Curriculum P - Frankly Chemistry · as topic 3 acid-base titrations molar solutions preparation of a standard solution simple volumetric calculations calculations requiring molar

ACID-BASE TITRATIONSAS TOPIC 3

MOLAR SOLUTIONS

PREPARATION OF A STANDARD SOLUTION

SIMPLE VOLUMETRIC CALCULATIONS

CALCULATIONS REQUIRING MOLAR CONCENTRATIONOF A PRIMARY STANDARD

DILUTION QUESTIONS

WATER OF CRYSTALLISATION

PERCENTAGE PURITY

TEST QUESTION I

TEST QUESTION II

TEST QUESTION III

EXPERIMENT TO FIND THE EQUATION FOR A REACTION

ESTIMATING THE CONCENTRATION OF SALTS IN SEA WATER

EXPERIMENT TO FIND THE EQUATION FOR A REACTIONDECOMPOSITION OF POTASSIUM HYDROGEN CARBONATE

Questionsheet 1

Questionsheet 2

Questionsheet 3

Questionsheet 4

Questionsheet 5

Questionsheet 6

Questionsheet 7

Questionsheet 8

Questionsheet 9

Questionsheet 10

Questionsheet 11

Questionsheet 12

Questionsheet 13

QUESTIONSHEETS AS LevelCHEMISTRY

15 marks

14 marks

12 marks

16 marks

15 marks

14 marks

14 marks

15 marks

16 marks

16 marks

12 marks

9 marks

16 marks

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AuthorsTrevor Birt John BrockingtonDonald E Caddy Kevin FrobisherAndrew Jones Andy ShepherdAdrian Bond Stuart Barker

EditorsJohn BrockingtonStuart Barker

Do notwrite inmargin

TOTAL /

AS Level TOPIC 3 Questionsheet 1

a) Define the terms molar concentration (or ‘molarity’) and molar solution.

Molar concentration ..........................................................................................................................................

....................................................................................................................................................................... [2]

Molar solution ...................................................................................................................................................

....................................................................................................................................................................... [2]

b) What is the molar concentration of each of the following?

(i) Hydrochloric acid containing 146 g dm-3 of HCl.

....................................................................................................................................................................... [1]

(ii) Potassium iodide solution containing 300 g dm-3 of KI.

....................................................................................................................................................................... [1]

c) A solution of ethanedioic acid-2-water, H2C

2O

4 .2H

2O, was made by dissolving 1.48 g of the pure solid in

water and making up to the graduation mark in a 250 cm3 graduated flask. Calculate the concentration of thesolution:

(i) in g dm-3

....................................................................................................................................................................... [1]

(ii) in mol dm-3

...........................................................................................................................................................................

....................................................................................................................................................................... [2]

d) How much anhydrous sodium carbonate must be weighed out in order to prepare 250 cm3 of solution ofconcentration 0.05 mol dm-3?

...........................................................................................................................................................................

...........................................................................................................................................................................

....................................................................................................................................................................... [3]

e) What mass of sodium hydroxide is present in 500 cm3 of a solution of concentration 0.0982 mol dm-3?

...........................................................................................................................................................................

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....................................................................................................................................................................... [3]

MOLAR SOLUTIONS

15

TOPIC 3 ANSWERS & MARK SCHEMES

QUESTIONSHEET 1

AS Level

MOLAR SOLUTIONS

a) Molar concentration The number of moles of solute (1)dissolved in 1 dm3 of solution (1)

Molar solution One which contains 1 mol of solute (1)in 1 dm3 of solution (1)

b) (i) Mr (HCl) = 36.5 Concentration = 146/36.5 = 4.00 mol dm-3 (1)

(ii) Mr (KI) = 166 Concentration = 300/166 = 1.81 mol dm-3 (1)

c) (i) Concentration = 1.48 × 4 = 5.92 g dm-3

(1)

(ii) Mr (H

2C

2O

4 .2H

2O) = 126 (1)

Concentration = 5.92/126 = 0.0470 mol dm-3 (1)

d) 1 M Na2CO

3 contains 106 g dm-3 (1)

0.05 M Na2CO

3 contains 106 × 0.05 = 5.30 g dm-3 (1)

∴in 250 cm3 there is 5.30/4 = 1.325 g Na2CO3 (1)

e) Number of moles in solution = molarity × volume (in dm3)= 0.0982 × 0.5 = 0.0491 (1)

Mr (NaOH) = 40 (1)

Mass = 0.0491 × 40 = 1.96 g (1)


TOTAL /


14


a) (i) What is meant by a primary standard in volumetric analysis?

...........................................................................................................................................................................

....................................................................................................................................................................... [1]

(ii) State three requirements of a primary standard.

...........................................................................................................................................................................

