Chem ch3

Worked solutions to student book questions

Chapter 3 Volumetric analysis Q1. Calculate the molarity of these solutions: a 1.5 mol of HCl dissolved in 3.0 L of solution b 0.64 g of H2SO4 dissolved in 500 mL of solution c 2.1 g of NaHCO3 dissolved in 1.00 L of solution A1. a

b

c

Molarity is concentration in mol L–1. Step 1 Write the appropriate formula. n c= V Step 2 Calculate the concentration to the correct number of significant figures. 1.5 mol c(HCl) = 3.0 L = 0.50 M Step 1 Write the appropriate formula. n c= V m c= MV Step 2 Calculate the concentration, remembering that the volume must be in litres, to the correct number of significant figures. 0.64 g/98.076 g mol −1 c(H2SO4) = 0.500 L = 0.01305 M = 0.013 M Step 1 Write the appropriate formula. n c= V m = MV Step 2 Calculate the concentration to the correct number of significant figures. 2.1 g/84.008 g mol −1 c(NaHCO3) = 1.00 L = 0.02500 M = 0.025 M

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Worked solutions to student book questions

Chapter 3 Volumetric analysis Q2.

Calculate the amount, in mol, of solute present in: a 20.00 mL of 0.255 M KOH solution b 2.50 L of 1.05 M sucrose (C12H22O11) solution c 25.00 mL of 0.0200 M AgNO3 solution. A2. a

Step 1 Step 2

b

Step 1 Step 2

c

Step 1 Step 2

Write the appropriate formula. n=c×V Calculate the amount of KOH, remembering that volume must be in litres. Express the answer with the correct number of significant figures. n(KOH) = 0.255 M × 0.02000 L = 5.10 × 10–3 mol Write the appropriate formula. n=c×V Calculate the amount of sucrose. Express the answer with the correct number of significant figures. n(sucrose) = 1.05 M × 2.50 L = 2.625 mol = 2.63 mol Write the appropriate formula. n=c×V Calculate the amount of AgNO3 to the correct number of significant figures. n(AgNO3) = 0.0200 M × 0.02500 L = 5.00 × 10–4 mol

Q3.

Calculate the mass of solute present in these solutions: a 100.0 mL of 0.50 M NaOH b 20.00 mL of 1.50 M CuSO4 c 10.00 mL of 0.10 M KSCN A3. a

Step 1

Step 2

Write the appropriate formula. m n= =c×V M m=c×V×M Calculate the mass of NaOH to the correct number of significant figures. m(NaOH) = 0.50 M × 0.1000 L × 39.99 g mol–1 = 1.9995 g = 2.0 g

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Worked solutions to student book questions

Chapter 3 Volumetric analysis b

Step 1

Step 2 Step 3 c

Step 1

Step 2

Write the appropriate formula. m n= =c×V M m=c×V×M Calculate the mass of CuSO4. m(CuSO4) = 1.50 M × 0.02000 L × 159.54 g mol–1 = 4.7862 g Express the answer with the correct number of significant figures. m(CuSO4) = 4.79 g Write the appropriate formula. m n= =c×V M m=c×V×M Calculate the mass of KSCN to the correct number of significant figures. m(KSCN) = 0.10 M × 0.01000 L × 97.102 g mol–1 = 0.097102 g = 0.097 g

Q4.

Calculate the concentration, in g L–1, of the following solutions: a sports drink containing 6.0 g of sucrose in each 100 mL (i.e. 6.0% w/v sucrose) b mouthwash containing 1.50 mg of benzoic acid per mL c vinegar that is a 0.95 M solution of ethanoic acid (CH3COOH) A4. a

Step 1 Step 2

b

Step 1

Step 2

Rewrite 6.0% w/v as g L–1. 6.0% w /v means 6.0 g per 100 mL ? g per 1000 mL (1 L) Calculate the concentration of sucrose in g L–1. 1000 c(sucrose) = 6.0 × g L–1 100 = 60 g L–1 Rewrite 1.50 mg per 1 mL as g L–1. 1.50 mg per 1 mL 0.00150 g per 1 mL ? g per 1000 mL (1 L) Calculate the concentration of benzoic acid in g L–1. 1000 c(benzoic acid) = 0.00150 × g L–1 1 = 1.5 g L–1

