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Worked solutions to student book questions Chapter 4 Analysing acids and bases

Heinemann Chemistry 2 4th edition Enhanced Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 1

Q1. Antacid tablets should normally be chewed before they are swallowed. Why?

A1. Antacid tablets are normally chewed to provide a larger surface area for faster reaction with stomach acids.

Q2. A laboratory test to determine how much hydrochloric acid is neutralised by a brand of antacid does not give a complete picture of its effectiveness in the stomach. What other factors might be important?

A2. Other factors to consider when deciding an antacid’s effectiveness include the neutralising action of the antacid over a prolonged period (30 minutes, for example), and whether or not the antacid upsets the acid balance in the stomach. The presence and nature of food in the stomach may also affect the neutralisation reaction. Furthermore, some brands claim to have a coating action on the stomach wall which might be unrelated to the neutralising action.

Q3. Examine the range of antacids shown in Figure 4.1 on page 37 of the student book. If you were to choose one of these from all the others, what features, apart from its ability to neutralise acid, might influence your choice?

A3. The consumer may have a preference for tablets, gels or solutions. He or she could also be influenced by price, attractiveness of packaging and whether or not the medication can be conveniently carried.

Q4.

Many antacids fizz when dissolved in a glass of water. One such brand lists among its ingredients, sodium hydrogen carbonate and citric acid. Write an ionic equation for the reaction between HCO3

–(aq) and H3O+(aq) responsible for the ‘fizz’.

A4.

HCO3–(aq) + H3O+(aq) → CO2(g) + 2H2O(l)

Q5.

In each of the following equations: i identify the acids and bases ii name the conjugate acid–base pairs a NH3(aq) + H2O(l) → NH4

+(aq) + OH–(aq) b HSO4

–(aq) + H2O(l) → H3O+(aq) + SO42–(aq)

c NH4+(aq) + S2–(aq) → NH3(aq) + HS–(aq)

d CH3COO–(aq) + H3O+(aq) → H2O(l) + CH3COOH(aq)



A5.

(Acid is listed first.) a i H2O; NH3

ii NH4+, NH3 and H2O, OH-

b i HSO4–; H2O

ii HSO4–, SO4

2– and H3O+, H2O c i NH4

+; S2–

ii NH4+, NH3 and HS–, S2–

d i H3O+; CH3COO–

ii CH3COOH, CH3COO– and H3O+, H2O

Q6. The graphs in Figure 4.7 show the pH curves for titrations involving combinations of acids and bases of various strengths. You have a choice of phenolphthalein and methyl orange indicator. Phenolphthalein changes colour over a pH range 8.2 to 10.0. Methyl orange changes colour between pH 3.2 and 4.4. Decide which indicator(s) would be suitable to identify the equivalence point for each reaction. Provide reasons for your selections. a b

c d

Figure 4.7 Change in pH during a titrations of: a a strong acid with a strong base; b a strong acid with a weak base; c weak acid with a strong base; d weak acid with a weak base.



A6. a The equivalence point occurs in the range pH 3 to pH 11. Both indicators will

change colour over this pH range. Both indicators will provide a sharp end point, i.e. they will change colour at the equivalence point with the addition a small volume, 1 drop, of acid.

b The equivalence point occurs in the pH range 3 to 7. Methyl orange provides a sharper end point over this pH range.

c The equivalence point occurs in the pH range 7 to 11. Phenolphthalein provides the sharper end point.

d Both indicators will provide a broad end point and neither would be suitable.

Q7. The ethanoic acid content of white vinegar was determined by titrating a 20.00 mL aliquot of the vinegar with 0.9952 M sodium hydroxide solution. The phenolphthalein indicator changed permanently from colourless to pink when 20.34 mL of sodium hydroxide solution was added from the burette. a Write an equation for the reaction. b Calculate the amount of sodium hydroxide, in mol, used in the titration. c Calculate the amount of ethanoic acid, in mol, used in the titration. d Calculate the concentration of ethanoic acid in the vinegar.