....................................................................................................................................................................... [3]

b) Anhydrous sodium carbonate can be used as a primary standard in acid-base titrations.In an experiment, a student weighed an empty weighing bottle on an analytical balance. She took the bottle offthe balance pan, added some anhydrous sodium carbonate and reweighed the bottle, after which she tipped thesodium carbonate into a 100 cm3 beaker and weighed the bottle again. Using a stirring rod, she dissolved thesodium carbonate in deionised water, and then poured the solution into a 250 cm3 graduated volumetric flask.She washed the beaker and stirring rod with deionised water and transferred the washings also to the graduatedflask. With more deionised water she made the solution up to the graduation mark, adding water dropwise atthe end. Finally, she inserted the stopper and slowly inverted the flask about six times.

(i) Why did the student use an analytical balance?

....................................................................................................................................................................... [1]

(ii) Why did she take the bottle off the balance pan before adding the sodium carbonate?

....................................................................................................................................................................... [1]

(iii) Why did she reweigh the bottle after tipping the sodium carbonate into a beaker?

....................................................................................................................................................................... [1]

(iv) Why did she use deionised water rather than tap water?

....................................................................................................................................................................... [1]

(v) Why did she wash out the beaker and transfer the washings?

....................................................................................................................................................................... [1]

(vi) When filling the flask to the graduation mark, why did she add deionised water dropwise at the end?

....................................................................................................................................................................... [1]

(vii) How did the student know when she had added the correct amount of water?

....................................................................................................................................................................... [1]

(viii)Why did she invert the flask about six times?

....................................................................................................................................................................... [1]

(ix) Why did she invert the flask slowly?

....................................................................................................................................................................... [1]

(x) What would the student have done if she had overfilled the graduated flask with deionised water?

....................................................................................................................................................................... [1]


QUESTIONSHEET 2

AS Level


a) (i) A substance which can be weighed out directly to give a solution of known concentration (1)

(ii) Must be readily available in a very pure state (1)Must be stable / must not deteriorate on standing (1)Must be non-hygroscopic / must not absorb water vapour (1)Should have a high M

r so that a substantial amount is weighed out / so that weighing errors are negligible (1)

(Accept any 3)

b) (i) For accuracy (1)

(ii) To prevent the sodium carbonate from soiling / damaging the balance pan if it was spilt (1)

(iii) To allow for particles of sodium carbonate which adhered to the weighing bottle / which were not transferred (1)

(iv) Tap water contains dissolved substances which could affect the titration (1)

(v) To avoid errors due to some sodium carbonate solution remaining in the beaker (1)

(vi) To avoid overfilling the flask (1)

(vii) When the bottom of the liquid meniscus coincided with the graduation mark (1)

(viii)To get a uniform / homogeneous solution (1)Or to avoid a concentration gradient in the flask (1)

(ix) To give the air bubble a chance to move up and down / to ensure that the solution became properly mixed (1)

(x) Discarded the solution / started again (1)


TOTAL /


12


NOTE Throughout this Questionsheet marks are awarded for correctly balanced chemical equations.

a) What volume of sulfuric acid containing 0.20 mol dm-3 is required to neutralise completely 20.0 cm3 of asolution of potassium hydroxide containing 0.50 mol dm-3?

...........................................................................................................................................................................

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....................................................................................................................................................................... [3]

b) 25.0 cm3 of a solution of sodium hydroxide is neutralised by 19.8 cm3 of a solution of hydrochloric acidcontaining 0.15 mol dm-3. Calculate the concentration of the sodium hydroxide solution.

...........................................................................................................................................................................

...........................................................................................................................................................................

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c) Calculate the volume of sodium hydroxide solution, containing 0.0990 mol dm-3, required to neutralise asolution containing 0.148 g of ethanoic acid, CH

3COOH.

...........................................................................................................................................................................

...........................................................................................................................................................................

...........................................................................................................................................................................

....................................................................................................................................................................... [3]

d) A sample of sodium carbonate solution was neutralised by 19.6 cm3 of hydrochloric acid of concentration0.103 mol dm-3. Calculate the mass of sodium carbonate in the sample solution.

...........................................................................................................................................................................

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QUESTIONSHEET 3

AS Level


a) H2SO

4(aq) + 2KOH(aq) → K

2SO

4(aq) + 2H

2O(1) (1)

n (H2SO

4) = n (KOH) / 2 = 20.0(10-3)0.5 / 2 = 5 × 10-3 mol H

2SO

4 (1)

Volume required = 5(10-3) / 0.2 = 25 x 10-3 dm3 = 25.0 cm3 (1)

b) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O (1) (1)

n (NaOH) = n (HCl) = 19.8(10-3) 0.15 = 2.97 × 10-3 mol NaOH (1)c (NaOH) = 2.97(10-3 ) / 25.0(10-3) = 0.1188 mol dm-3 = 0.119 mol dm-3 (1)

c) CH3COOH(aq) + NaOH(aq) → CH

3COO- Na+(aq) + H

2O(l) (1)

n (NaOH) = n (CH3COOH) = 0.148/60 = 2.47 × 10-3 mol (1)