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Worked solutions to student book questions

Chapter 3 Volumetric analysis c

Step 1

Step 2

Rewrite 0.95 M CH3COOH as g L–1, remembering that n =

m , M

so m = n × M 0.95 M CH3COOH means 0.95 mol CH3COOH per 1 L (0.95 mol × 60 g mol–1) g per 1 L Calculate the concentration of CH3COOH in g L–1 to the correct number of significant figures. c(CH3COOH) = (0.95 × 60) g L–1 = 57 g L–1

Q5.

Calculate the molarity of solutions with the following concentrations: a 5.884 g L–1 K2Cr2O7 b 11.6 g L–1 KCl c 1.50 mg of benzoic acid (C6H5COOH) per mL of solution A5. a

Step 1

Step 2

b

Step 1

Step 2

Rewrite 5.884 g L–1 K2Cr2O7 as mol L–1, using the formula for amount, m n= . M 5.884 g K2Cr2O7 per 1 L 5.884 g mol per 1 L 294.196 g mol −1 Calculate the concentration of K2Cr2O7 in mol L–1 to the correct number of significant figures. 5.884 g c(K2Cr2O7) = mol L–1 −1 294.196 g mol = 0.02000 mol L–1 Rewrite 11.6 g L–1 KCl as mol L–1, using the formula for amount, m n= . M 11.6 g KCl per 1 L 11.6 g mol per 1 L 74.602 g mol −1 Calculate the concentration of KCl in mol L–1 to the correct number of significant figures. 11.6 g c(KCl) = mol L–1 −1 74.602 g mol = 0.15549 mol L–1 = 0.155 M

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Chapter 3 Volumetric analysis c

Step 1

Step 2

Rewrite 1.50 mg mL–1 C6H5COOH as mol L–1, using the formula for m . amount, n = M 1.50 mg C6H5COOH per 1 mL 0.00150 g mol per 1 mL 122 g mol −1 ? mol per 1000 mL Calculate the concentration of C6H5COOH in mol L–1 to the correct number of significant figures. 0.00150 g c(C6H5COOH) = × 1000 mol L–1 −1 122 g mol = 0.01229508 mol L–1 = 0.0123 M

Q6.

For 250 mL of a 0.20 M solution of potassium sulfate, K2SO4, calculate the number of moles of: a potassium ions, K+ b sulfate ions, SO42– c oxygen atoms A6. a b c

n(K2SO4) = C × V = 0.20 × 0.250 = 0.050 mol n(K+) = 2 × n(K2SO4) = 2 × 0.050 = 0.10 mol n(SO42-) = n(K2SO4) = 0.050 mol n(O) = 4 × n(K2SO4) = 0.20 mol

Q7.

The seal enclosure at the Melbourne Zoo contains 455 000 L of salt water. The water in the pool is maintained at a sodium chloride concentration of 0.048 M. What total mass of sodium chloride is present in the water when the pool is completely full? A7.

Step 1

Step 2

Write the appropriate formula. m n= =c×V M m=c×V×M Calculate the mass of NaCl to the correct number of significant figures. m(NaCl) = 0.048 M × 455 000 L × 58.44 g mol–1 = 1276329.6 g = 1.3 ×106g = 1.3 × 103kg (two significant figures)

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Worked solutions to student book questions

Chapter 3 Volumetric analysis Q8.

During the preparation of the standard solution shown in Figure 3.4, why is water added to the level of the calibration mark on the flask after the solid has dissolved, rather than before? A8. When a substance is dissolved, there is often a slight change in volume due to the attractions between the solute and solvent particles. Since molar concentration is measured in mole of solute per litre of solution, it is necessary to accurately measure the volume of solution rather than the volume of water used. E1.

