Homework One 1. What is a standard solution? (1)

A solution of accurately known concentration

2. State three properties of a PRIMARY STANDARD. (1)

Stable in air (oxygen) and in solubility / readily soluble in a solvent (usually water) / high

purity / reasonably large gfm to minimise uncertainty when weighting samples

3. Name three primary standard chemicals. (1)

Any valid example. e.g. hydrated oxalic acid / anhydrous sodium carbonate / potassium

dichromate / sodium oxalate / disodium salt of EDTA

4. (i) Describe in detail how you would prepare 250 cm3 of a standard solution

containing 1.26g of oxalic acid {(COOH)2.2H2O} (3)

Accurately weigh approximately 1.26g of oxalic acid (weigh by difference in a beaker)

Dissolve using DI or distilled water, transfer into a 250cm3 volumetric flask

Repeat rinse the beaker and flush rinsings into flask

Make the solution up to the 250 mark with DI or distilled water

Stopper and invert to ensure thorough mixing

(ii) Calculate the concentration, in mol l-1, of this solution ? (1)

n = m/gfm C = n/v

= 1.26/126 = 0.01/025

= 0.01 moles = 0.04 mol l-1

5. Solutions which are not primary standards require standardisation before they

are used. What does standardisation mean? (1)

Determining the concentration of a solution by titrating it against a primary standard of

known concentration.

6. Margaret carried out a titration to find the concentration of a solution of

hydrochloric acid by reacting it with 0.001mol l-1 sodium hydroxide. Margaret’s

results are shown below:

1st titre = 23.1cm3 2nd titre = 23.4cm3

3rd titre = 23.0cm3 4th titre = 22.8cm3

Calculate the average titre from these results (1)

Titres must be +/- 0.1cm3 so..... (23.1 + 23.0)/2 = 23.05cm3

7. In an experiment to standardise dilute hydrochloric acid, 25.0 cm3 of 0.200 mol l-1

sodium carbonate solution was neutralised by 20.0 cm3 of the dilute hydrochloric acid.

Calculate the concentration, in mol l-1, of the hydrochloric acid.

Na2CO3 + 2HCl 2NaCl + CO2 + H2O

1 : 2

CaCO3 n = V X C HCl n = n/V

= 0.025 X 2 = 0.01/0.02

= 5 X 10-3 moles = 0.5 mol l-1

(3)

8. In an experiment to determine the concentration of a hydrogen peroxide solution,

H2O2, 25.0 cm3 of hydrogen peroxide solution was pipette into a conical flask. An

excess of dilute sulfuric acid was added to the flask using a measuring cylinder.

The solution in the flask was then titrated with 0.0200 mol l-1 potassium permanganate

solution, KMnO4. 28.1 cm3 of potassium permanganate solution was required to reach

the end-point.

5H2O2 + 2MnO4- + 6H+ 2Mn2+ + 5O2 + 8H2O

(i) Why was sulfuric acid added to the flask? (1)

To provide hydrogen ions

(ii) Explain why an indicator is not required for this titration. (1)

The permanagate ion is self indicating (purple to colourless)

(iii) Explain why the hydrogen peroxide was measured out with a pipette

but the sulfuric acid was measured out with a measuring cylinder. (1)

The pipette was used for hydrogen peroxide to ensure accuracy.

The measuring cylinder was used for sulphuric acid as it is required in excess and therefore

accuracy is not as important (should this be accuracy or precision????)

(iv) Calculate the concentration, in mol l-1 of the hydrogen peroxide

solution (3)

5H2O2 : 2MnO4-

1.405 X 10-3 : 5.62 X 10-3

MnO4- n = V X C H2O2 C = n/V

= 0.0281 X 0.02 = 1.405 X 10-3 / 0.025

= 5.62 X 10-3 = 0.0562 mol l-1

9. Andy was asked to dilute 500 cm3 of a 2.0 mol l-1 solution of sulfuric acid with

250 cm3 of deionised water.