Volume required = 2.47(10-3)/0.0990 = 0.0249 dm3 = 24.9 cm3 (1)

d) Na2CO

3(aq) + 2HCl(aq) → 2NaCl(aq) + CO

2(g) + H

2O (1)

n (Na2CO

3) = n (HCl)/2 = 19.6(10-3) 0.103/2 = 1.01 × 10-3 mol (1)

m (Na2CO

3) = 1.01(10-3) 106 = 0.107 g (1)


TOTAL /


CALCULATIONS REQUIRING MOLAR CONCENTRATION OF A PRIMARY STANDARD

16


a) 1.547 g of anhydrous sodium carbonate were dissolved in deionised water and the solution was made up to250 cm3 in a graduated flask. 25.0 cm3 of this solution neutralised 24.6 cm3 of dilute sulfuric acid. Calculate,first, the concentration of sodium carbonate, then that of the sulfuric acid, both in moles per dm3.

...........................................................................................................................................................................

...........................................................................................................................................................................

...........................................................................................................................................................................

...........................................................................................................................................................................

....................................................................................................................................................................... [5]

b) (i) Why is sodium hydroxide unsuitable as a primary standard?

...........................................................................................................................................................................

....................................................................................................................................................................... [2]

(ii) In order to standardise a solution of sodium hydroxide, a chemist first prepared a solution of ethanedioicacid-2-water, H

2C

2O

4 .2H

2O, by dissolving 14.6 g of crystals in water and making the solution up to 250

cm3 in a graduated flask. He then pipetted 25.0 cm3 of this solution into a conical flask, addedphenolphthalein solution as indicator, and titrated it against the sodium hydroxide solution: 24.1 cm3 ofthe latter were required. Calculate, first, the molar concentration of the ethanedioic acid solution, andthen that of the sodium hydroxide.

...........................................................................................................................................................................

...........................................................................................................................................................................

...........................................................................................................................................................................

...........................................................................................................................................................................

....................................................................................................................................................................... [5]

(iii) The chemist then used the standardised sodium hydroxide to estimate the concentration of sulfuric acidfrom a car battery. He found that 2.00 cm3 of battery acid were neutralised by 20.7 cm3 of the sodiumhydroxide solution. Calculate the concentration of sulfuric acid in the battery acid in grams per dm3.

...........................................................................................................................................................................

...........................................................................................................................................................................

...........................................................................................................................................................................

....................................................................................................................................................................... [4]


QUESTIONSHEET 4

AS Level

CALCULATIONS REQUIRING MOLAR CONCENTRATION OF PRIMARY STANDARD

a) n (Na2CO

3) = 1.547/106 = 0.0146 mol (1)

c (Na2CO

3) = 0.0146(103)/250 = 0.0584 mol dm-3 (1)

Na2CO

3(aq) + H

2SO

4(aq) → Na

2SO

4(aq) + CO

2(g) + H

2O(l) (1)

n (H2SO

4) = n (Na

2CO

3) = 25.0(10-3) 0.0584 = 1.46 × 10-3 mol (1)

c (H2SO

4) = 1.46(10-3)/24.6(10-3) = 0.0593 mol dm-3 (1)

b) (i) Solid sodium hydroxide is deliquescent / absorbs water vapour from the air (1)Hence solutions will be more dilute than calculated (1)

(ii) n (H2C

2O

4 .2H

2O) = 14.6/126 = 0.116 mol (1)

c (H2C

2O

4 .2H

2O) = 0.116(103)/250 = 0.464 mol dm-3 (1)

H2C

2O

4(aq) + 2NaOH(aq) → Na

2C

2O

4(aq) + 2H

2O(l) (1)

n (NaOH) = n (H2C

2O

4 .2H

2O) ×2 = 25.0(10-3) (0.464) 2 = 0.0232 mol (1)

c (NaOH) = 0.0232/24.1(10-3) = 0.963 mol dm-3 (1)

(iii) H2SO


2SO

4(aq) + 2H

2O(l) (1)

n (H2SO

4) = n (NaOH)/2 = 20.7(10-3) 0.963/2 = 9.97 × 10-3 mol (1)

c (H2SO

4) = 9.97(10-3)/2.00(10-3) = 4.98 mol dm-3 (1)

⇒ 4.98 × 98 = 488 g dm-3 (1)


TOTAL /


DILUTION QUESTIONS

15

NOTE Throughout this Questionsheet, marks are awarded for correctly balanced chemical equations.

a) Why must concentrated solutions be diluted before being titrated by normal laboratory standard solutions?

....................................................................................................................................................................... [1]

b) 10.0 cm3 of concentrated hydrochloric acid were pipetted into a 1dm3 graduated flask and diluted to thegraduation mark with deionised water. 25.0 cm3 of this diluted solution were titrated by 23.6 cm3 of aqueoussodium hydroxide containing 0.099 mol dm-3. Calculate, first, the molar concentration of the dilutedhydrochloric acid, and then that of the concentrated hydrochloric acid.