A student noted that her 50 mL burette was dirty and that droplets of liquid stuck to the inside surface of the tube. She decided to investigate if this would affect the burette’s accuracy. She filled the burette to the zero mark and drained a 50.00 mL titre into a flask that had previously been weighed. The flask was weighed again and the mass of water calculated. The student then calculated the volume of water by assuming that the density of water is 1.000 00 g mL–1. This procedure was repeated another four times. The masses of water for the five trials were 49.95 g, 49.93 g, 49.90 g, 49.93 g and 49.92 g. a What is the average mass of the five water samples? b Calculate the average volume of these samples. c Is the use of a dirty burette a source of random error or systematic error? d How can the source of this error be removed? The burette was cleaned and the experiment repeated. This time the following masses of water were obtained: 50.02 g, 50.00 g, 49.98 g, 49.99 g, 50.01 g. e Calculate the average volume of these samples. f Does the variation in volumes indicate a random error or systematic error? Another burette was selected. Before testing, it was cleaned thoroughly. The same procedure was used and the average titre was found to be 49.50 g rather than the expected 50.00 g. The error was thought to be due to a poor standard of calibration. g Is this a systematic error or random error? h How should the measurements taken from this burette be adjusted? AE1. a b c d e f g h

The average mass is 49.93 g. The average volume is 49.93 mL. systematic error Cleaning the burette will remove any error caused by dirty glass. 50.00 mL random error systematic error Add 0.50 mL for every 50 mL or 0.01 mL for every 1 mL delivered by the burette.

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Worked solutions to student book questions

Chapter 3 Volumetric analysis Q9.

Explain the difference between the following: a standard solution and primary standard b equivalence point and end point c burette and pipette d aliquot and titre. A9. a

b c d

A standard solution is a solution of accurately known concentration. A primary standard is a substance that is readily obtained in a pure form, has a known formula and can be stored without deteriorating or reacting with the atmosphere. It should also be cheap and have a high molar mass. The equivalence point in a titration occurs when the reactants have been mixed in the mole ratio shown by the reaction equation. The end point occurs when the indicator changes colour. A burette is a piece of equipment capable of delivering variable volumes of a liquid accurately (generally up to 50.00 mL), while pipettes usually deliver only a fixed volume of liquid (e.g. 20.00 mL). An aliquot is the volume of liquid delivered from a pipette, while a titre is delivered by a burette and is the volume needed to reach the end point of a titration.

Q10.

The following paragraph describes an acid–base titration. Some of the key words are missing. Use the list of words in the box to fill in the gaps. pipette primary standard burette

measuring cylinder indicator indicator titre

beaker base desiccator

volumetric flask standard solution aliquot

A sample of anhydrous sodium carbonate of approximately 2 g is weighed accurately. (The solid must be dry if it is to be used as a _______________.) The solid is tipped into a _____________ and shaken with about 50 mL of distilled water until the solid dissolves. More water is added to make the solution to a volume of exactly 100.0 mL. A 20.00 mL ____________ of the solution is taken by using a _____________ and placed in a conical flask. A few drops of methyl orange are added and the mixture is titrated against dilute hydrochloric acid. A10.

A sample of anhydrous sodium carbonate of approximately 2 g is weighed accurately. (The solid must be dry if it is to be used as a primary standard.) The solid is tipped into a volumetric flask and shaken with about 50 mL of distilled water until the solid dissolves. More water is added to make the solution to a volume of exactly 100.0 mL. A 20.00 mL aliquot of the solution is taken by pipette and placed in a conical flask. A few drops of methyl orange are added and the mixture is titrated against dilute hydrochloric acid.

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Chapter 3 Volumetric analysis Chapter review Q11.

Calculate the molarity of: a 250.0 mL solution containing 1.42 g of potassium chloride (KCl) b 100.0 mL solution containing 0.63 g of anhydrous sodium carbonate (Na2CO3) c 250.0 mL solution containing 0.400 g of sodium oxalate (Na2C2O4) A11. a