Calculate the concentration of the diluted solution. (2)

Final volume is 750cm3 so...

C1V1 = C2V2

C2 = (C1V1)/V2

= (2.0 X 0.5) / 0.75

= 1.33 mol l-1

Total Marks (20)

Homework Two 1. Use the options (i) to (v) below to answer questions a,b,c, and d

(i) Mg2+ (ii) S2O32- (iii) NaOH (iv) I- (v) MnO4

-

a. Which substance can be estimated in an acid/base titration? (iii) NaOH

b. Which substance will reduce iodine, I2? (iii) NaOH check this out – OH- present but

also need H2?????

c. Which substance can be estimated in a complexometric titration? (i) Mg2+ using EDTA

d. Which substance is a powerful oxidising agent? (v) MnO4- (check data book – at bottom

LHS of ion electron table)

2. EDTA forms a 1:1 complex with Ni 2+(aq). What is the concentration, in mol l-1, of a

nickel(II) solution, if 20 cm3 of it reacts with 2 x 10 -3 moles of EDTA?

A 0.10 B 0.01 C 0.02 D 0.002

EDTA n = 2X10-3 1:1 ratio Ni 2+ C = n/v

= 2X10-3 /0.02

= 0.1 mol l-1 A

3. A 125 cm3 sample of a 3.0 mol l-1 HCl is diluted by adding 250 cm3 of water. The new

concentration of the HCl solution is

A 8.0 mol l-1 B 1.5 mol l-1 C 1.0 mol l-1 D 0.5 mol l-1

C1V1 = C2V2 C2 = C1V1 / V2

= 3 X 0.125 / 0.375

= 1.0 mol l-1 C

4. Which of the substances below would not be suitable as a primary standard?

A Oxalic acid B Potassium iodate

C Sodium hydroxide D Potassium hydrogenphthalate

C Sodium hydroxide because it absorbs moisture from air

5. What is the concentration of a solution obtained by dissolving 8.40 g of sodium

hydrogencarbonate (NaHCO3) in 500 cm3 of solution?

A 0.1 mol l-1 B 0.2 mol l-1 C 0.5 mol l-1 D 1.0 mol l-1

n = m / gfm C = n/V

= 8.40 / 84 = 0.1 / 0.5

= 0.1 moles = 0.2 mol l-1 B

6. What volume of 1.50 mol l-1 calcium chloride is required to make, by dilution with

water, one litre of a solution with a chloride ion concentration of 0.15 mol l-1?

a 50 cm3 b 75 cm3 c 100 cm3 d 200 cm3

CaCl2 C1V1 = C2V2

1: 2 V2 = C1V1 / V2

1.5 : 3.0 = (0.15 X 1) / 3.0

= 0.05 litres A

7. What volume of 0.25 mol l-1 calcium nitrate is required to make, by dilution with water,

500 cm3 of a solution with a nitrate ion concentration of 0.10 mol l-1?

A 50 cm3 B 100 cm3 C 200 cm3 D 400 cm3

Ca (NO3)2 C1V1 = C2V2

1 : 2 V2 = C1V1 / V2

0.25 : 0.5 = (0.1 X 0.5) / 0.5

= 0.1 litres B

8. A salt has the formula Pt(NH3)xCl2

0.02 moles of this salt required 40 cm3 of 2.0 mol l-1 hydrochloric acid for exact

neutralisation of the NH3. The value of X is

A 2 B 4 C 6 D 8.

HCl n = V X C

Salt : HCl = 0.04 X 2

0.02 : 0.08 = 0.08 moles

1 : 4 Pt (NH3)4 Cl2 B

9. To determine the concentration of acetic acid (CH3COOH) in a sample of vinegar,

25.00 cm3 of vinegar was pipetted into a 250 cm3 volumetric flask. Deionised water

was added to fill the flask up to the mark. 20.00 cm3 samples of this diluted vinegar

were titrated with 0.105 mol l-1 sodium hydroxide solution.