...........................................................................................................................................................................

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....................................................................................................................................................................... [4]

c) In an experiment to determine the ethanoic acid content of vinegar, 25.0 cm3 of vinegar were diluted to 200cm3 with deionised water. 25.0 cm3 of the diluted solution were titrated by 25.5 cm3 of sodium hydroxidesolution of concentration 0.102 mol dm-3. Calculate the following.

(i) The mass of ethanoic acid in a 350 cm3 bottle of vinegar.

...........................................................................................................................................................................

...........................................................................................................................................................................

...........................................................................................................................................................................

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....................................................................................................................................................................... [6]

(ii) The mass of ethanoic acid in a 1.14 dm3 bottle of vinegar.

...........................................................................................................................................................................

....................................................................................................................................................................... [2]

(iii) The volume of 0.102 M NaOH(aq) which would have been needed to titrate 25.0 cm3 of the originalvinegar.

...........................................................................................................................................................................

....................................................................................................................................................................... [2]


QUESTIONSHEET 5

AS Level

DILUTION QUESTIONS

a) A very large volume of titrant would otherwise be required / the burette would have to be refilled during the titration (1)

b) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) (1)

n (HCl) = n (NaOH) = 23.6(10-3) 0.099 = 2.34 × 10-3 mol (1)c (diluted HCl) = 2.34(10-3)/25.0(10-3) = 0.0936 mol dm-3 (1)c (concentrated HCl) = 0.0936(103)/10.0 = 9.36 mol dm-3 (1)

c) (i) CH3COOH(aq) + NaOH(aq) → CH

3COO- Na+(aq) + H

2O (l) (1)

n (CH3COOH) = n (NaOH) = 25.5(10-3) 0.102 = 2.60 × 10-3 mol (1)

c (diluted CH3COOH) = 2.60(10-3)/25.0(10-3) = 0.104 mol dm-3 (1)

c (original CH3COOH) = 0.104(200)/25.0 = 0.832 mol dm-3 (1)

⇒ 0.832(350)/103 = 0.291 mol per 350 cm3 (1)⇒ 0.291(60) = 17.5 g per 350 cm3 (1)

(ii) From (i), c = 0.832 mol dm-3

⇒ 0.832(1.14) = 0.948 mol per 1.14 dm3 (1)⇒ 0.948(60) = 56.9 g per 1.14 dm3 (1)

(iii) n (NaOH) = n (CH3COOH) = 0.832(25.0)/103 = 0.0208 mol (1)

Volume required = 0.0208/0.102 = 0.204 dm3 = 204 cm3 (1)


TOTAL /



14


a) 1.33g of hydrated ethanedioic acid, H2C

2O

4 .nH

2O, were dissolved in deionised water and the solution made

up to 250 cm3 in a graduated flask. 25.0 cm3 of this solution were titrated by 21.1 cm3 of aqueous sodiumhydroxide of concentration 0.100 mol dm-3. Calculate the number of molecules of water of crystallisation inthe hydrated ethanedioic acid by answering the following questions.

(i) How many moles of anhydrous H2C

2O

4 are there in 1 dm3 of solution?

...........................................................................................................................................................................

...........................................................................................................................................................................

....................................................................................................................................................................... [3]

(ii) Convert this to a mass of anhydrous H2C

2O

4 in 1 dm3.

....................................................................................................................................................................... [1]

(iii) What is the mass of water of crystallisation in 1 dm3?

...........................................................................................................................................................................

....................................................................................................................................................................... [2]

(iv) Convert this to the number of moles of water of crystallisation in 1 dm3.

....................................................................................................................................................................... [1]

(v) How many moles of water are associated with 1 mole of anhydrous H2C

2O

4?

....................................................................................................................................................................... [1]

b) 2.475 g of hydrated potassium carbonate, K2CO

3.xH

2O, were dissolved in deionised water to make 1 dm3 of

solution. 25.0 cm3 of this solution required 30.0 cm3 of a solution of sulfuric acid, containing 0.0125 moldm-3, for neutralisation. Calculate the value of x.

