# Moles and Solutions 1.1.7 2008 SPECIFICATIONS 1.1.7 Moles Moles and Solutions 1.1.7 2008 SPECIFICATIONS 1.1.7 Moles and solutions Calculate the amount of substance in moles using solution volume and concentration. Describe a solutions concentration using the terms concentrated and dilute. Procedure Fill your burette with : 0.100 mol dm3 sulfuric acid, H2SO4 : B2. Pipette 25.0 cm of sodium hydroxide solution : A2 into a conical flask. Add a few drops of phenolphthalein indicator to the conical flask. Titrate the contents of the conical flask with sulfuric acid, H2SO4 : B2. 3 Repeat the titration until you have concordant results (within 0.10 cm3). Record all titration results in a table showing initial and final burette readings. Equipment/materials

A2: sodium hydroxide solution NaOH B2: 0.100 mol dm3 sulfuric acid, H2SO4 Phenolphthalein indicator 25.0 cm3 pipette 50.0 cm3 burette and stand Funnel Conical flask Wash bottle White tile use white paper Why ? Data 2NaOH + H2SO4 Na2SO4 + 2H2O H = 1.0, O = 16.0, Na = 23.0 Analysis of results Indicate which titres you use in your average. Calculate the average of your concordant results. Calculate the amount, in mol, of H2SO4 reacting with the NaOH. Calculate the amount, in mol, of NaOH in the pipette. Calculate the amount, in mol, of NaOH in 1.00 dm3 of A2. Calculate the mass of NaOH in 1.00 dm3 of A2. THE MOLARITY CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION MOLES UNITS =

CONCENTRATION x VOLUME concentration volume mol dm-3 dm3 MOLES MOLES = CONCENTRATION (mol dm-3-3) x VOLUME (dm33) CONC x VOLUME BUT IF... concentration mol dm-3 -3 -3) 3x VOLUME (cm33) MOLES = CONCENTRATION (mol dmcm volume 1000 COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED From the examiner It is important to present your results (with units) in a table showing

initial and final burette readings, using appropriate number of decimal places. Show clearly how the average titre is obtained. Explain each line of your calculation as shown in the Analysis of Results section above. Consider how many significant figures to use within the calculation. Consider how many significant figures to use in your final answer . Questions 1.Sulfuric acid is known as a dibasic acid. Explain what this means. 2.Explain why a titration reading of 23.58 cm3 would generally be unacceptable. 3.Under what conditions would the acid salt, NaHSO4, be produced? 4.Why is phenolphthalein such a good indicator to use in this case? 5.Explain why more than three significant figures for the final answer would not be appropriate. All values in cm3 rough 1st 2nd 3rd initial 0.00 20.20

0.00 0.05 final 22.15 41.30 22.20 21.20 titre 22.15 21.10 22.20 21.15 Mean value Mean value to 1 decimal place Concordancy is 0.10 cm3 Mean value ( that the mean of concordant values ) 4th

cm3 All values in cm3 rough 1st 2nd 3rd initial 0.00 20.20 0.00 0.05 final 22.15 41.30 22.20

21.20 titre 22.15 21.10 22.20 21.15 Mean value Mean value to 1 decimal place Concordancy is 0.10 cm3 Mean value ( that the mean of concordant values ) 4th cm3 All values in cm3 rough 1st 2nd

3rd initial 0.00 20.20 0.00 0.05 final 22.15 41.30 22.20 21.20 titre 22.15 21.10 22.20 21.15

Mean value 21.1 Mean value to 1 decimal place Concordancy is 0.10 cm3 Mean value ( that the mean of concordant values ) 4th cm3 All values in cm3 rough 1st 2nd 3rd 4th initial 0.00 20.20

0.00 0.05 0.00 final 22.15 41.30 22.20 21.20 21.20 titre 22.15 21.10 22.20 21.15 21.20

Mean value Mean value to 1 decimal place Concordancy is 0.10 cm3 Mean value ( that the mean of concordant values ) cm3 All values in cm3 rough 1st 2nd 3rd 4th initial 0.00 20.20 0.00 0.05 0.00

final 22.15 41.30 22.20 21.20 21.20 titre 22.15 21.10 22.20 21.15 21.20 Mean value 21.2 Mean value to 1 decimal place Concordancy is 0.10 cm3

