Curriculum P - Frankly Chemistry · as topic 3 acid-base titrations molar solutions preparation of a standard solution simple volumetric calculations calculations requiring molar
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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
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?
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)
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AS Level TOPIC 3 Questionsheet 3
12
SIMPLE VOLUMETRIC CALCULATIONS
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?
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.
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
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.
CALCULATIONS REQUIRING MOLAR CONCENTRATION OF A PRIMARY STANDARD
16
NOTE Throughout this Questionsheet marks are awarded for correctly balanced chemical equations.
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.
(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.
(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.
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.
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.
NOTE Throughout this Questionsheet, marks are awarded for correctly balanced chemical equations.
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.
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.
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?
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.
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)
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AS Level TOPIC 3 Questionsheet 8
TEST QUESTION I
NOTE Throughout this Questionsheet marks are awarded for correctly balanced chemical equations.
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.
Reason 2 Any excess NaOH would pollute the environment (1)but CaCO
3 being insoluble would not do so (1)
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AS Level TOPIC 3 Questionsheet 9
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.
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.
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.
(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)
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AS Level TOPIC 3 Questionsheet 10
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.
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.
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.
(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
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AS Level TOPIC 3 Questionsheet 11
EXPERIMENT TO FIND THE EQUATION FOR A REACTION
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
....................................................................................................................................................................... [3](ii) How reliable is the result?
....................................................................................................................................................................... [1](iii) How could the experiment be improved?
....................................................................................................................................................................... [3]Quality of language [1]
12
TOPIC 3 ANSWERS & MARK SCHEMES
QUESTIONSHEET 11
AS Level
EXPERIMENT TO FIND THE EQUATION FOR A REACTION
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
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AS Level TOPIC 3 Questionsheet 12
ESTIMATING THE CONCENTRATION OF SALTS IN SEA WATER
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.
ESTIMATING THE CONCENTRATION OF SALTS IN SEA WATER
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)
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AS Level TOPIC 3 Questionsheet 13
EXPERIMENT TO FIND THE EQUATION FOR A REACTIONDECOMPOSITION OF POTASSIUM HYDROGEN CARBONATE
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]
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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
f) How could the experiment be improved?...........................................................................................................................................................................