The equation for the reaction is:

CH3COOH + NaOH CH3COONa + H2O

The results of the titrations are shown in the table.

Burette

readings (cm3)

Titrations

1 2 3 4

Final 12.50 24.65 36.75 49.90

Initial 0.00 12.50 24.65 36.75

Titre volume (i) (ii) (iii) (iv)

a. What are the values of the titre volumes (i) to (iv)? (1)

(i)

12.50

(ii)

12.15

(iii)

12.10

(iv)

13.15

b. Calculate the average titre volume. (1)

(12.15 + 12.10)/2 = 12.125 cm3

c. Calculate the concentration of ethanoic acid in the undiluted vinegar. (1)

NaOH n = V X C 20cm3 sample 1.273125 X 10 –3 moles

= 0.012125 X 0.105 250cm3 sample 0.0159140625 moles

= 1.273125 X 10 –3 moles

C = n/v

= 0.159140626 / 0.025

= 0.64 mol l-1

d. How would a control experiment be carried out for this analysis? (2)

use a pure sample of acetic acid with an accurately known mass or concentration and

titrating this against the 0.105 mol l-1 NaOH solution.

10. Human teeth contain the element calcium. To determine the percentage mass of calcium

in a tooth 1.40 g of a dried tooth was dissolved in a small quantity of hot conc. nitric

acid. The solution was neutralised by the addition of sodium hydroxide and then made up

to 250 cm3 in a volumetric flask.

10.0 cm3 of this solution was pipetted into a conical flask and 1 cm3 of a concentrated.

ammonia/ammonium chloride pH 10 buffer was added. The solution was titrated with

0.02 mol l-1 EDTA using Eriochrome Black T indicator. The indicator turned blue after

22.5 cm3 of EDTA was added.

Calculate the percentage mass of calcium in the tooth. (3)

Ca + HNO3 Ca (NO3)2 + H2O

Ca (NO3)2 + EDTA Ca 2+ EDTA complex

1 : 1 ratio

10cm3 sample 4.5 X 10 -4 moles

250cm3 sample 0.01125 moles

Ca2+ m = n X gfm % mass = (Ca 2+/ tooth) X 100

= 0.01125 X 40.1 = (0.451125 / 1.40) X 100

= 0.451125g = 32.2%

11. In an experiment to find the percentage of sodium chloride in a sample of rock salt,

Catherine weighed 1.50 g of rock salt which was crushed and dissolved in deionised

water in a beaker. The mixture was filtered to remove insoluble material and the

filtrate was added to a 250 cm3 standard flask. Deionised water was used to make the

volume up to 250 cm3. The flask was stoppered and shaken to ensure thorough mixing.

25.0 cm3 samples of this solution were titrated with 0.100 mol l-1 silver(I) nitrate.

On average 23.5 cm3 of silver(I) nitrate was needed.

a. Catherine omitted two important steps when making up the solution in the standard

flask. Suggest any one of these mistakes. (1)

Should have dried it first to remove any traces of water

Mass should be “weigh approximate but accurate”

Repeat rinse beaker and add to flask

b. The equation for the reaction is

NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)

Calculate the percentage by mass of sodium chloride in the rock salt. (3)

AgNO3 n = V X C NaCl n = 0.00235 moles

= 0.0235 X 0.1 m = n X gfm

= 0.00235 moles = 0.00235 x (23.0 + 35.5)

= 0.137474g

% mass = (NaCl / rocksalt) / 100

= (0.137474 / 1.50) X 100

= 9.2%

Total Marks (20)

Homework Three

1. An experiment was carried out to find the purity of chalk, which contains mainly

calcium carbonate, CaCO3.

A 3.63 g sample of impure chalk was added to 50.0 cm3 of 2.00 mol l-1 hydrochloric acid.

The resulting solution was filtered into a volumetric flask and made up to 250.0 cm3.