Mean value ( that the mean of concordant values ) cm3 THE MOLARITY CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION MOLES UNITS = CONCENTRATION x VOLUME concentration volume mol dm-3 dm3 MOLES MOLES = CONCENTRATION (mol dm-3-3) x VOLUME (dm33) CONC x VOLUME BUT IF... concentration mol dm-3 -3 -3) 3x VOLUME (cm33) MOLES = CONCENTRATION (mol dmcm

volume 1000 COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED MOLE OF SOLUTE IN A SOLUTION MOLES 1 = CONCENTRATION x VOLUME Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH MOLE OF SOLUTE IN A SOLUTION MOLES 1 = CONCENTRATION x VOLUME Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles =

= conc x volume in cm3 1000 2 mol dm-3 x 25cm3 1000 = 0.05 moles MOLE OF SOLUTE IN A SOLUTION MOLES 1 CONCENTRATION x VOLUME Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = = 2 = conc x volume in cm3 1000 2 mol dm-3 x 25cm3 1000

= 0.05 moles What volume in cm3 of 0.1M H2SO4 contains 0.002 moles ? MOLE OF SOLUTE IN A SOLUTION MOLES 1 CONCENTRATION x VOLUME Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH moles = = 2 = conc x volume in cm3 1000 2 mol dm-3 x 25cm3 1000 = 0.05 moles What volume in cm3 of 0.1M H2SO4 contains 0.002 moles ? volume = (in cm3) =

1000 x moles conc 1000 x 0.002 0.1 mol dm-3 (re-arrangement of above) = 20 cm3 STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? How many moles will be in 1 dm3 ? How many moles will be in 250cm3 ? = 0.100 mol dm-3 = 0.100 mol = 0.100/4 = 0.025 mol STANDARD SOLUTION How to work out how much to weigh out

A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? How many moles will be in 1 dm3 ? How many moles will be in 250cm3 ? = 0.100 mol dm-3 = 0.100 mol = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate? What is the relative formula mass? What is the molar mass? = Na2CO3 = 106 = 106g mol -1 STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? How many moles will be in 1 dm3 ? How many moles will be in 250cm3 ? = 0.100 mol dm-3 = 0.100 mol

= 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate? What is the relative formula mass? What is the molar mass? = Na2CO3 = 106 = 106g mol -1 What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650g of Na2CO3 ? (mass = moles x molar mass) STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? How many moles will be in 1 dm3 ? How many moles will be in 250cm3 ? = 0.100 mol dm-3 = 0.100 mol = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate? What is the relative formula mass?

What is the molar mass? = Na2CO3 = 106 = 106g mol -1 What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650g of Na2CO3 ? (mass = moles x molar mass) ANS. The chemist will have to weigh out 2.650g, dissolve it in water and then make the solution up to 250cm3 in a graduated flask. STANDARD SOLUTION How to work out how much to weigh out A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out? What concentration is the solution to be? How many moles will be in 1 dm3 ? How many moles will be in 250cm3 ? = 0.100 mol dm-3 = 0.100 mol = 0.100/4 = 0.025 mol What is the formula of anhydrous sodium carbonate? What is the relative formula mass? What is the molar mass?

= Na2CO3 = 106 = 106g mol -1 What mass of Na2CO3 is in 0.025 moles = 0.025 x 106 = 2.650g of Na2CO3 ? (mass = moles x molar mass) ANS. The chemist will have to weigh out 2.650g, dissolve it in water and then make the solution up to 250cm3 in a graduated flask. Exercise 1.1.7 . Calculations based on concentrations in solution Calculate the number of moles in the volume of solution stated. 1 2 3 4 5 6 7 8 9 10 25 50

250 500 25 50 50 100 25 25 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 of of of of of of of of of of

1.0 0.5 0.25 0.01 1.0 0.5 0.25 0.1 0.05 0.2 mol mol mol mol mol mol mol mol mol mol dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3

dm-3 HCl HCl HCl HCl NaOH KOH HNO3 H2SO4 KMnO4 FeSO4 odds or evens . . . Up to 30 !! Calculate the mass of material in the given volume of solution 11 12 13 14 15 16 17 18 19 20 25 50 100 100

25 50 20 50 25 25 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 of of of of of of of of of of 1 0.5

0.25 0.1 1 0.2 0.1 0.1 0.02 0.1 mol mol mol mol mol mol mol mol mol mol dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3

Now do all 31, 32, 33, 34, 35 HCl NaCl NH4NO3 AgNO3 BaCl2 H2SO4 NaOH K2CrO4 KMnO4 Pb(NO3)2 What is the concentration in moles dm-3 of the following? 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 3.65