Step 1

Step 2

b

Step 1

Step 2

c

Step 1

Step 2

Write the appropriate formula. m n= =c×V M m c= MV Calculate the molarity (concentration in mol L–1) to the correct number of significant figures. 1.42 g c(KCl) = 74.55 g mol −1 × 0.250 L = 0.07619 mol L–1 = 0.0762 M Write the appropriate formula. m n= =c×V M m c= MV Calculate the molarity (concentration in mol L–1) to the correct number of significant figures. 0.63 g c(Na2CO3) = 105.99 g mol −1 × 0.1000 L = 0.05944 mol L–1 = 0.059 M Write the appropriate formula. m n= =c×V M m c= MV Calculate the molarity (concentration in mol L–1) to the correct number of significant figures. 0.400 g c(Na2C2O4) = 133.96 g mol −1 × 0.2500 L = 0.01194 mol L–1 = 0.0119 M

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Worked solutions to student book questions

Chapter 3 Volumetric analysis Q12.

What mass of solute is required to prepare the following standard solutions? a 250 mL of 0.500 M sodium oxalate (Na2C2O4) b 100 mL of 0.250 M potassium hydrogen phthalate (KH(C8H4O4)) A12. a

Step 1

Step 2

b

Step 1

Step 2

Write the appropriate formula. m n = =c×V M m=c×V×M Calculate the mass required to the correct number of significant figures. m(Na2C2O4) = 0.500 M × 0.250 L × 133.96 g mol–1 = 16.745 g = 16.7 g Write the appropriate formula. m n = =c×V M m =c×V×M Calculate the mass required to the correct number of significant figures. m(KH(C8H4O4)) = 0.250 M × 0.100 L × 204.1 g mol–1 = 5.10 g

Q13.

40.4 g of Fe(NO3)3.9H2O is dissolved in sufficient water to make up 1 litre of solution. Find the concentrations of the following ions in the solution: a iron(III) ions b nitrate ions A13. a

Step 1

Step 2

Step 3 Step 4

Write the appropriate formula. m n= M Calculate the amount to the correct number of significant figures. 40.4 g n(Fe(NO3)3.9H2O) = 403.847 g mol −1 = 0.10003 mol From the formula, 1 mol of Fe3+ ions are formed from 1 mol of Fe(NO3)3.9H2O. n(Fe3+) = 0.10003 mol Calculate the concentration of Fe3+ ions in 1 L of water, to the correct number of significant figures. c(Fe3+) = 0.10003 mol/1 L = 0.100 mol

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Worked solutions to student book questions

Chapter 3 Volumetric analysis b

Step 1

Step 2

Step 3

Step 4

Write the appropriate formula. m n= M Calculate the amount. 40.4 g n(Fe(NO3)3.9H2O) = 403.847 g mol −1 = 0.10003 mol From the formula, 3 mol of NO3– ions are formed from 1 mol of Fe(NO3)3.9H2O. n(NO3–) = 3 × 0.10003 mol = 0.30009 mol Calculate the concentration of Fe3+ ions in 1 L of water, to the correct number of significant figures. c(NO3–) = 0.30009 mol/1 L = 0.300 M

Q14. a b c

Convert the following molar concentrations to units of g L–1. i 0.0024 M NaCl solution ii 6.3 × 10–5 M Pb(NO3)2 solution Convert the following concentrations in g L–1 to units of mol L–1. i 0.068 g L–1 dissolved O2 in tap water ii 0.32 g L–1 cadmium ions in seawater Convert the following molar concentrations to units of ppm: i 0.0036 M Ca(OH)2 solution ii 2.9 × 10–6 M Cd(NO3)2 solution

A14. a

i

Step 1

Step 2

ii

Step 1

Step 2

Rewrite 0.0024 M NaCl as g L–1, remembering that n =

m , M

so m = n × M. 0.0024 M NaCl means 0.0024 mol NaCl per 1 L = (0.0024 mol × 58.44 g mol–1) g per 1 L Calculate the concentration of NaCl in g L–1 to the correct number of significant figures. c(NaCl) = (0.0024 × 58.44) g L–1 = 0.140256 g L–1 = 0.14 g L–1 m Rewrite 6.3 × 10–5 M Pb(NO3)2 as g L–1, remembering that n = , M so m = n × M. 6.3 × 10–5 M Pb(NO3)2 means 6.3 × 10–5 mol Pb(NO3)2 per 1 L = (6.3 × 10–5 mol × 331.22 g mol–1) g per 1 L Calculate the concentration of Pb(NO3)2 in g L–1 to the correct number of significant figures.