25.00 cm3 samples of this solution were then titrated against 0.105 mol l-1

sodium hydroxide solution.

The results of the titrations are shown in the table below.

The equations for the reactions taking place are:

CaCO3 + 2HCl CaCl2 + CO2 + H2O

NaOH + HCl NaCl + H2O

a. This technique is a back titration. Why does calcium carbonate require to be analysed

by back titration? (1)

CaCO3 is insoluble, therefore a direct titration is not possible.

We react the CaCO3 with a known number of moles of HCl acid and then titrate the excess

HCl with known concentration of NaOH.

Burette

readings (cm3)

Titrations

1 2 3

Final 32.50 37.25 43.15

Initial 0.00 5.50 11.30

Titre volume 32.50 31.75 31.85

b. What is the average titre volume? (1)

Average titre = (31.75 + 31.85) / 2 = 31.80 cm3

c. Calculate the number of moles of hydrochloric acid added to the chalk. (1)

HCl n = V X C

= 0.05 X 2.00

= 0.1 moles

d. Calculate the number of moles of hydrochloric acid in the standard flask. (1)

NaOH n = V X C

= 0.0318 X 0.105

NaOH : HCl = 3.339 X 10-3 moles in 25.00 cm3 samples

1:1 ratio = 3.339 X 10-2 moles in 250 cm3 flask

e. Calculate the number of moles of calcium carbonate in the chalk. (1)

CaCO3 + 2HCl CaCl2 + CO2 + H2O

1 : 2

0.016695 : 3.339 X 10-2

f. Calculate the percentage mass of calcium carbonate in the chalk (1)

CaCO3 m = n X gfm %mass = (1.6711695/3.63) X 100

= 0.016695 X 106.1 = 46%

= 1.6711695g

2. In an experiment to find the percentage of iron in iron ore, 2.650 g of ore was dissolved

in hydrochloric acid. A powerful oxidising agent was then added to convert all the

iron(II) ions to iron (III) ions and the solution diluted to 250 cm3 in a standard flask.

25.0 cm3 of this solution was then transferred to a conical flask and when a few drops

of ammonium thiocyanate solution were added, a red colour appeared due to reaction

with iron(III) ions. The solution was titrated with 0.10 mol l–1 titanium(III) solution

until the red colour just disappeared. The titration results were as follows.

Note: titanium(III) is a powerful reducing agent, itself being oxidised to titanium(IV).

a. Write the ion–electron half-equations for the reactions that occur during

the titration. (1)

Ti 3+ Ti 4+ + e-

Fe 3+ + e- Fe 2+

b. Deduce the reacting mole ratio of iron(III) ions with titanium(III) ions from the

ion-electron equations in a. (1)

1 : 1 ratio

c. (i) What volume of titanium(III) solution should be used for any calculation? (1)

titres must be +/- 0.1 cm3 so..

(22.65 + 22.60 + 22.65) / 3 = 22.63 cm3

Titration Titre/cm3

1 24.00

2 22.65

3 22.60

4 22.65

(ii) Calculate the percentage of iron in the ore. (3)

1 : 1 ratio of Ti 3+ : Fe 3+

Ti 3+ n = V X C Fe 3+ m = n X gfm

= 0.02263 X 0.10 = 0.02263 X 55.8

= 0.002263 moles in 25 cm3 sample = 1.262754 g

= 0.02263 moles in 250 cm3 original sample

% Fe in ore = (Fe/ore) X 100

= (1.262754 / 2.650) X 100

= 47.65 % 48% of iron in ore

3. The chloride of an element Z reacts with water according to the following equation.

ZCl4(l) + 2 H2O(l) ZO2(s) + 4 HCl(aq)

1.304 g of ZCl4 was added to water. The solid ZO2 was removed by filtration and the

resulting solution was made up to 250 cm3 in a volumetric flask. A 25.0 cm3 portion of

this solution was titrated against a 0.112 mol l-1 solution of sodium hydroxide, of

which 21.7 cm3 were required to reach the end point.

a. Assuming each reaction is 100% efficient. Calculate the number of moles of ZCl4

that reacted with the water. (2)

NaOH n = V X C NaOH + HCl NaCl + H2O

= 0.0217 X 0.112

= 2.4304 X 10 -3 moles in 25 cm3 samples

= 2.4304 X 10 -2 moles in 250 cm3 original sample

NaOH + HCl 1 : 1 ratio

....so we know the no moles of HCl, we can use this to deduce the no moles of ZCl4.

