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Date November 2016 Marks available 2 Reference code 16N.2.sl.TZ0.2
Level SL Paper 2 Time zone TZ0
Command term Suggest and Outline 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).

Distinguish between a weak acid and a strong acid.

Weak acid:

Strong acid:

[1]
a.

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

[1]
b.

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.

[5]
c.

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.

[2]
d.

Markscheme

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.

a.

«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”

b.

i

«n(NaOH) =  ( 14.0 1000 )  dm-3 x  0.100 mol dm-3 =» 1.40 x 10-3 «mol»

 

ii

« 1 2 × 1.40 × 10 3 =   7.00 × 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 = « 25 1000 ×5.00g=»0.125«g»

«% purity =  0.0883 g 0.125 g × 100 =» 70.6 «%»

ALTERNATIVE 2:
«mol of pure hydrated ethanedioic acid in 1 dm3 solution = 7.00 × 10-4 ×  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 =  3.53 g 5.00 g × 100 =» 70.6 «%»

ALTERNATIVE 3:
mol of hydrated ethanedioic acid (assuming sample to be pure) =  5.00 g 126.08 g m o l 1 = 0.03966 «mol»
actual amount of hydrated ethanedioic acid = «7.00 × 10-4 ×  1000 25 =» 2.80 × 10-2 «mol»

«% purity =  2.80 × 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.

c.

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”.

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Core » Topic 4: Chemical bonding and structure » 4.3 Covalent structures
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