A7. a Step 1 Write a balanced equation. CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) b Step 2 Calculate the amount of NaOH used in the titration, using n = c × V. n(NaOH) = 0.9952 M × 0.02034 L = 0.02024 mol c Step 3 Use the ratio of amounts of substances to calculate the amount of

CH3COOH that was present in the 20.00 mL aliquot. From the equation in part a, 1 mol CH3COOH reacts with 1 mol NaOH.

(NaOH)

COOH)(CH3

nn

= 11

n(CH3COOH) in 20.00 mL aliquot = 0.02024 mol d Step 4 Calculate the concentration of the CH3COOH in the 20.00 mL aliquot

which is the same as the concentration in the vinegar, to the correct number of significant figures.

c(CH3COOH) = L0.02000

mol 0.02024

= 1.012 M



Q8. The ammonia content of cloudy ammonia was determined first diluting a 25.0 mL sample to 250 mL in a volumetric flask. A 20.0mL aliquot of this solution was titrated with 0.0987 M hydrochloric acid. The volume of the acid used was 22.18 mL. The equation for the reaction is:

NH3(aq) + HCl(aq) → NH4Cl(aq) a Calculate the amount of hydrochloric acid, in mol, used in the titration. b Calculate the amount of ammonia, in mol, used in the titration. c Calculate the concentration of ammonia in the diluted solution used in the

titration. d Calculate the concentration of ammonia in the original sample.

A8. Remember to give the answer for each part with the correct number of significant figures, but keep all digits in your calculator for further calculations. a Step 1 Calculate the amount of HCl used in the titration. n(HCl) = 0.0987 M × 0.02218 L = 0.002189166 mol = 2.19 × 10–3 mol b Step 2 Write a balanced equation. NH3(aq) + HCl(aq) → NH4Cl(aq) Step 3 Use the ratio of amounts of substances to calculate the amount of NH3

that was present in the 20.00 mL aliquot. From the equation, 1 mol NH3 reacts with 1 mol HCl.

(HCl)

)(NH3

nn

= 11

n(NH3) in 20.00 mL aliquot = 0.002189166 mol = 2.19 × 10–3 mol c Step 4 Calculate the amount of NH3 in the diluted solution.

c(NH3) in the diluted solution = L0.0200

mol 60.00218916

= 0.1094583 M = 0.109 M d Step 5 Calculate the amount of NH3 in 250 mL of the dilute solution. n = c × V n(NH3) in the diluted solution = 0.1094583 M × 0.250 L = 0.0273645 mol As this equals the amount of NH3 in 25.0 mL of the original concentrated

NH3 solution, the concentration can be calculated.

c(NH3) in original solution = L0.025

mol 0.0273645

= 1.094583 M = 1.09 M



Q9. A 0.4376 g aspirin tablet was heated gently with 50.00 mL of 0.196 M sodium hydroxide solution. The aspirin reacts according to the equation:

C6H4(OCOCH3)COOH(aq) + 2NaOH(aq) → C6H4(OH)COONa(aq) + CH3COONa(aq) + H2O(l)

After cooling, the resulting solution was titrated against 0.298 M hydrochloric acid to determine the amount of unreacted sodium hydroxide. A titre of 18.64 mL was obtained. Calculate: a the amount, in mol, of sodium hydroxide initially added to the aspirin b the amount of hydrochloric acid used c the amount of sodium hydroxide in excess after reaction with the aspirin d the amount of sodium hydroxide that reacted with the aspirin e the amount of aspirin in the tablet f the percentage, by mass, of aspirin in the tablet. A9. a n(NaOH)initial = 0.196 M × 0.05000 L = 0.00980 mol (three significant figures) b Step 1 Write the balanced equation for the titration of HCl against the excess

NaOH. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Step 2 Calculate the amount of HCl needed to neutralise the excess NaOH. n(HCl) = 0.298 M × 0.01864 L = 0.00555 mol (three significant figures) c From the equation, 1 mol of NaOH reacted with 1 mol of HCl. Calculate the

amount of NaOH that reacted with the HCl, which was the amount of NaOH in excess after the aspirin reaction.

