Questions & Answers

Q1.
Short Answer 1 mark

Aqueous silver ions, Ag

Q1(aq).
Short Answer 1 mark

, react slowly with aqueous iron(II) ions, Fe

Q1(aq).
Short Answer 1 mark

An equilibrium is established.

Q1(aq).
Short Answer 1 mark

Ag(s) + Fe

Q1(aq).
Short Answer 1 mark

The concentration of Ag

Q1(aq).
Short Answer 1 mark

at equilibrium can be determined by titration with a standard solution of aqueous potassium thiocyanate, KSCN(aq). During the titration, the remaining Ag

Q1(aq).
Short Answer 1 mark

ions react with SCN

Q1(aq).
Short Answer 1 mark

ions to form a precipitate of AgSCN(s).

Q1(aq).
Short Answer 1 mark

+ SCN

Q1(aq).
Short Answer 1 mark

AgSCN

Q1(aq)(s).
Short Answer 1 mark

When all Ag

Q1(aq).
Short Answer 1 mark

ions have been removed from solution, excess SCN

Q1(aq).
Short Answer 1 mark

ions react with Fe

Q1(aq).
Short Answer 1 mark

to form a complex ion, FeSCN

Q1(aq).
Short Answer 1 mark

, which has a red colour.

Q1(aq).
Short Answer 1 mark

+ SCN

Q1(aq).
Short Answer 1 mark

FeSCN

Q1(aq).
Short Answer 1 mark

The appearance of the red colour indicates the end-point. A student carries out an experiment to determine the equilibrium constant, K [Fe

Q1(aq).
Short Answer 1 mark

]eqm [Fe

Q1(aq).
Short Answer 1 mark

]eqm [Ag

Q1(aq).
Short Answer 1 mark

]eqm The student makes 250.0 cm of 0.0200 mol dm KSCN(aq) to use in the titration.

Q1(a).
Short Answer 1 mark

Calculate the mass of solid potassium thiocyanate, KSCN

Q1(a)(s).
Short Answer 1 mark

, needed to make 250.0 cm 0.0200 mol dm KSCN(aq). mass of KSCN

Q1(a)(s).
Short Answer 1 mark

= g [1] * 0000800000002 *

Q1(b).
Short Answer 1 mark

Describe how the student should make 250.0 cm of 0.0200 mol dm KSCN(aq) starting from the mass of KSCN

Q1(b)(s).
Short Answer 1 mark

calculated in

Q1(b)(a).
Short Answer 1 mark

in a 50 cm beaker. Give the name and size of any key apparatus used. W rite your answer using a series of numbered steps.

Q1(c).
Short Answer 1 mark

The student uses the following method to determine K step 1 Add 25.0 cm of 0.100 mol dm aqueous silver nitrate, AgNO

Q1(aq).
Short Answer 1 mark

, into a dry conical flask. Label this flask A. step 2 Add 25.0 cm of 0.100 mol dm aqueous iron(II) sulfate, FeSO

Q1(aq).
Short Answer 1 mark

, into flask A. step 3 Seal flask A, using a bung. step 4 Allow flask A to stand for twelve hours. step 5 Transfer 10.0 cm of the mixture from flask A into another conical flask, flask B, without disturbing the precipitate in flask A. step 6 Titrate the sample in flask B with 0.0200 mol dm aqueous potassium thiocyanate, KSCN(aq). step 7 Repeat steps 5 and 6 until concordant values are obtained.

Q1(i).
Short Answer 1 mark

Suggest why flask A is sealed with a bung in step 3.

Q1(ii).
Short Answer 1 mark

Suggest why flask A is left to stand for twelve hours in step 4.

Q1(iii).
Short Answer 1 mark

Identify the precipitate in flask A in step 5.

Q1(d).
Short Answer 1 mark

The student's results are shown in Table 1.1 Table 1.1 rough titration titration 1titration 2titration 3 final burette reading / cm 3 22.50 21.75 31.65 32.20 initial burette reading / cm 3 0.00 0.00 9.75 10.20 titre / cm 22.50 21.75 21.90 22.00

Q1(i).
Short Answer 1 mark

State if concordant titres have been achieved. Explain your answer.

Q1(ii).
Short Answer 1 mark

Calculate the percentage error in the titre volume in titration 2. Show your working. percentage error =

Q1(iii).
Short Answer 1 mark

The student repeats the experiment using KSCN(aq) at a higher concentration. The student obtains smaller titres. Suggest one reason why a larger titre is better than a smaller titre.

Q1(e).
Short Answer 1 mark

Another student calculates a mean titre of 21.85 cm . Use this value to complete the following calculation.

