Assessment Schedule 2009 Chemistry: Describe properties of aqueous systems (90700) Evidence Statement NCEA Level 3 Chemistry (90700) 2009 page 1 of 5 Question Evidence Achievement Achievement with Merit Achievement with Excellence ONE CH 3 COOH H 2 O CH 3 COO H 3 O = [H 3 O ][CH 3 COO Ğ ] [CH 3 COOH] Equation and correct. = 1.74 10 5 = [H 3 O ][CH 3 COO Ğ ] [CH 3 COOH] 1.74 10 5 = [H 3 O ] 0.0500 [H 3 O ] = 9.32 10 4 Method correct with minor arithmetic error (carry through error from 1 accepted). ph correct. ph = log(9.32 10 4 ) = 3.03 p of methanoic acid is smaller so it is a stronger acid (more acid is dissociated) and therefore higher [H 3 O ]. High [H 3 O ] = low ph Relates p or to strength of acid or degree of dissociation or [H 3 O ] Calculates ph of methanoic acid = 2.52 Correctly accounts for lower ph by relating p or to strength of acid or degree of dissociation AND [H 3 O ]
NCEA Level 3 Chemistry (90700) 2009 page 2 of 5 (d) poh = 14 ph poh = log[oh ] = 14 8.65 = 5.35 5.35 = log[oh ] [OH ] = 4.47 10 6 K b = K w = 1 10 14 1.82 10 4 = 5.49 10 11 K b = [HCOOH][OH Ğ ] [HCOO Ğ ] 5.49 10 11 = [4.47 2 10 6 ] [HCOO Ğ ] [HCOO Ğ ] = 0.363ÊmolÊL Ğ1 = 10 3.74 = 1.82 10 4 = [H 3 O ][HCOO Ğ ] [HCOOH] = [H 3 O ][HCOOH] 1 10 14 [H 3 O ] = 2.24 10 9 [HCOO Ğ ] = 1 10 14 1.82 10 4 = 0.363ÊmolÊL Ğ1 (2.24 10 9 ) 2 Relates strength of acid To degree of dissociation and [H 3 O ] One step of calculations accurate that correctly identifies the number. Correct procedure with minor errors. Correct concentration Sensible rounding and units are required. [H 3 O ] = K w x / [Base] [H 3 O ] 2 = 1 x 10-14 x / [Base] [Base] = 1 x 10-14 x / [H 3 O ] 2 = 1 x 10-14 x 1.8 x 10-4 / (2.24 x 10-9 ) 2 = 0.363 mol L -1 2M 1A 1M 1E
NCEA Level 3 Chemistry (90700) 2009 page 3 of 5 TWO K s = [Ag ][Cl ] Let s = solubility = 1.56 10 10 = s 2 s = 1.25 10 5 [Ag ] = 1.25 10 5 (mol L 1 ) Correct answer for [Ag ]. Units not required. K s = [Ag ][Cl ] 1.56 10 10 = [0.100][Cl ] [Cl ] = 1.56 10 9 n = c V = 1.56 10 9 5.00 mol = 7.80 10 9 mol [Cl ] correct. [Cl ] plus one other step correct. Correct answer for mass of NaCl Sensible rounding and units are required. m = n M = 7.80 10 9 mol 58.5 g mol 1 = 4.56 10 7 g Dilute NH 3 will react with Ag to produce the complex ion [Ag(NH 3 ) 2 ]. Ag 2NH 3 [Ag(NH 3 ) 2 ] This decreases [Ag ]. This moves the solubility equilibrium below to the right AgCl(s) Ag (aq) Cl (aq) increasing the solubility of AgCl / causing more AgCl to dissolve, therefore the precipitate disappears. Identifies formation of a complex ion. States ammonia reacts with / removes silver ions therefore causing AgCl to dissolve. Both equations correct with limited reasoning One equation correct with related sound reasoning Sound reasoning for both parts with no equations States a complex ion forms with a correct equilibrium or ionic product statement. Full discussion with both equations included. Full discussion with descriptions of both equations in words. 1A 1M 1M 1E
NCEA Level 3 Chemistry (90700) 2009 page 4 of 5 THREE C 6 H 5 COOH, C 6 H 5 COO, OH, Na (Only one species in each box) 3 out of 4 species correct All 4 species correct A buffer is a solution containing a weak acid and its conjugate base (or a weak base and its conjugate acid). It is able to resist changes in ph. When 9.80 ml of base has been added, some of the benzoic acid has been converted to the benzoate ion (the conjugate base). There is still unreacted benzoic acid in the reaction vessel, so both acid and conjugate base are present