)HCl(

(NaOH)excess

nn

= 11

n(NaOH)excess = 0.00555 mol d n(NaOH)reacted = n(NaOH)initial – n(NaOH)excess = (0.00980 – 0.00555) mol = 0.00425 mol e From the equation for the reaction between aspirin and NaOH, 1 mol of aspirin

reacts with 2 mol of NaOH.

)NaOH(

(aspirin)nn =

21

n(aspirin) = 2

0.00425 mol

= 0.00213 mol f Step 1 Calculate the mass of aspirin. m(aspirin) = 0.00213 mol × 180 g mol–1 = 0.3834 g Step 2 Find the percentage of aspirin in the tablet.

% aspirin = 4376.0

0.3834 × 100

= 87.6% (three significant figures)



Chapter review

Q10. Write full equations for these acid–base reactions: a nitric acid is added to sodium hydroxide solution b sulfuric acid is added to potassium hydroxide solution c hydrochloric acid is added to ammonia solution d ethanoic acid solution is added to potassium hydroxide solution.

A10.

a HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l) b H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l) c HCl(aq) + NH3(aq) → NH4Cl(aq) d CH3COOH(aq) + KOH(aq) → CH3COOK(aq) + H2O(l)

Q11.

Write ionic equations for the reactions in Question 10.

A11.

a H+(aq) + OH–(aq) → H2O(l) b H+(aq) + OH–(aq) → H2O(l) c H+(aq) + NH3(aq) → NH4

+(aq) d CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)



Q12. What mass of sodium sulfate is produced when 25.0 mL of 0.100 M sulfuric acid is added to 20.0 mL of 0.15 M sodium hydroxide solution?

A12. Step 1 Write a balanced equation. H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) Step 2 To determine which reactant is in excess, calculate amount of each reactant

divided by their respective coefficient. The smallest amount is the limiting reactant and the one from which to calculate amount of product formed. The other is the excess reactant.

Note: These calculations can only be used to determine the excess reactant. Continue the calculations, using original data.

)SOt(Hcoefficien

)SO(H

42

42n = 1

M 0.100L 0.025 ×

= 0.0025 mol

t(NaOH)coefficien

)(NaOHn = 2

M 0.15L 0.0200 ×

= 0.0015 mol Hence NaOH is the limiting reactant. Step 3 From the equation, 1 mol of Na2SO4 is produced by 2 mol of NaOH.

(NaOH)

)SO(Na 42

nn

= 21

n(Na2SO4) = 2

)(NaOH1 n×

= 2

M 0.15L 0.0200 ×

= 0.0015 mol Step 4 Calculate the mass of Na2SO4 to the correct number of significant figures. m(Na2SO4) = 0.0015 mol × 142.04 g mol–1 = 0.21306 g = 0.21 g



Q13. What volume of 0.100 M sulfuric acid would be required to neutralise a solution containing 0.500 g of sodium hydroxide and 0.800 g of potassium hydroxide?

A13. The NaOH and KOH react independently with the H2SO4. Treat them as separate reactions and write two balanced equations. Add the volumes of H2SO4 from each reaction to find the total volume needed to neutralise the solution. Step 1 For the NaOH solution, write a balanced equation. 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l) Step 2 Calculate amount of NaOH.

n(NaOH) = 1mol g 39.98g0.500

−

= 0.012506 mol Step 3 From the equation, 1 mol of H2SO4 is produced from 2 mol of NaOH.

(NaOH))SO(H 42

nn

= 21

n(H2SO4) = mol 201250601 .×

= 0.006253 mol Step 4 Calculate the volume of H2SO4 needed to neutralise the NaOH.