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QUESTIONSHEET 6

AS Level


a) (i) H2C

2O


2C

2O

4(aq) + 2H

2O(l) (1)

n (H2C

2O

4) = n (NaOH)/2 = 21.1(10-3) (0.100)/2 = 1.055 × 10-3 mol in 25 cm3 (1)

n (anhydrous H2C

2O

4) in 1 dm3 = 1.055(10-3) (103)/25 = 0.0422 mol (1)

(ii) Mr (H

2C

2O

4) = 90

∴ m (anhydrous H2C

2O

4) in 1 dm3 = 0.0422(90) = 3.80 g (1)

(iii) m (hydrated H2C

2O

4) in 1 dm3 = 1.33(4) = 5.32 g (1)

∴ m (H2O) in 1 dm3 = 5.32 - 3.80 = 1.52 g (1)

(iv) n (H2O) in 1 dm3 = 1.52/18 = 0.0844 mol (1)

(v) n (H2C

2O

4) : n (H

2O) = 0.0422 : 0.0844 = 1 : 2 (1)

Or no. of moles of water of crystallisation = 2 (1)

b) H2SO

4(aq) + K

2CO

3(aq) → K

2SO

4(aq) + H

2O(1) + CO

2(g) (1)

n (K2CO

3) = n (H

2SO

4)

= 30.0(10-3) (0.0125)= 3.75 × 10-4 mol in 25 cm3 (1)

∴in l dm3 there are 3.75(10-4)(40) = 0.015 mol K2CO

3 (1)

Mr (K

2CO

3) = 138

∴ in 1 dm3 there are 0.015(138) = 2.07 g K2CO

3 (1)

Hence, mass of water of crystallisation = 2.475 - 2.07= 0.405 g (1)

⇒0.405/18 = 0.0225 mol H2O with 0.015 mol K

2CO

3

∴ x = 0.0225/0.015 = 1.5 (1)


TOTAL /


PERCENTAGE PURITY

14


a) On storage, calcium hydroxide (agricultural ‘slaked lime’) is slowly converted into calcium carbonate.

(i) What is the cause of this deterioration?

....................................................................................................................................................................... [1]

(ii) Write a chemical equation for the change.

....................................................................................................................................................................... [1]

b) In an experiment to determine the purity of a batch of slaked lime, 0.204 g was weighed out, transferred to abeaker, and sufficient water added to dissolve the calcium hydroxide. The insoluble calcium carbonate wasfiltered off and discarded. The solution was then titrated by 20.8 cm3 of hydrochloric acid of concentration0.210 mol dm-3.

(i) Why was the calcium carbonate filtered off before the solution was titrated?

...........................................................................................................................................................................

....................................................................................................................................................................... [2]

(ii) Calculate the percentage purity of the slaked lime.

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c) A bottle of sodium chloride had become contaminated by sodium carbonate. 5.07 g of the mixture weredissolved in deionised water and made up to 250 cm3. 25.0 cm3 of this soution were titrated with 18.6 cm3 ofhydrochloric acid of concentration 0.105 mol dm-3. Calculate the percentage purity of the sodium chloride.

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QUESTIONSHEET 7

AS Level

PERCENTAGE PURITY

a) (i) Absorption of carbon dioxide from the air (1)

(ii) Ca(OH)2(s) + CO

2(g) → CaCO

3(s) + H

2O(l) (1)

b) (i) CaCO3 would have reacted with HCl(aq) (1)

Hence too much HCl(aq) would have been used in the titration (1)

(ii) Ca(OH)2(aq) + 2HCl(aq) → CaCl

2(aq) + 2H

2O(l) (1)

n (Ca(OH)2) = n (HCl)/2 = 20.8(10-3) 0.210/2 = 2.18 × 10-3 mol (1)

Mr (Ca(OH)

2) = 74

m (Ca(OH)2) = 2.18(10-3) 74 = 0.162 g (1)

Purity = 0.162(100)/0.204 = 79.4% (1)

c) Na2CO

3(aq) + 2HCl(aq) → 2NaCl(aq) + CO

2(g) + H

2O(l) (1)

Note NaCl does not react with hydrochloric acid.

n (Na2CO

3) = n (HCl)/2 = 18.6(10-3)0.105/2 = 9.77 × 10-4 mol in 25 cm3 (1)

⇒ 9.77 × 10-3 mol Na2CO

3 in 250 cm3 (1)

Mr (Na

2CO

3) = 106

m (Na2CO

3) = 9.77(10-3)106 = 1.04 g (1)

m (NaCl) = 5.07 - 1.04 = 4.03 g (1)Purity = 4.03(100)/5.07 = 79.5% (1)


TOTAL /


TEST QUESTION I


a) 1.15 g of a monobasic (monoprotic) organic acid were dissolved in water to make 250 cm3 of solution. 25.0cm3 of this solution required 25.0 cm3 of sodium hydroxide solution containing 0.100 mol dm-3 for completeneutralisation. Calculate the molar mass of the organic acid.

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b) Find the volume of a solution of nitric acid, containing 0.40 mol dm-3, which will react exactly with 1.00 g ofmagnesium carbonate.

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c) A road tanker has crashed on a motorway, spilling 1000 dm3 of 10 M hydrochloric acid (i.e. of concentration10 mol dm-3).

(i) Calculate the minimum mass of calcium carbonate that the fire brigade would need to neutralise thisacid.

...........................................................................................................................................................................

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(ii) Suggest two reasons why the fire brigade neutralised the acid with calcium carbonate rather than sodiumhydroxide (‘caustic soda’).

Reason 1 ....................................................................................................................................................