together in reasonable / sufficient amounts. Hence the solution has buffering properties. When 25 ml of base has been added, the acid molecules have been converted to the conjugate base. The amount of benzoic acid is too low to have buffering properties. Defines a buffer Identifies presence of reasonable amounts of (benzoic) acid and conjugate base (benzoate) when 9.80 ml base added Identifies lack of (benzoic) acid when 25.0 ml base added Defines a buffer and correct discussion for 9.80 ml or 25.0 ml. Reasonably correct discussion for 9.80 ml and 25.0 ml without defining a buffer Defines a buffer and correctly compares 9.80 ml and 25.0 ml in Thymol blue or thymol blue and phenolphthalein would be best. (May state thymol blue is best both thymol blue and phenolphthalein would be best depending on justification and ph value chosen) A suitable indicator is one that has a p and therefore its colour change at the equivalence point of the reaction indicated by the steep / vertical part of the curve. From the graph, the equivalence point is about 8.3. Both thymol blue and phenolphthalein are suitable because their p values are within 1 of the equivalence point. Methyl orange is unsuitable because it will change colour before the equivalence point is reached / volume indicated will incorrect / give wrong answer. Identifies correct or incorrect indicator(s) with limited reasoning. Eg Thymol blue is best because its p / colourchange is near the equivalence point. Methyl orange is unsuitable because its p /colour change is not at /near the equivalence point. Identifies / classifies correct and incorrect indicators and effect of wrong choice but gaps in discussion. 2M Correct discussion. Must state equivalence point value and use correctly. 1M 1E
NCEA Level 3 Chemistry (90700) 2009 page 5 of 5 FOUR NaOH Na OH CH 3 NH 2 H 2 O CH 3 NH 3 OH CH 3 NH 3 Cl CH 3 NH 3 Cl CH 3 NH 3 H 2 O CH 3 NH 2 H 3 O Any 2 equations correct. Equations correct for all solutions. NaOH is a strong base and completely dissociates into its ions. Strong base, so produces more OH ions in solution than CH 3 NH 2. ph > 7 and greater than for CH 3 NH 2. CH 3 NH 2 is a weak base, so only some ions react with water to produce OH ions. ph > 7. CH 3 NH 3 Cl is an acidic salt that completely dissociates into its ions producing CH 3 NH 3. CH 3 NH 3 is a weak acid, so partially dissociates in water to produce H 3 O ions. So ph < 7. Conductivity relates to the number of ions in a solution. Recognises reasons for ph variation are due to differences in H 3 O or OH concentrations Recognises conductivity is related to the concentration of ions in solution. Recognises reasons for variations in ph and conductivity and makes one valid comparison. Difference in conductivity correctly discussed relating to the specific solutions Discussion addresses variation in BOTH ph and conductance using correct reasons. NaOH and CH 3 NH 3 Cl completely dissociate, so produce a large number of ions, so high conductivity. CH 3 NH 2 has an equilibrium reaction with water, so only a few ions are produced, so low conductivity. 1A Difference in ph correctly discussed relating to the specific solutions. 1M 1E Judgement Statement Achievement Achievement with Merit Achievement with Excellence 3 A 3 M 2 E 1 A 3 E NOTE: Lower case a, m, e may be used throughout the paper to indicate contributing evidence for overall grades for questions. Only upper case A, M and E grades shown at the end of each full question are used to make the final judgement.