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Worked solutions to student book questions

Chapter 3 Volumetric analysis

b

i

Step 1

Step 2

ii

Step 1

Step 2

c

i

Step 1

Step 2 Step 3

Step 4

c(Pb(NO3)2) = (6.3 × 10–5 × 331.22) g L–1 = 0.02086686 g L–1 = 0.021 g L–1 Rewrite 0.068 g L–1 O2 as mol L–1, using the formula for amount, m n= . M 0.068 g O2 per 1 L 0.068 g mol per 1 L = 32 g mol −1 Calculate the concentration of O2 in mol L–1 to the correct number of significant figures. 0.068 g c(O2) = mol L–1 −1 32 g mol = 0.002125 mol L–1 = 0.0021 M Rewrite 0.32 g L–1 Cd2+ ions as mol L–1, using the formula for m . amount, n = M 0.32 g Cd2+ ions per 1 L 0.32 g mol per 1 L = 112.40 g mol −1 Calculate the concentration of Cd2+ ions in mol L–1 to the correct number of significant figures. 0.32 g c(Cd2+ ions) = mol L–1 112.40 g mol −1 = 0.00285 mol L–1 = 0.0029 M Rewrite 0.0036 M Ca(OH)2 as g L–1, using the formula for mass, m = n × M. 0.0036 M means 0.0036 mol per 1 L = (0.0036 mol × 74.08 g mol–1) g per 1 L Calculate the concentration of Ca(OH)2 in g L–1. c(Ca(OH)2) = (0.0036 × 74.08) g L–1 = 0.266688 g L–1 Rewrite 0.266688 g L–1 Ca(OH)2 in ppm, which means grams per million (106) grams. Assume that 1 g = 1 mL. 0.266688 g per L = 0.266688 g per 1000 mL = 0.266688 g per 103 g = ? per 106 g Calculate the concentration in ppm to the correct number of significant figures. 10 6 c(Ca(OH)2) = 0.266688 × 3 10 = 266.688 ppm = 270 ppm

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Chapter 3 Volumetric analysis ii

Step 1

Step 2

Step 3

Step 4

Rewrite 2.9 × 10–6 M Cd(NO3)2 as g L–1, using the formula for mass, m = n × M. 2.9 × 10–6 M means 2.9 × 10–6 mol per 1 L = (2.9 × 10–6 mol × 236.4 g mol–1) g per 1 L Calculate the concentration of Cd(NO3)2 in g L–1. c(Cd(NO3)2) = (2.9 × 10–6 × 236.4) g L–1 = 0.68556 × 10–3 g L–1 Rewrite 0.67396 × 10–3 g L–1 Cd(NO3)2 in ppm, which means grams per million (106) grams. Assume that 1 g = 1 mL. 0.68556 × 10–3 g per L = 0.68556 × 10–3 g per 1000 mL = 0.68556 × 10–3 g per 103 g = ? per 106 g Calculate the concentration in ppm to the correct number of significant figures. 10 6 –3 c(Cd(NO3)2) = 0.68556 × 10 × 3 10 = 0.68556 ppm = 0.69 ppm

Q15.

A carton of orange juice lists among its contents: Total carbohydrates: 8.5 g/100 mL Vitamin C: 40 mg/100 mL a Express the concentration of each solute in g L–1. b Why is it also possible to express the concentration of vitamin C as a molarity, but not possible to do so for the carbohydrates? A15. a

b

Rewrite 8.5 g/100 mL as g L–1. 8.5 g per 100 mL = ? g per 1000 mL (1 L) Step 2 Calculate the concentration of carbohydrates g L–1. 1000 g L–1 c(carbohydrates)= 8.5 × 100 = 85 g L–1 Step 3 Rewrite 40 mg/100 mL as g L–1. 40 mg per 100 mL = 0.040 g per 100 mL = ? g per 1000 mL (1 L) Step 4 Calculate the concentration of vitamin C in g L–1. 1000 g L–1 c(vitamin C) = 0.040 × 100 = 0.40 g L–1 Vitamin C is a single chemical compound, whereas the term ‘carbohydrates’ refers to a group of compounds including glucose (C6H12O6), sucrose (C12H22O11) and others. Thus the amount, in mol, of vitamin C present in 40 mg can be calculated but this is not possible for the 8.5 g mixture of carbohydrates.