............................................................................................................................................................... [2]

Reason 2 ....................................................................................................................................................

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15


QUESTIONSHEET 8

AS Level

TEST QUESTION I

a) Let the acid be represented by HA; then, since it is monobasic,n (HA) = n (NaOH) (1)

= 25.0(10-3)0.1 = 2.5 × 10-3 mol in 25.0 cm3 (1)

∴ there were 10(2.5 × 10-3 ) = 2.5 × 10-2 mol HA in 250 cm3 (1)M

r (HA) = 1.15 / (2.5 × 10-2 ) = 46.0 ⇒ 46.0 g mol-1 (1)

b) MgCO3(s) + 2HNO

3(aq) → Mg(NO

3)

2 (aq) + CO

2(g) + H

2O(l) (1)

Mr (MgCO

3) = 84

∴ n (MgCO3) = 1.00/84 = 0.0119 mol (1)

n (HNO3) = 2n (MgCO

3) = 2(0.0119) = 0.0238 mol (1)

Volume required = 0.0238/0.40 = 0.0595 dm3 = 59.5 cm3 (1)

c) (i) CaCO3(s) + 2HCl(aq) → CaCl

2(aq) + CO

2(g) + H

2O(l) (1)

n (CaCO3) = n (HCl)/2 = 10(103)/2 = 5000 mol (1)

Mr (CaCO

3) = 100

m (CaCO3) = 5000 (100) = 500,000 g = 500 kg (1)

(ii) Reason 1 CaCO3 is naturally occurring (1)

and hence cheap (1)

Reason 2 Any excess NaOH would pollute the environment (1)but CaCO

3 being insoluble would not do so (1)


TOTAL /


TEST QUESTION II

16

Two brands of washing soda, “Deluxwash” and “Econosoda”, are available in a supermarket. Both brands containsodium carbonate but in different percentages. The percentage of sodium carbonate in each brand was determinedby titration against hydrochloric acid.

a) The concentration of the hydrochloric acid to be used was first determined by titration against a sodiumhydroxide solution of known concentration.

(i) Write a balanced equation for the reaction between hydrochloric acid and sodium hydroxide.

....................................................................................................................................................................... [1]

(ii) It required 19.2 cm3 of 0.5 M sodium hydroxide to neutralise 25.0 cm3 of hydrochloric acid. Calculatethe molarity of the hydrochloric acid.

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b) 20.0 g of Econosoda was fully dissolved in 1 dm3 of water. 25.0 cm3 of this solution was pipetted into aconical flask. It required 19.1 cm3 of the hydrochloric acid to neutralise the solution of Econosoda in theconical flask.

(i) Write a balanced equation for the reaction between sodium carbonate and hydrochloric acid.

....................................................................................................................................................................... [2]

(ii) Calculate the molarity of the sodium carbonate solution in the conical flask.

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....................................................................................................................................................................... [3]

(iii) Calculate the percentage by mass of sodium carbonate contained in Econosoda.

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c) In a similar experiment it was found that Deluxwash contained 98% sodium carbonate by mass. 500 g ofEconosoda cost £1.75 and 500 g of Deluxwash cost £2.45. Determine by calculation which brand of washingsoda is better value for money.

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....................................................................................................................................................................... [4]


QUESTIONSHEET 9

AS Level

TEST QUESTION II

a) (i) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) (1)

(ii) n (NaOH) = 0.5(19.2/1000) = 9.6 x 10-3 mol (1)n (HCl) = n (NaOH) = 9.6 x 10-3 mol (1)c (HCl) = 9.6 x 10-3 / (25.0/1000) = 0.384 M (1)

b) (i) Na2CO

3(aq) + 2HCl(aq) → 2NaCl(aq) + H

2O(l) + CO

2(g) (2)

Award (1) for correct reactants and products and (1) for correct balancing.

(ii) n (HCl) = 0.384(19.1/1000) = 7.334 x 10-3 mol (1)n (Na

2CO

3) = n (HCl) / 2 = 7.334 x 10-3 / 2 = 3.667 x 10-3 mol (1)

∴ c (Na2CO

3) = 3.667 x 10-3 / (25/1000) = 0.147 M (1)

(iii) Moles Na2CO

3 in 25 cm3 = 3.667 x 10-3

∴ moles Na2CO

3 in 1 dm3 = (3.667 x 10-3 / 25) x 1000 = 0.1467 (1)

Mass Na2CO

3 in 1 dm3 = 0.1467 x 106 = 15.6 g (1)

∴ percentage Na2CO

3 = (15.6/20.0) x 100 = 78% (1)

c) 500 g Econosoda contains 500(78/100) = 390 g Na2CO

3 (1)

500 g Deluxwash contains 500(98/100) = 490 g Na2CO

3 (1)

Econosoda costs 175/390 = 0.45p per gram Na2CO

3 (½)

Deluxwash costs 245/490 = 0.50p per gram Na2CO

3 (½)

∴ Econosoda is the better value for money (1)


TOTAL /


TEST QUESTION III

16

The human stomach contains a weak solution of hydrochloric acid. If excess hydrochloric acid is produced, itmay lead to painful indigestion. This can be treated by solid indigestion tablets which contain sodium hydrogen-carbonate to neutralise the excess hydrochloric acid.

a) Write a balanced equation for the reaction between hydrochloric acid and an aqueous solution of sodiumhydrogencarbonate.

