DP Chemistry (last assessment 2024)

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Question 20N.2.hl.TZ0.5

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Date November 2020 Marks available [Maximum mark: 12] Reference code 20N.2.hl.TZ0.5
Level hl Paper 2 Time zone TZ0
Command term Calculate, Identify, Justify, Predict, State, Suggest, Write Question number 5 Adapted from N/A
5.
[Maximum mark: 12]
20N.2.hl.TZ0.5

A student performs a titration to determine the concentration of ethanoic acid, CH3COOH, in vinegar using potassium hydroxide.

(a)

Write a balanced equation for the reaction.

[1]

Markscheme

CH3COOH(aq)+KOH(aq)CH3COOK (aq)+H2O(l) ✔

Accept the ionic equation.

Examiners report

Most candidates could write a balanced neutralization equation.

The pH curve for the reaction is given.

(b(i))

Identify the major species, other than water and potassium ions, at these points.

[2]

Markscheme

B: CH3COOH  AND CH3COO ✔

C: CH3COO

Accept names.

Accept CH3COOK for CH3COO

Examiners report

Identifying species present at various points along a pH titration curve was one of the most poorly answered questions in the exam. Very few candidates realized there were two major species at point B even when they were able in general to realize that B was a buffer zone.

(b(ii))

State a suitable indicator for this titration. Use section 22 of the data booklet

[1]

Markscheme

phenolphthalein ✔

Accept “phenol red” or “bromothymol blue”.

Examiners report

Almost all candidates could identify a suitable indicator to use in a titration of a weak acid with a strong base.

(b(iii))

Suggest, giving a reason, which point on the curve is considered a buffer region.

[1]

Markscheme

B AND the region where small additions «of the base/KOH » result in little or no
change in pH
OR
B AND the flattest region of the curve «at intermediate pH/before equivalence
point »
OR
B AND half the volume needed to reach equivalence point
OR
B AND similar amounts of weak acid/CH3COOH/ethanoic acid AND conjugate base/CH3COO/ethanoate ✔

Examiners report

Most students could identify a buffer zone region in a titration but very few (50%) could coherently explain why.

(c)

State the Ka expression for ethanoic acid.

[1]

Markscheme

Ka=CH3COO-H3O+CH3COOH

Accept H+ instead of H3O+.

Examiners report

Poorly answered with only 50% correctly writing a Ka expression. The major error was in candidates trying to calculate a Ka rather than write an expression for it.

(d)

Calculate the Kb of the conjugate base of ethanoic acid using sections 2 and 21 of the data booklet.

[1]

Markscheme

«Ka=104.76=1.7×105»
«Kw=Ka·Kb=1.0×1014=1.7×105×Kb»
«Kb=»5.8×1010 ✔

Accept answers between 5.75.9×1010.

(e)

In a titration, 25.00cm3 of vinegar required 20.75cm3 of 1.00moldm-3 potassium hydroxide to reach the end-point.

Calculate the concentration of ethanoic acid in the vinegar.

[2]

Markscheme

«n(KOH)=0.02075dm3×1.00moldm3=»0.0208«mol» ✔

«n(KOH)=n(CH3COOH)»
«[CH3COOH]=0.0208mol0.02500dm3=»0.830«moldm3» ✔

Award [2] for correct final answer.

Examiners report

Like with other calculations in this exam, the majority of candidates could correctly determine a concentration from titration data.

(f(i))

Potassium hydroxide solutions can react with carbon dioxide from the air. The solution was made one day prior to using it in the titration.

State the type of error that would result from the student’s approach.

[1]

Markscheme

systematic «error» ✔

Examiners report

80% of candidates could identify the method used as a systematic error, with some stating human or random error.

(f(ii))

Potassium hydroxide solutions can react with carbon dioxide from the air. The solution was made one day prior to using it in the titration.

Predict, giving a reason, the effect of this error on the calculated concentration of ethanoic acid in 5(e).

 

[2]

Markscheme

CH3COOH would be higher ✔

actual KOH is lower «than the value in calculation»
OR
larger volume of KOH «solution» needed to neutralize the acid ✔

Accept KOH partially neutralised by CO2 from air.

Examiners report

Most candidates identified that the systematic error would result in the concentration of the alkali being lowered but then failed to propagate this through to the effect on the concentration of the acid.