Step 1

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Chapter 3 Volumetric analysis Q16.

Each 200 mL of an electrolyte solution designed for treating dehydration contains 0.47 g of sodium chloride (NaCl), 0.30 g of potassium chloride (KCl) and 3.56 g of glucose (C6H12O6). a Calculate the concentration of each compound, in g L–1. b Calculate the molarity of each compound. c Calculate the concentration of potassium ions in the solution, in mol L–1. d Calculate the concentration of chloride ions in the solution, in mol L–1. A16. a

Step 1 Step 2

Step 3 Step 4

Step 5 Step 6

Rewrite 0.47 g/200 mL NaCl as g L–1. 0.47 g per 200 mL = ? g per 1000 mL (1 L) Calculate the concentration of NaCl g L–1. 1000 g L–1 c(NaCl) = 0.47 × 200 = 2.35 g L–1 = 2.4 g L–1 Rewrite 0.30 g/200 mL of KCl as g L–1. 0.30 g per 200 mL = ? g per 1000 mL (1 L) Calculate the concentration of KCl g L–1. 1000 g L–1 c(KCl) = 0.30 × 200 = 1.5 g L–1 Rewrite 3.56 g/200 mL of C6H12O6 as g L–1. 3.56 g per 200 mL = ? g per 1000 mL (1 L) Calculate the concentration of C6H12O6 as g L–1. 1000 g L–1 c(C6H12O6) = 3.56 × 200 = 17.8 g L–1

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Chapter 3 Volumetric analysis b

c d

Step 1

Calculate the concentration of NaCl in M, remembering to use n =

m . M

c(NaCl) = 2.4 g L–1 2.4 g = mol L–1 58.44 g mol −1 = 0.041068 mol L–1 = 0.041 M Step 2 Calculate the concentration of KCl in M. c(KCl) = 1.5 g L–1 1.5 g = mol L–1 −1 74.602 g mol = 0.0201067 mol L–1 = 0.020 M Step 3 Calculate the concentration of C6H12O6 in M. c(C6H12O6) = 17.8 g L–1 17.8 g = mol L–1 −1 180 g mol = 0.09889 mol L–1 = 0.0989 M As the concentration of KCl is 0.020 M and as there is 1 mol of K+ ions in 1 mol of KCl, the concentration of K+ ions will be 0.020 M. c(K+ ions) = 0.020 M There are Cl– ions from KCl and NaCl. The concentration of KCl is 0.020 M and NaCl is 0.041 M. As there is 1 mol of Cl– ions in 1 mol of KCl, and 1 mol of Cl– ions in 1 mol of NaCl, then calculate the concentration of Cl– ions. c(Cl– ions) = (0.020 + 0.041) M = 0.061 M

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Chapter 3 Volumetric analysis Q17.

To what volume of water must 10 mL of 8.0 M HCl be added in order to prepare a 0.50 M HCl solution? A17.

Step 1 Step 2 Step 3

Step 4

Write the appropriate formula. n=c×V Calculate the amount of HCl used to make the solution. n(HCl) = 0.010 L × 8.0 mol L–1 = 0.080 mol Calculate the total volume of the dilute HCl solution. 0.080 mol V(HCl) = 0.50 mol L−1 = 0.16 L = 160 mL Calculate the volume of water needed to be added to the initial 10 mL of concentrated HCl. V(water) = (160 – 10) mL = 150 mL = 0.15 L

Q18.

The label on a laundry stain remover indicates that it contains 50 g/L hydrogen peroxide (H2O2). What is the molarity of the hydrogen peroxide? A18.

m M

Step 1

Convert the mass of H2O2 to moles using n =

Step 2

M(H2O2) = 2 ×1.0 + 2 × 16.0 = 34.0 g mol–1 50 = 1.47 mol n(H2O2) = 34 Calculate the concentration of H2O2 in mol L–1. 1.47 c(H2O2) = 1.0 = 1.47 = 1.5 M (two significant figures)

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Worked solutions to student book questions

Chapter 3 Volumetric analysis Q19.