V(H2SO4) = M0.100

mol00625313.0

= 0.06253 L Step 5 For the KOH solution, write a balanced equation. 2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l) Step 6 Calculate amount of KOH.

n(KOH) = 1mol g 56.108g0.800

−

= 0.01426 mol Step 7 From the equation, 1 mol of H2SO4 is produced from 2 mol of KOH.

(KOH)

)SO(H 42

nn

= 21

n(H2SO4) = 2

0.014261× mol

= 0.007129 mol Step 8 Calculate the volume of H2SO4 needed to neutralise the KOH.

V(H2SO4) = M0.100mol007129.0

= 0.07129 L Step 9 Calculate the total volume of H2SO4 needed to neutralise the KOH and the

NaOH. V(H2SO4) = (0.07129 + 0.06253) L = 0.13382 L = 0.134 L



Q14. The cleaning agent in a household window cleaner is an aqueous solution of ammonia. Draw a flow chart to show how you would perform a titration to find the concentration of ammonia in the window cleaner.

A14. 1 Place a standard solution of hydrochloric acid in a burette. Record the initial

volume. ↓

2 Place an aliquot of the window cleaner in a conical flask. ↓

3 Add two or three drops of indicator to the cleaner. (Methyl orange is a suitable indicator.)

↓

4 Titrate the window cleaner with the hydrochloric acid. Record the volume of the acid used to reach the end point.

↓

5 Repeat steps 1–4 to obtain three concordant titres (titres within 0.1 mL). In practice, it may be necessary to dilute the window cleaner, using a pipette and volumetric flask, in order to obtain titres that are not excessively large.

Q15. Read the description of the analysis of concrete cleaner on page 41 again. What safety instructions would you give to a student intending to perform an analysis such as this?

A15.

Hydrochloric acid is very corrosive and its vapour irritates the skin, eyes and respiratory system. You should instruct the student to wear safety glasses and avoid skin contact when dealing with this substance. A pipette filler should always be used when dispensing aliquots of this liquid.



Q16. The concentration of sodium hydroxide in waste water from an alumina refinery was found by titrating 20.00 mL aliquots of the water against 0.150 M hydrochloric acid, using phenolphthalein as indicator. The average titre of several titrations was 11.40 mL. a Why is an indicator used? b Write an equation for the reaction that occurred. c What was the molarity of NaOH in the waste water? d What mass of NaOH would be present in 100 L of the waste water?

A16. a An indicator is necessary to detect the equivalence point as both the reactants and

products in the reaction are colourless. b NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) c Step 1 n(HCl) = 0.150 M × 0.0114 L = 0.00171 mol Step 2 From the equation, 1 mol NaOH reacts with 1 mol HCl.

(HCl)

)(NaOHn

n = 11

n(NaOH) = 0.00171 mol Step 3 Calculate the concentration of NaOH. c(NaOH) = 0.00171 mol/0.0200 L = 0.0855 M d Step 1 Calculate the mass of NaOH present in 1 L to the correct number of

significant figures. c(NaOH) = 0.0855 mol L–1 m(NaOH) = (0.0855 mol × 39.98 g mol–1) g L–1 = 3.4183 g L–1 Step 2 Convert this to the mass of NaOH present in 100 L to the correct number

of significant figures. m(NaOH) = 3.4183 g L–1 = (3.4183 × 100) g in 100 L of water = 341.83 g = 342 g



Q17. A 42.7 mL volume of a hydrochloric acid solution is required to react completely with 20.0 mL of 0.612 M sodium carbonate solution. a Write an equation for the reaction. b Calculate the concentration of the HCl, in mol L–1.

A17.

a 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g) b Step 1 Calculate the amount of Na2CO3. n(Na2CO3) = 0.612 M × 0.0200 L = 0.01224 mol Step 2 From the equation, 2 mol HCl reacts with 1 mol of Na2CO3.