....................................................................................................................................................................... [1]

b) A 5.00 g indigestion tablet, containing 85% sodium hydrogencarbonate, was crushed and dissolved in 250cm3 of water.

(i) Calculate the concentration of the sodium hydrogencarbonate solution formed.

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(ii) Calculate the volume of 0.15 M hydrochloric acid that 25 cm3 of the sodium hydrogencarbonate solutionwould neutralise.

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c) Whilst doing some car repairs, a mechanic spilt 10 cm3 of car battery acid on his hand. The battery acid was2 mol dm-3 sulfuric acid. He decided to neutralise the acid using the indigestion tablets referred to in b).Calculate the number of indigestion tablets required to neutralise the acid fully.

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d) Another popular indigestion cure is Milk of Magnesia, which contains a suspension of magnesium hydroxide.Give two reasons why Milk of Magnesia is better as an indigestion cure than solid sodium hydrogencarbonatetablets.

Advantage 1 ......................................................................................................................................................

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Advantage 2 ......................................................................................................................................................

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QUESTIONSHEET 10

AS Level

TEST QUESTION III

a) HCl(aq) + NaHCO3(aq) → NaCl(aq) + H

2O(l) + CO

2(g) (1)

b) (i) m (NaHCO3) = 5.00(85/100) = 4.25 g (1)

n (NaHCO3) = 4.25/84 = 0.0506 mol (1)

∴ c = 0.0506 / (250/1000) = 0.2024 mol dm-3 (1)

(ii) n (NaHCO3) = 0.2024(25/1000) = 5.06 x 10-3 mol (1)

n (HCl) = n (NaHCO3) = 5.06 x 10-3 mol (1)

∴ V (HCl) = 5.06 x 10-3 / 0.15 = 0.0337 dm3 = 33.7 cm3 (1)

c) n (H2SO

4) = 2(10/1000) = 0.02 mol (1)

2 mol NaHCO3 ≡ 1 mol H

2SO

4 (1)

∴ n (NaHCO3) = 0.02 x 2 = 0.04 mol (1)

m (NaHCO3) = 0.04 x 84 = 3.36 g (1)

∴ one 5.0 g tablet is sufficient (1)

d) 1 In a suspension particles have a larger surface area than in a tablet (1)∴reaction with acid occurs more rapidly (1)

2 1 mol of Mg(OH)2 produces 2 mol of OH-(aq) (1)

∴ Mg(OH)2 is twice as effective as NaHCO

3 (1)

3 Mg(OH)2, unlike NaHCO

3, does not produce CO

2 (1)

which can cause flatulence (1)Maximum 4 marks


TOTAL /



Weigh a clean dry test tube, add about 0.5 g of iron filings, and reweigh. Weigh another clean dry test tube, addabout 2.5 g of copper(II) sulfate, and reweigh. Add distilled water to the tube containing copper(II) sulfate untilit is about a third full, then heat until almost boiling to dissolve the crystals. Add this solution of copper(II) sulfateto the iron filings in two stages, mixing together thoroughly. Leave for one minute for the reaction to complete.Allow the solid to settle and remove the solution with a teat pipette. Wash the copper with distilled water and thepropanone. Dry by putting in an oven for a few minutes. Finally reweigh the tube.

Results

Mass of tube and iron 21.301 gMass of tube empty 20.789 gMass of iron 0.512 gMass of tube and copper 21.403 gMass of tube empty 20.789 gMass of copper 0.614 g

a) Calculate the:(i) moles of iron used; ................................................................................................................................ [1]

(ii) moles of copper formed; ....................................................................................................................... [1]

(iii) ratio of moles of copper formed: moles of iron reacted

....................................................................................................................................................................... [1]

(iv) Hence write the equation for the reaction. ............................................................................................ [1]

b) (i) What are the most significant errors in the experiment?

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....................................................................................................................................................................... [3](ii) How reliable is the result?

....................................................................................................................................................................... [1](iii) How could the experiment be improved?