Q1(i).
Short Answer 1 mark

Calculate [Ag

Q1(aq).
Short Answer 1 mark

] in the equilibrium mixture in flask A. [Ag

Q1(aq).
Short Answer 1 mark

]eqm = mol dm

Q1(ii).
Short Answer 1 mark

Calculate [Fe

Q1(aq).
Short Answer 1 mark

] in the equilibrium mixture in flask A. [Fe

Q1(aq).
Short Answer 1 mark

]eqm = mol dm

Q1(iii).
Short Answer 1 mark

The formula for the equilibrium constant, K , is shown. [Fe

Q1(aq).
Short Answer 1 mark

]eqm [Fe

Q1(aq).
Short Answer 1 mark

]eqm [Ag

Q1(aq).
Short Answer 1 mark

]eqm Determine the value of K Give the units of K = units [2] * 0000800000005 *

Q1(f).
Short Answer 1 mark

Several other students perform the same experiment at dif ferent temperatures. The K values that they obtain are used to produce the graph in Fig. 1.1. 2.5 2.7 2.9 / units not shown 3.1 3.3 2.6 2.8 3.0 3.2 3.4 3.5 300 320 340 360 temperature / K

Q1(i).
Short Answer 1 mark

One student suggests that K is directly proportional to temperature. State and explain if the results displayed in Fig. 1.1 support this suggestion.

Q1(ii).
Short Answer 1 mark

Another student suggests that the data represented in the graph in Fig. 1.1 is reliable. Explain how the graph supports this suggestion.

Q1(iii).
Short Answer 1 mark

Use the data displayed in Fig. 1.1 to state if the forward reaction is exothermic or endothermic.

Q1(aq).
Short Answer 1 mark

Ag(s) + Fe

Q1(aq).
Short Answer 1 mark

Explain your answer. forward reaction explanation [1] [Total: 17] * 0000800000007 * BLANK PAGE * 0000800000008 *

Q2.
Short Answer 1 mark

A student carries out an experiment to determine the concentration of aqueous sulfate ions,

Q2(aq).
Short Answer 1 mark

, in a sample of lake water.

Q2(a).
Short Answer 1 mark

The student uses the following method. step 1 Transfer 25.00 cm of the lake water sample to a beaker and record its conductivity step 2 Add 5.00 cm of 0.100 mol dm aqueous barium hydroxide, Ba(OH)

Q2(aq).
Short Answer 1 mark

, to the beaker. step 3 Stir the mixture and record the conductivity of the contents of the beaker as shown in Fig. 2.1. step 4 Repeat steps 2 and 3 until a total of 40.00 cm of 0.100 mol dm Ba(OH)

Q2(aq).
Short Answer 1 mark

has been added to the beaker. 0.0 conductivity meter

Q2(i).
Short Answer 1 mark

Suggest a suitable piece of apparatus to transfer 25.00 cm of the lake water sample to the beaker in step 1.

Q2(ii).
Short Answer 1 mark

0.100 mol dm Ba(OH)

Q2(aq).
Short Answer 1 mark

is an irritant to skin and eyes. Other than wearing safety goggles, state one safety precaution that the student should take when conducting this experiment.

Q2(b).
Short Answer 1 mark

The student's results are given in Table 2.1. A correction can be applied to the conductivity values to take into account dilution of the solution as its volume increases using the following equation. corrected conductivity = measured conductivity × (total volume in beaker) 25.00 Table 2.1 reading number volume of 0.100 mol dm Ba(OH)

Q2(aq).
Short Answer 1 mark

added to beaker / cm total volume in beaker / cm measured conductivity corrected conductivity 1 0.00 25.00 37 000 37 000 2 5.00 23 000 27 600 3 10.00 12 000 16 800 4 15.00 2 300 3 680 5 20.00 5 000 6 25.00 12 000 7 30.00 18 000 8 35.00 21 000 9 40.00 24 500

Q2(i).
Short Answer 2 marks

Complete T able 2.1.

Q2(ii).
Short Answer 1 mark

Identify the independent variable in this experiment.

Q2(c).
Short Answer 1 mark

Plot a graph on the grid in Fig. 2.2 to show the relationship between corrected conductiv ity and volume of 0.100 mol dm Ba(OH) added to beaker. Use a cross (×) to plot each data point. Draw a line of best fit using readings 1 to 4 and another line of best fit using readings 5 to 9. Extend the lines so that they intersect. 0 10 20 305 15 25 35 40 10 000 20 000 30 000 40 000 50 000 60 000 70 000 corrected conductivity / μS cm volume of 0.100 mol dm Ba(OH)

Q2(aq).
Short Answer 1 mark

added to beaker / cm [2] * 0000800000011 *

Q2(d).
Short Answer 1 mark

The point on the graph where the two lines intersect indicates the volume of 0.100 mol dm Ba(OH)

Q2(aq).
Short Answer 1 mark

required to react exactly with the SO

Q2(aq).
Short Answer 1 mark

present in 25.00 cm of lake water being tested.