)CO(Na

)(HCl

32nn =

12

n(HCl) = 2 × 0.01224 mol = 0.02448 mol Step 3 Calculate the concentration of HCl to the correct number of significant

figures. c(HCl) = 0.02448 mol/0.0427 L = 0.5733 M = 0.573 M



Q18. A 1.20 g antacid tablet contains 80.0% by mass of Mg(OH)2 as the active ingredient. What volume of 0.1500 M HCl could the antacid tablet neutralise?

A18. Step 1 Write a balanced equation. Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + 2H2O(l) Step 2 Calculate the mass of Mg(OH)2 in tablet.

m(Mg(OH)2) = 1.20 g × 10080

= 0.960 g Step 3 Calculate the amount of Mg(OH)2.

n(Mg(OH)2) = 1mol g 58.326g0.960

−

= 0.01646 mol Step 4 From the equation, 2 mol HCl reacts with 1 mol Mg(OH)2.

)(Mg(OH)

)(HCl

2nn =

12

n(HCl) = 2 × 0.01646 mol = 0.03292 mol Step 5 Calculate the volume of HCl required for neutralisation, to the correct

number of significant figures.

V(HCl) = M0.1500mol03292.0

= 0.2196 L = 220 mL



Q19. Washing soda is added to hard water to allow soap to lather. A brand of washing soda contains partially hydrated sodium carbonate solid. A 0.300 g sample completely reacts with 20.0 mL of 0.250 M hydrochloric acid. a What mass of sodium carbonate was present? b Calculate the percentage by mass of sodium carbonate in the washing soda.

A19.

a 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g) Step 1 Calculate the amount of HCl. n(HCl) = 0.250 M × 0.0200 L = 0.00500 mol Step 2 From the equation, 1 mol Na2CO3 reacts with 2 mol HCl.

(HCl)

)CO(Na 32

nn

= 21

n(Na2CO3) = 2

mol 0.00500

= 0.00250 mol Step 3 Calculate the mass of Na2CO3, to the correct number of significant

figures. m(Na2CO3) = 0.00250 mol × 105.99 g mol–1 = 0.264975 g = 0.265 g b Express as a percentage to the correct number of significant figures.

% Na2CO3 = g 300.0g 2650.0 × 100%

= 88.33% = 88.3%



Q20. A 50.0 mL sample of vinegar was diluted to 250.0 mL in a volumetric flask. A 20.00 mL aliquot of this solution required the addition of 27.98 mL of 0.134 M sodium hydroxide solution in order to be neutralised. a Write an equation for the neutralisation reaction. b What is the molarity of ethanoic acid in the original vinegar? c Express your answer to part b in g L–1.

A20.

a CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) b Step 1 Calculate the amount of NaOH used in neutralisation. n(NaOH) = 0.134 M × 0.02798 L = 0.003749 mol Step 2 From the equation, 1 mol CH3COOH reacts with 1 mol NaOH in the

20.00 mL aliquot.

(NaOH)

COOH)(CH3

nn

= 11

n(CH3COOH) in 20.00 mL = 0.003749 mol Step 3 Calculate the amount of CH3COOH in the 250.0 mL flask, remembering

that only 20.00 mL was removed from the flask.

n(CH3COOH) in 250.0 mL = 0.003749 mol × 20.00250.0

= 0.04687 mol Step 4 The amount of CH3COOH in the original 50.0 mL of concentrated

vinegar is the same amount as in the 250.0 mL flask. Calculate the concentration of original vinegar solution.

c(CH3COOH) in 50.00 mL sample = 0.04687 mol/0.0500 L = 0.9373 M = 0.937 M c Convert mol L–1 to concentration in g L–1, using m = n × M. c(CH3COOH) = (0.9373 mol × 60.052 g mol–1) g L–1 = 56.269 g L–1 = 56.3 g L–1



Q21. A student titrated an aliquot of standard sodium carbonate solution with hydrochloric acid in a burette. State whether the concentration determined for the hydrochloric acid would be likely to be higher, lower or unchanged compared with the actual value if the student had previously washed with water, but not dried, the following apparatus: a the pipette used to deliver the aliquot of sodium carbonate solution b the flask containing the aliquot c the burette.

