Date | November 2016 | Marks available | 1 | Reference code | 16N.2.hl.TZ0.2 |
Level | HL | Paper | 2 | Time zone | TZ0 |
Command term | Draw | Question number | 2 | Adapted from | N/A |
Question
The concentration of a solution of a weak acid, such as ethanedioic acid, can be determined
by titration with a standard solution of sodium hydroxide, NaOH (aq).
5.00 g of an impure sample of hydrated ethanedioic acid, (COOH)2•2H2O, was dissolved in water to make 1.00 dm3 of solution. 25.0 cm3 samples of this solution were titrated against a 0.100 mol dm-3 solution of sodium hydroxide using a suitable indicator.
(COOH)2 (aq) + 2NaOH (aq) → (COONa)2 (aq) + 2H2O (l)
The mean value of the titre was 14.0 cm3.
(i) Suggest a suitable indicator for this titration. Use section 22 of the data booklet.
(ii) Calculate the amount, in mol, of NaOH in 14.0 cm3 of 0.100 mol dm-3 solution.
(iii) Calculate the amount, in mol, of ethanedioic acid in each 25.0 cm3 sample.
(iv) Determine the percentage purity of the hydrated ethanedioic acid sample.
Draw the Lewis (electron dot) structure of the ethanedioate ion, –OOCCOO–.
Outline why all the C–O bond lengths in the ethanedioate ion are the same length and suggest a value for them. Use section 10 of the data booklet.
Explain how ethanedioate ions act as ligands.
Markscheme
i
phenolphthalein
OR
phenol red
ii
«n(NaOH) = dm3 × 0.100 mol dm-3 =» 1.40 × 10-3 «mol»
iii
« × 1.40 × 10-3 =» 7.00 × 10-4 «mol»
iv
ALTERNATIVE 1:
«mass of pure hydrated ethanedioic acid in each titration = 7.00 × 10-4 mol × 126.08 g mol-1 =» 0.0883 / 8.83 × 10-2 «g»
mass of sample in each titration = « × 5.00 g =» 0.125 «g»
«% purity = × 100 =» 70.6 «%»
ALTERNATIVE 2:
«mol of pure hydrated ethanedioic acid in 1 dm3 solution = 7.00 × 10-4 × =» 2.80 × 10-2 «mol»
«mass of pure hydrated ethanedioic acid in sample = 2.80 × 10-2 mol × 126.08 g mol-1 =» 3.53 «g»
«% purity = × 100 =» 70.6 «%»
ALTERNATIVE 3:
mol of hydrated ethanedioic acid (assuming sample to be pure) = = 0.03966 «mol»
actual amount of hydrated ethanedioic acid = «7.00 × 10-4 × =» 2.80 × 10-2 «mol»
«% purity = × 100 =» 70.6 «%»
Award suitable part marks for alternative methods.
Award [3] for correct final answer.
Award [2 max] for 50.4 % if anhydrous ethanedioic acid assumed.
Accept single negative charges on two O atoms singly bonded to C.
Do not accept resonance structures.
Allow any combination of dots/crosses or lines to represent electron pairs.
electrons delocalized «across the O–C–O system»
OR
resonance occurs
Accept delocalized π-bond(s).
No ECF from (d).
122 «pm» < C–O < 143 «pm»
Accept any answer in range 123 «pm» to 142 «pm».
Accept “bond intermediate between single and double bond” or “bond order 1.5”.
coordinate/dative/covalent bond from O to «transition» metal «ion»
OR
acts as a Lewis base/nucleophile
can occupy two positions
OR
provide two electron pairs from different «O» atoms
OR
form two «coordinate/dative/covalent» bonds «with the metal ion»
OR
chelate «metal/ion»