Schools normally purchase concentrated (14 M) nitric acid and then dilute it for use. What volume is required to prepare 2.0 L of 0.15 M acid? A19.

Step 1 Step 2

Calculate the amount of dilute HNO3 required. n(HNO3)diluted = 0.15 M × 2.0 L = 0.30 mol Calculate the volume of concentrated acid which contains this amount of HNO3, to the correct number of significant figures. 0.30 mol V(HNO3)conc. = 14 M = 0.0214 L = 21.4 mL = 21 mL

Q20.

A student wishes to prepare 500 mL of a standard solution of any base of concentration 0.2500 M. a Would it be better to prepare the solution using solid sodium hydroxide or anhydrous sodium carbonate? Explain. b Write instructions to the student for making up the solution. A20. a

b

A standard solution can be prepared using anhydrous sodium carbonate because it acts as a primary standard. Unlike sodium hydroxide, it is readily obtained in pure form. Sodium hydroxide is deliquescent (absorbs water from the atmosphere) and also reacts with carbon dioxide in air. 1 Calculate the mass of anhydrous sodium carbonate required to make 500 mL of 0.2500 M solution. M(Na2CO3) = 106.0 g mol–1 Mass Na2CO3 required = 106.0 × 0.2500 × 0.500 = 13.25 g = 13.3 g 2 Weigh accurately a mass of 13.25 g of anhydrous sodium carbonate. 3 Transfer the solid to a 500 mL volumetric flask, washing all traces of the solid from the container used for weighing into the flask. 4 Add distilled water until the flask is about half full and shake until the solid dissolves. 5 Add more distilled water until the bottom of the meniscus is exactly level with the calibration mark. 6 Shake the flask to thoroughly mix the solution.

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Chapter 3 Volumetric analysis Q21.

Fluorine compounds are added to Melbourne’s supplies of drinking water to give a concentration of fluoride ions of about 0.90 ppm (1 ppm = 1 g in l06 g). a What amount, in mol, of fluoride is present in 1.0 g of Melbourne water? b How many fluoride ions would you swallow if you drank a 200 mL glass of Melbourne water? (1 mL of water has a mass of 1 g.) A21. a

Step 1

Step 2

b

Step 1 Step 2

Step 3

Rewrite 0.90 ppm fluoride ions as mole per 1.0 g of water, using m n= . M 0.90 ppm means 0.90 g F– ions per 106 g of water 0.90 g mol per 106 g of water = 18.9984 g mol −1 = ? mol per 1.0 g of water Calculate the concentration of F– ions in mol g–1 to the correct number of significant figures. 1.0 0.90 c(F–) = × 6 mol g–1 18.9984 10 = 4.73724 × 10–8 mol g–1 = 4.7 × 10–8 mol g–1 Rewrite 4.7 × 10–8 mol g–1 as mole per 200 g. 4.7 × 10–8 mol g–1 means 4.7 × 10–8 mol of F– ions per 1.0 g water = ? mol of F– ions per 200.0 g water Calculate the concentration of F– ions in mol per 200 g. 200 mol per 200 g c(F–) = 4.7 × 10–8 × 1.0 = 9.40 × 10–6 mol per 200 g Convert this concentration to particles per 200 g, to the correct number of significant figures, using the formula: Number of particles = n × NA Number of F– ions = 9.40 × 10–6 × 6.023 × 1023 = 5.66162 × 1018 = 5.7 × 1018

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Worked solutions to student book questions

Chapter 3 Volumetric analysis Q22.

Sketch a pipette, showing the calibration line. Clearly indicate the positions of the curved surface (meniscus) of a liquid in the pipette before and after the liquid is delivered. A22.

Q23.

