Question
The data below is from an experiment used to determine the percentage of iron present in a sample of iron ore. This sample was dissolved in acid and all of the iron was converted to \({\text{F}}{{\text{e}}^{{\text{2 + }}}}\). The resulting solution was titrated with a standard solution of potassium manganate(VII), \({\text{KMn}}{{\text{O}}_{\text{4}}}\). This procedure was carried out three times. In acidic solution, \({\text{MnO}}_4^ – \) reacts with \({\text{F}}{{\text{e}}^{2 + }}\) ions to form \({\text{M}}{{\text{n}}^{2 + }}\) and \({\text{F}}{{\text{e}}^{3 + }}\) and the end point is indicated by a slight pink colour.
Deduce the balanced redox equation for this reaction in acidic solution.
Identify the reducing agent in the reaction.
Calculate the amount, in moles, of \({\text{MnO}}_4^ – \) used in the titration.
Calculate the amount, in moles, of Fe present in the \(3.682 \times {10^{ – 1}}{\text{ g}}\) sample of iron ore.
Determine the percentage by mass of Fe present in the \(3.682 \times {10^{ – 1}}{\text{ g}}\) sample of iron ore.
Answer/Explanation
Markscheme
\[{\text{MnO}}_4^ – {\text{(aq)}} + {\text{5F}}{{\text{e}}^{2 + }}{\text{(aq)}} + {\text{8}}{{\text{H}}^ + }{\text{(aq)}} \to {\text{M}}{{\text{n}}^{2 + }}{\text{(aq)}} + {\text{5F}}{{\text{e}}^{3 + }}{\text{(aq)}} + {\text{4}}{{\text{H}}_2}{\text{O(l)}}\]
Award [2] if correctly balanced.
Award [1] for correctly placing H+ and H2O.
Award [1 max] for correct balanced equation but with electrons shown.
Ignore state symbols.
\({\text{F}}{{\text{e}}^{2 + }}\) / iron(II);
Do not accept iron.
\({\text{n}} = 2.152 \times {10^{ – 2}} \times 2.250 \times {10^{ – 2}}\);
\(4.842 \times {10^{ – 4}}{\text{ (mol)}}\);
Award [1] for correct volume
Award [1] for correct calculation.
1 mol of \({\text{MnO}}_4^ – \) reacts with 5 mol of \({\text{F}}{{\text{e}}^{2 + }}\);
\(5 \times 4.842 \times {10^{ – 4}} = 2.421 \times {10^{ – 3}}{\text{ (mol)}}\);
(same number of moles of Fe in the iron ore)
Allow ECF from part (a) and (c) provided some mention of mole ratio is stated.
\(2.421 \times {10^{ – 3}} \times 55.85 = 0.1352{\text{ (g)}}\);
\(\frac{{0.1352}}{{0.3682}} \times 100 = 36.72\% \);
Allow ECF from part (d).
Examiners report
Most G2 comments on Section A were about this question. Many commented that titration is not part of the SL syllabus; however it is the expectation that students would cover this and other basic chemical techniques as part of their practical programme. Question 1 in all papers is meant to be data response and students will be expected to be familiar with experimental techniques. Also, there was some confusion caused because there was one sample and three titres. However this unfortunate cause of confusion did not seem to impact on candidate performance as poor performance was found throughout the question even with some very routine questions. Generally this question was poorly answered. In a) not many candidates managed to write the correct balanced equation, however many gave the correct species that were missing, \({{\text{H}}^ + }\) and \({{\text{H}}_{\text{2}}}{\text{O}}\).
Most candidates were able to identify the reducing agent although a few candidates just mentioned “iron” or Fe, but metallic iron was not in the equation.
In (c) candidates were not familiar with the process of selecting the 2 best titres and averaging them. Some chose the first written, some averaged all three and some weaker candidates merely added the 3 titres and used this. Some candidates also forgot to convert cm3 to dm3.
[N/A]
In 1(e), a few candidates scored ECF marks for the % based on n(Fe) calculated in (d). A couple of candidates realised that their answer to (d) did not help and followed on from (c) to find the number of moles of \({\text{F}}{{\text{e}}^{2 + }}\) and hence the % of Fe, scoring ECF marks.
Question
The percentage by mass of calcium carbonate in eggshell was determined by adding excess hydrochloric acid to ensure that all the calcium carbonate had reacted. The excess acid left was then titrated with aqueous sodium hydroxide.
(a) A student added \({\text{27.20 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.200 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) HCl to 0.188 g of eggshell. Calculate the amount, in mol, of HCl added.
(b) The excess acid requires \({\text{23.80 c}}{{\text{m}}^{\text{3}}}\) of \({\text{0.100 mol}}\,{\text{d}}{{\text{m}}^{ – 3}}\) NaOH for neutralization. Calculate the amount, in mol, of acid that is in excess.
(c) Determine the amount, in mol, of HCl that reacted with the calcium carbonate in the eggshell.
