81 × 104 = 3.24 / Correct process for calculation / pH of CH3CH2COOH correctly calculated
(b) (i) / A buffer is a solution that undergoes minimal change of pH when small amounts of acid orbase are added. / Function of a buffer solution described
(ii) / [CH3CH2COO] = [CH3CH2COOH] x Ka / 10pH
[CH3CH2COO] = 0.0250 mol L1×1.35 × 105 / 104.50
[CH3CH2COO] = 0.0107 mol L1 / Correct process for calculation of [CH3CH2COO] / Correct answer
(iii) / Added acid will react with CH3CH2COO so that there is essentially no change in [H3O+]:
CH3CH2COO + H3O+ → CH3CH2COOH + H2O
Added base will react with CH3CH2COOH so that there is essentially no change in [OH]:
CH3CH2COOH + OH → CH3CH2COO + H2O
(These equations show that the ratio of CH3CH2COOH: CH3CH2COO changes slightly, but this does not significantly affect the pH).
Since the pH of the buffer is lower than the pKa of CH3CH2COOH, the [CH3CH2COOH] will be higher than the [CH3CH2COO]. This means the buffer will be more effective against added base. / Limited discussion of buffer action against added acid OR base / Ability of solution in (b) (ii) to act as a buffer discussed for addition of either acid OR base, including appropriate equation / Ability of solution in (b) (ii) to act as a buffer discussed for addition of acid AND base, with appropriate equations, and an evaluation of the effectiveness of buffer against added acid and base.
Q / Evidence / Achievement / Achievement with Merit / Achievement with Excellence
THREE
(a) (i) / Ag2CrO4(s) 2Ag+(aq) + CrO42(aq) / Three out of five:
Correct equation / Achieved plus two of: / Merit plus:
(ii) / Ks = [Ag+]2 [CrO42] / Correct Ks expression
(iii) / Let s be the solubility:
[Ag+] = 2s
[CrO42] = s
Ks = (2s)2×s
1.9 x 1012 = 4s3
s = 7.80 × 105 mol L1
Solubility of Ag2CrO4(s) = 7.80 × 105 mol L1 / Correct process for calculation / Correct answer
(b) (i)
(ii) / Since equal volumes of each solution are mixed, the concentrations will be halved.
Qs = [Ag+]2[CrO42] = (6 x 104)2× 2.25 x 104 = 8.1 ×1011
Since QsKs, a precipitate of Ag2CrO4 will form.
Ag2CrO4(s) 2Ag+(aq) + CrO42(aq)
Upon addition of AgNO3, the added Ag+ (common ion) causes the equilibrium to shift in the reverse direction to use up some of the added Ag+, so more Ag2CrO4 precipitates. / Correct process for calculation
Recognises common ion effect / Correct calculation OR correct prediction based on calculation
Relates precipitation to common ion / Discussion relates precipitation to equilibrium
Q / Evidence / Achievement / Achievement with Merit / Achievement with Excellence
FOUR
(a) / The equation for the dissociation of HOCl is:
HOCl + H2O OCl + H3O+
Ka = [OCl][H3O+] / [HOCl]
Half way to equivalence (10 mL), the NaOH has reacted with half the HOCl, so [HOCl] = [OCl].
Ka = [H3O+] pKa = pH
So the pH half way to the equivalence point = pKa of HOCl / Three out of four:
Identifies that after 10 mL of NaOH has been added, pH = pKa / Achieved plus two of:
Full explanation for pKa / Merit plus one of:
(b) / [NaOCl] = 0.0250 × 20/30 = 0.0167 mol L1
Equation 1: OCl + H2O HOCl + OH
Equation 2: HOCl + H2O OCl + H3O+
Ka = [OCl][H3O+] / [HOCl]
From Equation 1, [HOCl] = [OH] = Kw/[H3O+]
Ka = [OCl][H3O+] / (Kw/[H3O+])
[H3O+] = √2.95 x 108× 1 × 1014 / 0.0167 pH = 9.88
OR using Kb method:
OCl + H2O HOCl + OH
Ka = 107.53 = 2.95 × 108 Kb = [OH]2 / [OCl] =Kw / Ka = 3.39 × 107
[OCl] = 0.0250 × 20/30 = 0.0167 mol L1
[OH] = √ 3.39 × 107× 0.0167 = 7.51 × 105 mol L1
[H3O+] = Kw/[OH] = 1.33 × 1010 mol L1 pH = 9.88 / Correct calculation
Correct process for calculating pH at equivalence point / Method for calculation of equivalence point pH generally correct with only one error, e.g. concentration of OCl incorrect
(c) / The larger the pKa, the weaker the acid. Since HOBr has a larger pKa, it will be a weaker acid and will therefore have a higher initial pH than HOCl.
Since HOBr is a weaker acid, its conjugate base OBrwill be stronger than OCl. At equivalence point:
OBr + H2O HOBr + OH will lie further to the right and so more OH- is formed and the pH will be higher. / Correct initial pH approximated for HOBr compared to HOCl
Correct equivalence point pH approximated for HOBr compared to HOCl / Correct explanation for approximate initial pH OR equivalence point pH of HOBr compared to HOCl / Correct explanation for approximate initial pH AND equivalence point pH of HOBr compared to HOCl

Judgement Statement

Achieved / Merit / Excellence
Total of THREE opportunities answered at Achievement level or higher.
3 x A / Total of at least THREE opportunities answered with TWO at Merit level or higher.
2 x M + 1 x A / Total of at least FOUR opportunities answered with TWO at Excellence level and ONE at Merit level or higher.
2 x E + 1 x M + 1 x A