A21. a higher b unchanged c lower

Q22.

In order to standardise a solution of hydrochloric acid, a student titrated the solution against 20.00 mL aliquots of a standard solution of sodium carbonate. Methyl orange indicator was used to identify the end point of the reaction:

2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g) The sodium carbonate solution had been prepared by dissolving 1.236 g of anhydrous Na2CO3 in water and making the solution up to 250.0 mL in a volumetric flask. The titres recorded were 21.56 mL, 20.98 mL, 20.96 mL and 21.03 mL. a What value for the titre of hydrochloric acid solution should the student use in the

calculation of the acid concentration? Explain your answer. b What is the molarity of the hydrochloric acid solution? c Calculate the concentration of the HCl, in mol L–1.

A22. a 20.99 mL. Use the average of the last three titres. The student appears to have

passed the end point with the first titre. b Step 1 Write a balanced equation. 2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g) Step 2 Calculate the concentration of the Na2CO3 standard solution.

c(Na2CO3) = MVm

= L2500.0mol g 105.96

g1.2361 ×−

= 0.046 659 M = 0.046 66 M (four significant figures)



c Step 1 From the equation, 2 mol of HCl reacts with 1 mol of Na2CO3 in the titration.

)CO(Na

(HCl)

32nn =

12

n(HCl) = 2 × 0.04666 M × 0.02000 L = 0.00186636 mol Step 2 Calculate the concentration of HCl. c(HCl) = 0.001 6636 mol/0.020 99 L = 0.088 918 M = 0.088 92 M

Q23. A solution of a metal carbonate with the formula M2CO3 was prepared by dissolving 2.80 g of the solid in 250 mL of water. 20.00 mL aliquots of this solution were then titrated with 0.150 M H2SO4, using methyl orange indicator. The average titre was 10.8 mL. a Write an equation for the reaction between the metal carbonate and sulfuric acid. b What amount of sulfuric acid, in mol, was needed to reach the end point of the

titration? c What amount of the metal carbonate was present in each 20.00 mL aliquot? d What amount of metal carbonate was present in the original sample? e Determine the molar mass of the metal carbonate. f What is the identity of the metal M?

A23.

a H2SO4(aq) + M2CO3(aq) → M2SO4(aq) + H2O(l) + CO2(g) b n(H2SO4) = 0.150 M × 0.0108 L = 0.001 62 mol (three significant figures) c From the equation, 1 mol M2CO3 reacts with 1 mol H2SO4 in the titration.

)SO(H)CO(M

42

32

nn

= 11

n(M2CO3) in 20.00 mL = 0.001 62 mol d Calculate the amount of M2CO3 in the 250.0 mL flask, remembering that only

20.00 mL was removed from the 250.0 mL flask for the titration.

n(M2CO3) in 250.0 mL = 0.001 62 × 20.00250.0 mol

= 0.020 25 mol = 0.0203 mol (three significant figures) e The amount of M2CO3 in the 2.80 g is the same amount as is in the 250.0 mL

flask. Calculate the molar mass of M2CO3, using M = nm .

M(M2CO3) = mol0.02025g 2.80

= 138.272 g mol–1 = 138 g mol–1 (three significant figures)



f By calculating molar mass of CO32–, determine which element is M.

M(CO32–) = 60 g mol–1

M(M) = 2

)(CO)CO(M 2332

−− MM

= 2

60138 −

= 39 g mol–1 This is the molar mass of potassium.

Q24. Sketch the pH curve that would be obtained when a strong base is titrated with a weak acid.

A24.



Q25. A 50 mL volume of 0.10 M nitric acid is mixed with 60 mL of 0.10 M calcium hydroxide solution. What volume of 0.050 M sulfuric acid is required to neutralise the mixture?