(d) State the equation for the reaction of HCl with the calcium carbonate in the eggshell.
(e) Determine the amount, in mol, of calcium carbonate in the sample of the eggshell.
(f) Calculate the mass and the percentage by mass of calcium carbonate in the eggshell sample.
(g) Deduce one assumption made in arriving at the percentage of calcium carbonate in the eggshell sample.
Answer/Explanation
Markscheme
(a) \(n{\text{(HCl) (}} = 0.200{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}} \times {\text{0.02720 d}}{{\text{m}}^{\text{3}}}{\text{)}} = 0.00544/5.44 \times {10^{ – 3}}{\text{ (mol)}}\);
(b) \(n{\text{(HCl) excess (}} = 0.100{\text{ mol}}\,{\text{d}}{{\text{m}}^{ – 3}} \times 0.02380{\text{ d}}{{\text{m}}^{\text{3}}}{\text{)}} = 0.00238/2.38 \times {10^{ – 3}}{\text{ (mol)}}\);
Penalize not dividing by 1000 once only in (a) and (b).
(c) \(n{\text{(HCl) reacted }}( = 0.00544 – 0.00238) = 0.00306/3.06 \times {10^{ – 3}}{\text{ (mol)}}\);
(d) \({\text{2HCl(aq)}} + {\text{CaC}}{{\text{O}}_3}{\text{(s)}} \to {\text{CaC}}{{\text{l}}_2}{\text{(aq)}} + {{\text{H}}_2}{\text{O(l)}} + {\text{C}}{{\text{O}}_2}{\text{(g)}}/\)
\({\text{2}}{{\text{H}}^ + }{\text{(aq)}} + {\text{CaC}}{{\text{O}}_3}{\text{(s)}} \to {\text{C}}{{\text{a}}^{2 + }}{\text{(aq)}} + {{\text{H}}_2}{\text{O(l)}} + {\text{C}}{{\text{O}}_2}{\text{(g)}}\);
Award [1] for correct reactants and products.
Award [1] if this equation correctly balanced.
Award [1 max] for the following equations:
2HCl(aq) + CaCO3(s) \( \to \) CaCl2(aq) + H2CO3(aq)
2H+(aq) + CaCO3(s) \( \to \) Ca2+(aq) + H2CO3(aq)
Ignore state symbols.
(e) \({n{\text{(CaC}}{{\text{O}}_3}{\text{)}} = \left( {\frac{1}{2}n{\text{(HCl)}}} \right) = \frac{1}{2} \times 0.00306}\);
\({ = 0.00153/1.53 \times {{10}^{ – 3}}{\text{ (mol)}}}\);
Award [2] for correct final answer.
(f) \({M_{\text{r}}}{\text{(CaC}}{{\text{O}}_3}{\text{) (}} = 40.08 + 12.01 + 3 \times 16.00) = 100.09/100.1/M = 100.09/100.1{\text{ (g}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}}\);
Accept 100.
\(m{\text{(CaC}}{{\text{O}}_3}{\text{) }}( = nM) = 0.00153{\text{ (mol)}} \times 100.09{\text{ (g}}\,{\text{mo}}{{\text{l}}^{ – 1}}{\text{)}} = 0.153{\text{ (g)}}\);
\({\text{\% CaC}}{{\text{O}}_3}{\text{ }}\left( { = \frac{{0.153}}{{0.188}} \times 100} \right) = 81.4\% /81.5\% \);
Accept answers in the range 79.8 to 81.5%.
Award [3] for correct final answer.
(g) only \({\text{CaC}}{{\text{O}}_3}\) reacts with acid / impurities are inert/non-basic / impurities do not react with the acid / nothing else in the eggshell reacts with acid / no other carbonates;
Do not accept “all calcium carbonate reacts with acid”.
Examiners report
Responses to this question were mixed. Many candidates were able to calculate the amount of HCl given its volume and concentration; however some failed to convert the volume from \({\text{c}}{{\text{m}}^{\text{3}}}\) to \({\text{d}}{{\text{m}}^{\text{3}}}\). Some calculated the amount of acid which had reacted rather than the excess asked for. A significant number of candidates gave carbonic acid as a product of the reaction and some were not able to write the formula of calcium carbonate. Although candidates correctly determined the amount and percentage of \({\text{CaC}}{{\text{O}}_{\text{3}}}\) in the egg sample; many struggled with the assumption made. Only a small number realising that one had to assume that only the \({\text{CaC}}{{\text{O}}_{\text{3}}}\) reacted with the acid, nothing else in the sample would react. Some of the incorrect answers were: “it contained no contaminants”, “it is 100% calcium carbonate” or “the eggshell was pure”. There were a significant number of candidates however who received 0 marks for the whole question. Clearly, as was pointed out in the most recent November 2009 subject report, it appears that many schools are not covering core laboratory areas such as volumetric chemistry.