Distinguish between a weak acid and a strong acid.

Weak acid:

Strong acid:

Suggest why it is more convenient to express acidity using the pH scale instead of using the concentration of hydrogen ions.

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)   Calculate the amount, in mol, of NaOH in 14.0 cm³ of 0.100 mol dm^-3 solution.

(ii)  Calculate the amount, in mol, of ethanedioic acid in each 25.0 cm³ sample.

(iii) Determine the percentage purity of the hydrated ethanedioic acid sample.

The Lewis (electron dot) structure of the ethanedioate ion is shown below.

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.

Weak acid: partially dissociated/ionized «in solution/water»
AND
Strong acid: «assumed to be almost» completely/100% dissociated/ionized «in solution/water»

Accept answers relating to pH, conductivity, reactivity if solutions of equal concentrations stated.

«log scale» reduces a wide range of numbers to a small range
OR
simple/easy to use
OR
converts exponential expressions into linear scale/simple numbers

Do not accept “easy for calculations”

i

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

ii

«\(\frac{1}{2} \times 1.40 \times {10^{ - 3}} = \)  \(7.00 \times {10^{ - 4}}\) «mol»

iii
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 = «\(\frac{{25}}{{1000}}\)×5.00g=»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) = \(\frac{{5.00{\rm{g}}}}{{126.08{\rm{gmo}}{{\rm{l}}^{ - 1}}}}\) = 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.

electrons delocalized «across the O–C–O system»
OR
resonance occurs
Accept delocalized π-bond(s).

122 «pm» < C–O < 143 «pm»

Accept any answer in the range 123 «pm» to 142 «pm». Accept “bond intermediate between single and double bond” or “bond order 1.5”.