A25. Step 1 Write a balanced equation for the first reaction occurring. 2HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2H2O(l) Step 2 There is no need to determine which reactant is in excess. You are told that

H2SO4 is used to neutralise the solution, which clearly indicates that the base, Ca(OH)2, is the excess reactant. From the equation, 1 mol of Ca(OH)2 reacts with 2 mol of HNO3.

)(HNO

)(Ca(OH)

3

2

nn =

21

n(Ca(OH)2)reacted = 2

)(HNO3n

= 2

M100.0L050.0 ×

= 0.00250 mol Step 3 Calculate the amount of excess Ca(OH)2. n(Ca(OH)2)excess = n(Ca(OH)2)initial – n(Ca(OH)2)reacted = 0.0060 mol – 0.00250 mol = 0.0035 mol Step 4 Write a balanced equation for the second (neutralisation) reaction. Ca(OH)2(aq) + H2SO4(aq) → CaSO4(aq) + 2H2O(l) Step 5 Knowing that 0.0035 mol of Ca(OH)2 needs to be neutralised, calculate the

amount of H2SO4 needed for the neutralisation. From the equation, 1 mol of H2SO4 reacts with 1 mol of Ca(OH)2.

)(Ca(OH)

)SO(H

2

42

nn =

11

n(H2SO4) = n(Ca(OH)2) = 0.0035 mol Step 6 Calculate the volume of H2SO4.

V(H2SO4) = M 050.0mol0035.0

= 0.070 L = 70 mL (two significant figures)



Q26. Lawn fertiliser contains ammonium ions (NH4

+). A 1.234 g sample of lawn fertiliser was dissolved in water to make a 250.0 mL solution. A 20.00 mL aliquot of this solution was added to a flask containing 20.00 mL of 0.1022 M sodium hydroxide solution. The flask was heated until the reaction:

NH4+(aq) + OH–(aq) → NH3(aq) + H2O(l)

was complete. Excess sodium hydroxide in the resulting solution was titrated with 0.1132 M hydrochloric acid, using phenolphthalein as indicator. The end point was reached when 9.97 mL had been added. Calculate: a the amount, in mol, of HCl used in the titration b the amount of NaOH in excess after reaction with the fertiliser c the amount of NaOH that reacted with the NH4

+ ions d the amount of NH4

+ ions in the 1.234 g fertiliser sample e the percentage by mass of nitrogen in the fertiliser, assuming nitrogen is only

present as ammonium ions.

A26. a Step 1 Write a balanced equation for the titration. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Step 2 Calculate the amount of HCl used in the titration. n(HCl) = 0.1132 M × 0.009 97 L = 0.001 129 mol = 0.001 13 mol (three significant figures) b From the equation, 1 mol of NaOH reacts with 1 mol of HCl. This is the amount

of NaOH in excess after reaction with the fertiliser.

)HCl(

(NaOH)excess

nn

= 11

n(NaOH)excess = n(HCl) = 0.001 13 mol c Step 1 Write the balanced equation for the reaction of 20.00 mL of fertiliser and

NaOH. NH4

+(aq) + OH–(aq) → NH3(aq) + H2O(l) Step 2 Calculate the amount of NaOH added initially. n(NaOH)initial = 0.1022 M × 0.0200 L = 0.002 044 mol Step 3 Calculate the amount of NaOH that reacted with the fertiliser in the

20.00 mL aliquot. n(NaOH)reacted = n(NaOH)initial – n(NaOH)excess = (0.002 044 – 0.001 129) mol = 0.000 915 mol



d Step 1 From the equation, 1 mol of fertiliser reacted with 1 mol of NaOH.