A student is to perform an analysis of sodium hydroxide solution by titrating it with standard hydrochloric acid, as shown in Figure 3.5. Before beginning, the student rinses the glassware that is to be used in the analysis. However, the student does not wish to wait until the glassware has dried before using it. For each of the following apparatus, a, b and c, state if it should be rinsed with: i de-ionised water only ii sodium hydroxide solution only iii hydrochloric acid only. a pipette b burette c conical flask A23. a b c

ii iii i

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Worked solutions to student book questions

Chapter 3 Volumetric analysis Q24.

The concentration of a solution of sodium hydroxide was found by titration with a standard solution of hydrochloric acid. The following steps (not in the correct order) were used when carrying out the titration: I The burette was filled with hydrochloric acid. II The conical flask was rinsed with water. III The burette was rinsed with hydrochloric acid. IV An aliquot of sodium hydroxide was placed in the conical flask. V The level of acid in the burette was read. VI The level of acid in the burette was read. VII A pipette was rinsed with sodium hydroxide solution. VIII Hydrochloric acid was discharged from burette until the end point was reached. IX An indicator was added to the sodium hydroxide solution. The titration was repeated several times and the following values were obtained: 25.32, 25.30, 25.34, 25.90, 25.30 mL. a What is meant by the term ‘standard’ in ‘a standard solution of hydrochloric acid’? b Sodium hydroxide solution should not be left standing in a glass burette. Explain why. c From the list above, list the steps required to complete a titration in the correct order. d A ‘rough’ titration is usually carried out before the first accurate titration. Explain what is meant by a rough titration and why it is used. e Titrations are repeated until concordant results are obtained. Explain what is meant by the term ‘concordant’ in this context. f What is the best value to use for the titre of acid, considering the values given above? g For the titration values 25.32, 25.30, 25.34 mL, what sort of error explains the differences—random or systematic? Explain your answer. A24. a b c d

e f

A standard solution of hydrochloric acid is one of accurately known concentration. Concentrated solutions of sodium hydroxide react with the silica of the glass, reducing the accuracy of the glassware. Burettes with glass taps may fuse or ‘freeze’. For example, III, I, VII, II, IV, IX, V, VIII, VI. Other correct answers are possible but I comes after III, IV after VII and II, IX after IV, VIII after IX and V. A rough titration is carried out if the volume of the titrant is not known. It allows the titration to be carried out faster. The titrant is added quickly to the conical flask until end point is reached. This gives an inaccurate or rough value. For subsequent titrations the titrant is added quickly to within 1–2 mL of the end point and then the titrant is added dropwise until the end point is reached. Concordant results are titres that are within a 0.10 mL range, from highest to lowest. 25.32, 25.30, 25.34 and 25.30 mL should be used, giving an average titre of 25.32 mL. The 25.90 mL titre may have been a rough titre or due to a mistake in the titration.

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

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Worked solutions to student book questions

Chapter 3 Volumetric analysis g

The variation is due to random errors giving answers above and below the true value. These may result from slightly different views when reading the burette, filling of the pipette, judgement of the end point, etc.

Q25.

Anhydrous sodium carbonate is used as a primary standard in determining the concentration of hydrochloric acid by volumetric analysis. a What criteria are used to determine whether or not a substance is suitable for use as a primary standard? b What mass of anhydrous sodium carbonate is required to prepare 200 mL of a standard solution of sodium carbonate that has a concentration of 0.20 M? c How would you prepare this standard solution? A25. a

A primary standard: - Has a very high level of purity - Has a known formula - Is stable, e.g. will not react with atmospheric gases (e.g. carbon dioxide, water vapour) - Has a high molar mass to minimise errors in weighing - Is readily available - Is relatively inexpensive

b

c

n(Na2CO3) = cV = 0.20 × 0.20 = 0.040 mol m(Na2CO3) = nMr = 0.040 ×106 = 4.24 g Accurately weigh an empty weighing bottle, add the primary standard and reweigh. Transfer the weighed sample to a volumetric flask using a dry glass funnel. Rinse out the weighing bottle and glass funnel using a wash bottle. Half fill the volumetric flask with water and shake to dissolve the sample. When the sample has dissolved add water to the calibration mark and shake the flask again. Determine the concentration of the primary standard.

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

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