Q2(i).
Short Answer 1 mark

Use the graph in Fig. 2.2 to determine the volume of 0.100 mol dm Ba(OH)

Q2(aq).
Short Answer 1 mark

required to react exactly with SO

Q2(aq).
Short Answer 1 mark

in the lake water sample. volume required = cm

Q2(ii).
Short Answer 1 mark

The equation for the reaction taking place in the beaker is shown.

Q2(s).
Short Answer 1 mark

Use your answer in (d)(i) to calculate the concentration of SO

Q2(aq).
Short Answer 1 mark

in the lake water sample. concentration of SO

Q2(aq).
Short Answer 1 mark

= mol dm [1] * 0000800000012 *

Q2(e).
Short Answer 1 mark

The concentration of SO

Q2(aq).
Short Answer 1 mark

in a sample of water can also be determined by measuring the mass of precipitate produced when excess Ba(OH)

Q2(aq).
Short Answer 1 mark

is added to the sample. The student suggests the following method. step 1 Place 25.0 cm of the water sample in a conical flask. step 2 Add excess 0.100 mol dm Ba(OH)

Q2(aq).
Short Answer 1 mark

to the flask. step 3 Filter the contents of the flask. step 4 Dry the residue in a warm oven. step 5 Measure the mass of residue.

Q2(i).
Short Answer 1 mark

Draw a labelled diagram to describe the arrangement of apparatus that would be needed to complete step 3.

Q2(ii).
Short Answer 1 mark

Suggest a step that the student should add between steps 3 and 4 to improve this method.

Q2(iii).
Short Answer 1 mark

Describe what the student can do to ensure that the residue weighed in step 5 is completely dry.

Q2(iv).
Short Answer 1 mark

The mass of residue is used to calculate the concentration of SO

Q2(aq).
Short Answer 1 mark

Suggest the ef fect, if any , on the concentration of SO

Q2(aq).
Short Answer 1 mark

that is calculated if the residue is not completely dried in step 4. Explain your answer. effect on concentration of SO

Q2(aq).
Short Answer 1 mark

calculated explanation [1] [Total: 13] * 0000800000013 * BLANK PAGE * 0000800000014 * Important values, constants and standards molar gas constant R = 8.31 J Kmol Faraday constant F = 9.65 × 10 C mol Avogadro constant L = 6.02 × 10mol electronic charge e = -1.60 × 10-19 molar volume of gas V = 22.4 dmmol at s.t.p. (101 kPa and 273 K) = 24.0 dmmol at room conditions ionic product of water K = 1.00 × 10-14mol (at 298 K (25 °C)) specific heat capacity of waterc = 4.18 kJ kg (4.18 J g ) * 0000800000015 * To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Group The Periodic Ta ble of Elements hydrogen 1.0 helium 4.0 12 13 14 15 16 17 18 3456789 10 11 12 lithium 6.9 beryllium 9.0 atomic number atomic symbolKey name relative atomic mass sodium 23.0 magnesium 24.3 potassium 39.1 calcium 40.1 rubidium 85.5 strontium 87.6 caesium 132.9 barium 137.3 francium radium boron 10.8 aluminium 27.0 gallium 69.7 indium 114.8 thallium 204.4 carbon 12.0 silicon 28.1 germanium 72.6tin 118.7 lead 207.2 titanium 47.9 zirconium 91.2 hafnium 178.5 rutherfordium vanadium 50.9 niobium 92.9 tantalum 180.9 dubnium chromium 52.0 molybdenum 95.9 tungsten 183.8 seaborgium manganese 54.9 technetium rhenium 186.2 bohrium iron 55.8 ruthenium 101.1 osmium 190.2 hassium cobalt 58.9 rhodium 102.9 iridium 192.2 meitnerium nickel 58.7 palladium 106.4 platinum 195.1 darmstadtium copper 63.5 silver 107.9 gold 197.0 roentgenium zinc 65.4 cadmium 112.4 mercury 200.6 copernicium flerovium livermorium nitrogen 14.0 phosphorus 31.0 arsenic 74.9 antimony 121.8 bismuth 209.0 oxygen 16.0 sulfur 32.1 selenium 79.0 tellurium 127.6 polonium fluorine 19.0 chlorine 35.5 bromine 79.9 iodine 126.9 neon 20.2 argon 39.9 krypton 83.8 xenon 131.3 radon nihonium moscovium tennessine oganesson scandium 45.0 yttrium 88.9 57-71 lanthanoids 89-103 actinoids lanthanum 138.9 lanthanoids actinoids actinium cerium 140.1 thorium 232.0 praseodymium 140.9 protactinium 231.0 neodymium 144.2 uranium 238.0 promethium neptunium samarium 150.4 plutonium europium 152.0 americium gadolinium 157.3 curium terbium 158.9 berkelium dysprosium 162.5 californium holmium 164.9 einsteinium erbium 167.3 fermium thulium 168.9 mendelevium ytterbium 173.1 nobelium lutetium 175.0 lawrencium - * 0000800000016 *