)OH()(NH4

−

+

nn

= 11

n(NH4+)in 20.00 mL = 0.000915 mol

Step 2 Calculate the amount of fertiliser, NH4+, in the 250.0 mL flask,

remembering that only 20.00 mL was removed from the 250.0 mL flask for the titration. This is the same amount of NH4

+ as is present in 1.234 g of fertiliser.

n(NH4+)in 250.0 mL = 0.000915 ×

00.200.250 mol

= 0.0114 mol (three significant figures) e Step 1 Calculate the mass of N in 1.234 g of fertiliser. n(N) = n(NH4

+) m(N) = 0.01144 mol × 14.01 g mol–1 = 0.1603 g Step 2 Convert to a percentage by mass of N in the fertiliser.

% N = 234.11603.0 × 100%

= 13.0% (three significant figures)

Q27.

The drain cleaner Drainol contains 13% m/v NaOH. The original solution is diluted, by taking 10 mL and making it up to 500 mL with distilled water. a What is the molarity of the original solution? b Concentrated solutions of NaOH are not used in a titration. Give two reasons

why. c The concentration of the diluted NaOH is determined by titration. Give a suitable

reactant and indicator to use in the titration with NaOH. d Standard NaOH solutions cannot be prepared directly from the solid. Explain

why.

A27. a Determine the moles of NaOH in 1 L. 13% m/v NaOH ≡ 13 g NaOH in 100 mL solution = 130 g NaOH L–1

n(NaOH) = 1mol g 0.04g 301

− = 3.25 M

b Any two of the following (reduce errors, safety, cost, solubilising silicate). - Measurement errors are higher if solutions are concentrated. - Concentrated sodium hydroxide can cause severe burns. - Concentrated solutions are expensive as more reagent is used. - Concentrated sodium hydroxide will solubilise silicate (glass). This will reduce

the accuracy of the glassware. c Any strong acid and acid–base indicator, for example hydrochloric acid and

methyl orange. If a weak acid is chosen such as acetic acid, phenolphthalein should be selected as the indicator.

d Sodium hydroxide pellets readily absorb water and carbon dioxide from the atmosphere so cannot be readily obtained as a pure solid.



Q28. An impure sample of limestone, mainly calcium carbonate, was analysed by using a back titration. Approximately 1 g of the finely powdered limestone was weighed accurately into a conical flask. An excess of HCl, exactly 50.00 mL, was added to the limestone. The mixture was stirred for 15 min with a magnetic stirrer to allow the reaction to be completed. The mixture was then titrated with a standard solution of NaOH and the following results were obtained:

- Mass of watch glass = 8.7954 g - Mass of limestone and watch glass = 9.8460 g - Concentration of standard NaOH solution = 0.0489 M - Titration value of NaOH obtained = 22.32 mL - Concentration of HCl = 0.395 M

a Write balance equations for the two reactions that occur. b Determine the moles of HCl in excess after the reaction with the limestone. c Calculate the total moles of HCl added to the limestone. d How many moles of HCl reacted with the limestone? e Calculate the number of moles of calcium carbonate in the limestone. f What is the percentage of calcium carbonate in the limestone. g In this experiment the whole sample of limestone was used in one titration. How

could the precision of the titration have been improved?

A28.

a 2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) b Calculate the number of moles of sodium hydroxide used.

n(NaOH) = 0.0489 M × L1000

32.22 = 0.001 09 mol

Use mole ratios in the balanced equation to determine number of moles of HCl(aq).

n(HCl(aq)) = n(NaOH) = 0.001 09 mol

c Total n(HCl) = 0.395 M × mL1000mL 00.50 = 0.019 75 mol = 0.0198 mol

d n(HCl) reacted with limestone = 0.019 75 – 0.001 091 = 0.018 66 mol = 0.0187 mol

e n(CaCO3) = 2

mol .018660 = 0.009 330 mol

m(CaCO3) = 0.009 330 mol × 100.1 g mol–1 = 0.9339 g f m(limestone ) = 9.8460 – 8.7954 = 1.0506 g

% calcium carbonate = 0506.19339.0 × 100% = 88.9%

g Dissolve an accurately known mass of limestone, for example 10 g, in exactly 500 mL of hydrochloric acid. Take aliquots of the solution and titrate until concordant results are obtained.

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