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....................................................................................................................................................................... [3]Quality of language [1]

12


QUESTIONSHEET 11

AS Level


a) (i) moles of iron used = 0.512/55.8 = 0.00918 (1)

(ii) moles of copper formed = 0.614/63.5 = 0.00967 (1)

(iii) ratio of moles of copper formed: moles of iron reacted = 0.00967/0.00918 = 1.05 (1)

(iv) Hence write the equation for the reactionFe + CuSO

4 → FeSO

4 + Cu

Or Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)

b) (i) Any three of:only carried out oncenot all iron reactedsolid not drysolid lost when liquid removed

(ii) Any one of:value is close to 1:1only one value

(iii) Any three of:repeat and averageuse finer iron filingsfilter off soliddry to constant mass

Quality of language: at least two sentences in which the meaning is clear


TOTAL /



A litre of sea water was given to a student, who carried out the following experiment to determine theconcentration of salts in it:

1. A large evaporating basin was weighed.2. A 25 cm3 sample of sea water was delivered into the evaporating basin from a measuring cylinder.3. The evaporating basin was placed on a tripod and gauze. The solution was evaporated to dryness by heating

using a Bunsen burner.4. The evaporating basin was reweighed when it had cooled.5. The procedure was repeated with another 25 cm3 sample.

Results Sample 1 Sample 2

Mass of evaporating basin + residue = 142.084 g 149.416 gMass of evaporating basin empty = 141.381 g 148.766 gMass of salt residue = 0.703 g 0.650 g

a) Calculate the concentration of salt in g per dm3 in each of the samples to three significant figures.

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....................................................................................................................................................................... [3]

b) Comment on the values obtained.

....................................................................................................................................................................... [1]

c) What do you think are the significant errors in the experiment?

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d) How could the experiment be modified to produce more accurate results?

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Quality of language [1]

9


QUESTIONSHEET 12

AS Level


a) Sample 1 0.703 x 1000/25 = 28.1 g (1)Sample 2 0.650 x 1000/25 = 26.0 g (1)Three significant figures (1)

b) Wide variation, result unreliable (1)

c) Incomplete evaporation of water (1)Loss of residue due to spitting (1)

d) Heat to constant mass (1)Heat in a flask (1)

Quality of language: at least two sentences in which the meaning is clear (1)


TOTAL /



A crucible containing a sample of potassium hydrogen carbonate, that had been kept dry in a desiccator, washeated gently at first, then more strongly, for 4 minutes. The crucible was allowed to cool in a desiccator and thenreweighed. The mixture was heated to constant mass.

ResultsMass of tube + potassium hydrogen carbonate = 16.246 gMass of tube empty = 16.216 gMass of potassium hydrogen carbonate = 0.030 gMass of tube + solid after first heating = 16.242 gMass of tube + solid after second heating = 16.239 gMass of tube + solid after third heating = 16.237 g

(Continued...)

a) Calculate the % loss in mass which occurred when potassium hydrogen carbonate was heated.

[2]

b) Using the formula mass of the compounds in the equations below, calculate the theoretical % loss in mass forall the equations:i. A. 2KHCO

3(s) → K

2CO

3(s) + H

2O(l) + CO

2(g)

[1]

ii. B. 2KHCO3(s) → K

2C

2O

5(s) + H

2O(l)

[1]

iii. C. 2KHCO3(s) → K

2CO

4(s) + H

2O(l) + CO(g)

[1]

iv. D. 2KHCO3(s) → K

2O(s) + H

2O(l) + 2CO

2(g)

[1]

v. E. 2KHCO3(s) → K

2C

2O

3(s) + H

2O(l) + O

2(g)

[1]


TOTAL /

AS Level TOPIC 3 Questionsheet 14 TOPIC 3 Questionsheet 13 Continued

16

c) Compare the values in parts a and b to deduce the correct equation for the reaction

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....................................................................................................................................................................... [1]

d) What are the most significant errors in the experiment?

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....................................................................................................................................................................... [3]

e) How reliable is the result, give reasons for your answer?

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....................................................................................................................................................................... [1]

f) How could the experiment be improved?...........................................................................................................................................................................

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....................................................................................................................................................................... [3]

Quality of language [1]



QUESTIONSHEET 13

AS Level


a) Loss in mass = 0.030 - 0.021 = 0.009 g (1)% Loss in mass = 0.009 x100 = 30% (1)

0.030b)

i. % Loss in mass = 62 x100 = 31% (1) 200

ii. % Loss in mass = 18 x100 = 9% (1) 200

iii. % Loss in mass = 46 x100 = 23% (1) 200

iv. % Loss in mass = 106 x100 = 53% (1) 200

v. % Loss in mass = 50 x100 = 25% (1) 200

c) A. 2KHCO3(s) → K

2CO

3(s) + H

2O(l) + CO

2(g) (1)

d) Any three of:

Only carried out once.Small mass % error high.Further decomposition.Spitting.(3)

e) Reliable because of the close agreement between experimental and theoretical values, or unreliable because there isonly one value. (1)

f) Any three of:Repeat.Use larger mass.Use an oven for temperature control.Put on lid(3)

Quality of language: at least two sentences in which the meaning is clear (1)

Documents

Curriculum P - Frankly Chemistry · as topic 3 acid-base titrations molar solutions preparation of a standard solution simple volumetric calculations calculations requiring molar