ZCl4(l) + 2 H2O(l) ZO2(s) + 4 HCl(aq)

1 : 4

6.076 X 10-3 moles 2.4304 X 10-2 moles

b. Identify element Z. (2)

....from question we now know that 6.076 X 10-3 moles of ZCl4 have a mass of 1.304g

1.304g 6.076 X 10-3 moles

214.6g 1 mole of ZCl4 4 Cl = 4 X 35.5 = 142g

Z : Cl mass of Z = mass ZCl4 – mass 4Cl

72.6g : 142g 72.6 = 214.6 – 4(35.5)

Element Z has mass 72.6g so it is germanium Ge

4. The mineral cerussite contains lead(II) carbonate, PbCO3 and some inert rocky

material. 1.00g of cerussite was crushed and added to 25.0 cm3 of 1.00 mol l-1 nitric acid

(an excess). When no more reaction was observed the mixture was filtered and the

filtrate was made up to 250 cm3 in a standard flask.

25.0 cm3 samples of this solution were titrated with 0.100 mol l-1 sodium hydroxide

solution. The average titre value was 20.5 cm3.

PbCO3 + 2HNO3 Pb(NO3)2 + CO2 + H2O

NaOH + HNO3 NaNO3 + H2O

a. Why is it considered good practice to carry out titrations more than once? (1)

To allow concordant results to be obtained (this improves reliability???)

b. Calculate the percentage by mass of lead(II) carbonate in the cerussite. (3)

{Hint: this is a back titration)

NaOH from titration.

From the equations we can see HNO3 : NaOH = 1:1.

So, using HNO3 we can deduce excess HNO3 n = V X C

= 0.0205 X 0.1

= 0.00205 moles in 25 cm3 sample

= 0.0205 moles in 250 cm3 sample

Original HNO3 from the equations we can see HNO3 : NaOH = 1:1

n = V X C

= 0.025 X 1.00

= 0.025 moles

HNO3 used = HNO3 original - HNO3 excess

= 0.025 - 0.0205

= 4.5 X 10 -3 moles

PbCO3 + 2HNO3 PbCO3 m = n X gfm

1 mole : 2 moles = 2.25 X 10-3 X 267.2

2.25 X 10-3 : 4.5 X 10-5 = 0.6012g

% mass = (PbCO3 / cerusite) X 100

= (0.6012 / 1.00 ) X 100

= 60%

Total marks (20)

Homework Four 1. A compound contains 4.6 g of sodium, 2.8 g of nitrogen and 9.6 g of oxygen. Use this

information to calculate the formula of the compound (2)

Empirical formulae Na : N : O

4.0g : 2.8g : 9.6g

No moles 0.2 : 0.2 : 0.6

Ratio 1 : 1 : 3 NaNO3

2. Bryce took two equal volumes of calcium iodide solution. To one he added and excess of

silver(I) nitrate which precipitated all the iodine out as silver(I) iodide, AgI.

