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.

i
phenolphthalein
OR
phenol red

ii
«n(NaOH) = \(\left( {\frac{{14.0}}{{1000}}} \right)\) dm³ × 0.100 mol dm^-3 =» 1.40 × 10^-3 «mol»

iii
«\(\frac{1}{2}\) × 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 = «\(\frac{{25}}{{1000}}\) × 5.00 g =» 0.125 «g»

«% purity = \(\frac{{0.0883{\rm{g}}}}{{0.125{\rm{g}}}}\) × 100 =» 70.6 «%»

ALTERNATIVE 2:
«mol of pure hydrated ethanedioic acid in 1 dm³ solution = 7.00 × 10^-4 × \(\frac{{1000}}{{25}}\)=» 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 = \(\frac{{3.53{\rm{g}}}}{{5.00{\rm{g}}}}\) × 100 =» 70.6 «%»

ALTERNATIVE 3:
mol of hydrated ethanedioic acid (assuming sample to be pure) = \(\frac{{5.00{\rm{g}}}}{{126.08{\rm{gmo}}{{\rm{l}}^{{\rm{ - 1}}}}}}\) = 0.03966 «mol»

actual amount of hydrated ethanedioic acid = «7.00 × 10^-4 × \(\frac{{1000}}{{25}}\) =» 2.80 × 10^-2 «mol»

«% purity = \(\frac{{2.80 \times {{10}^{ - 2}}}}{{0.03966}}\) × 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»