The precipitate was filtered, washed and dried and found to have a mass of 9.40 g

To the second volume Bryce added an excess of sodium carbonate solution which

precipitated all the calcium out as 2.00 g of calcium carbonate, CaCO3.

a. What type of quantitative analysis was carried out on the calcium iodide? (1)

gravimetric analysis

b. Why was the silver(I) nitrate added in excess? (1)

to ensure that the reaction has gone to completion

c. The silver(I) iodide was dried in an oven. Suggest a piece of apparatus that should

be used to store this compound as it cools. (1)

desiccators

(store sample in here as it cools down to ensure that no moisture is absorbed)

d. Use the results of the experiment to confirm that the formula of

calcium iodide is CaI2. (2)

CaI2 (aq) + 2AgNO3 (aq) 2AgI(s) + Ca(NO3)2 (aq)

1 : 2

239.9g 469.6g

(293.9 / 469.6) X 9.40g 9.40g

6g 9.40g

CaI2 + Na2CO3 CaCO3 + 2NaI

1 : 1

293.9g 100.1g

6g (100.1 / 293.9) X 6

2.04 = 2g

3. A 0.4500 g sample of impure potassium chloride, KCl, was dissolved in water and

treated with an excess of silver nitrate. 0.8402 g of silver chloride, AgCl, was

produced. Calculate the percentage of potassium chloride in the original sample. (2)

KCl (aq) + AgNO3 (aq) AgCl(s) + KNO3 (aq)

1 1

74.6g 143.4g

(74.6 / 143.4) X 0.8402 0.8402g

0.4370914923g 0.8402g

%purity = (KCl / original) X 100

= (0.4370914923 / 0.4500) X 100

= 97%

4. A 3.00 g sample of an alloy containing only Lead, Pb and tin, Sn, was dissolved in dilute

nitric acid. Sulfuric acid was added to this solution, which precipitated 1.69 g of

lead(II) sulfate, PbSO4. Assuming that all of the lead was precipitated, what is the

percentage of tin in the sample? (3)

Pb(NO3)2(aq) + H2SO4 (aq) PbSO4 (s) + 2HNO3 (aq)

1 1

Pb in PbSO4

207.2g 303.3g

1.69g (207.9 / 303.3) X 1.69

1.154526871g

Sn = Alloy - Pb %Sn = (1.845473129 / 3) X 100

= 3.00 - 1.15452687 = 61.5%

= 1.845473129g

5. Aluminium can be determined gravimetrically by reaction with a solution of

8-hydroxyquinoline (C9H7NO).

Al3+(aq) + 3C9H7NO(aq) Al(C9H6NO)3(s) + 3H+ (aq)

18.571 g of a mineral containing aluminium was dissolved in acid and the solution was

made up to 250 cm3 in a standard flask. An excess of 8-hydroxyquinoline was added to

25 cm3 of the aluminium solution. The precipitate formed was washed, dried and heated

to constant mass. 0.1248 g was obtained.

a. Outline the process of heating the precipitate to constant mass. (1)

Heat a crucible to drive off water. Cool in a desiccator.

Transfer the precipitate to the crucible and weigh in a +/- 0.001g balance

Heat the precipitate in the crucible, cool in desiccator, reweigh.

Repeat the process until the crucible and precipitate maintain a constant mass

(require 2 consecutive readings)

b. Calculate the percentage mass of aluminium in the mineral. (3)

Al3+(aq) + 3C9H7NO(aq) Al(C9H6NO)3(s) + 3H+ (aq)

1 : 3 1

27g 459g

(27/459) X 0.1248 0.1248g

0.00734g 0.1248g

0.00734g in 25cm3

0.0734g in 250cm3

% Al = (Al / mineral) X 100

= (0.0734 / 18.571) X 100

= 0.40%

6. A sample containing nickel was treated with dimethylglyoxime, resulting in the

precipitation of nickel(II) dimethylglyoximate. The sample had a mass of 1.247 g

and 3.761 g of precipitate was produced.

a. Find the percent purity of nickel in the sample. (3)

Ni 2+ precipitate complex

58.7g 288.7g

(58.7 / 288.7) X 3.761 0.7647062695g

%purity = (Ni2+ / sample) X 100

= (0.7647062695 / 1.247) X 100

= 61.32%

b. Suggest how nickel could be determined volumetrically. (1)

Titrate with EDTA (gives 1:1 ratio with 2+ ion)

Total marks (20)