3.65 6.62 1.00 1.96 1.58 25.0 25.0 4.80 10.0 240 480 120 24 100 g g g g g g g g g g cm3 cm3 cm3 cm3 cm3

of of of of of of of of of of of of of of of HCl HCl Pb(NO3)2 NaOH H2SO4 KMnO4 Na2S2O3.5H2O CuSO4.5H2O (COOH)2.2H2O FeSO4.(NH4)2SO4.6H2O NH3(g) HCl(g) SO2(g) HCl(g) NH3(g)

in in in in in in in in in in dissolved dissolved dissolved dissolved dissolved 1000 100 250 250 250 250 250 250 250 250 in in in in

in cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 1000 100 250 200 10 of of of of of of of of of of cm3 cm3 cm3

cm3 cm3 solution solution solution solution solution solution solution solution solution solution of of of of of solution solution solution solution solution Exercise 1.1.7 . Calculations based on concentrations in solution Calculate the number of moles of the underlined species in the volume of solution stated. 1

2 3 4 5 6 7 8 9 10 25 50 250 500 25 50 50 100 25 25 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3

of of of of of of of of of of 1.00 0.500 0.250 0.0100 1.00 0.500 0.250 0.100 0.0500 0.200 mol mol mol mol mol mol mol mol

mol mol dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 HCl HCl HCl HCl NaOH KOH HNO3 H2SO4 KMnO4 FeSO4 odds or evens . . . . Calculate the mass of material in the given volume of solution 11

12 13 14 15 16 17 18 19 20 25 50 100 100 25 50 20 50 25 25 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3

of of of of of of of of of of 1.00 0.500 0.250 0.100 1.00 0.200 0.100 0.100 0.0200 0.100 mol mol mol mol mol mol mol mol

mol mol dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 dm-3 odds or evens . . . HCl NaCl NH4NO3 AgNO3 BaCl2 H2SO4 NaOH K2CrO4 KMnO4 Pb(NO3)2 What is the concentration in moles dm-3 of the following? 21 22 23

24 25 26 27 28 29 30 31 32 33 34 35 3.65 3.65 6.62 1.00 1.96 1.58 25.0 25.0 4.80 10.0 240 480 120 24 100 g g

g g g g g g g g cm3 cm3 cm3 cm3 cm3 of of of of of of of of of of of of of of of HCl

HCl Pb(NO3)2 NaOH H2SO4 KMnO4 Na2S2O3.5H2O CuSO4.5H2O (COOH)2.2H2O FeSO4.(NH4)2SO4.6H2O NH3(g) HCl(g) SO2(g) HCl(g) NH3(g) in in in in in in in in in in dissolved dissolved dissolved dissolved dissolved

1000 100 250 250 250 250 250 250 250 250 in in in in in odds or evens . . . Up to 30 !! Now do all 31, 32, 33, 34, 35 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 cm3 1000 100

250 200 10 of of of of of of of of of of cm3 cm3 cm3 cm3 cm3 solution solution solution solution solution solution solution solution solution solution of

of of of of solution solution solution solution solution Exercise 1.1.7 Molarity Moles Mass ( 3 significant figures ) 1 0.0250 moles 19 0.0790 g 2 0.0250 moles 20 0.828 g

3 0.0625 moles 21 0.100 mol dm3 4 0.00500 moles 22 1.00 mol dm3 5 0.0250 moles 23 0.0300 mol dm3

6 0.0250 moles 24 0.100 mol dm3 7 0.0125 moles 25 0.0300 mol dm3 8 0.0100 moles 26 0.0400 mol dm3 9 0.00125

moles 27 0.400 mol dm3 10 0.00500 moles 28 0.400 mol dm3 11 0.9125 g 29 0.152 mol dm3 12 1.46

g 30 0.0102 mol dm3 13 2.00 g 31 0.0100 mol dm3 14 1.70 g 32 0.200 mol dm3 15 5.20 g

33 0.0200 mol dm3 16 0.980 g 34 0.00500 mol dm3 17 0.0800 g 35 0.417 18 0.970 g mol dm3 1.1.7 Tasks Write out these Key definitions

The concentration of a solution is . . A standard solution is . . . Examiner tip is . . . . Worksheets Questions : 1, 2, 3 Pages 16-17 SOLUTIONS Dissolving a SOLUTE in a SOLVENT makes a SOLUTION Volumetric solutions are made by dissolving a known amount of solute in a solvent (usually water) and then adding enough solvent to get the correct volume of solution. WRONG Dissolve 1g of solute in 250cm of de-ionised water 3 1g 250cm3 SOLUTIONS Dissolving a SOLUTE in a SOLVENT makes a SOLUTION Volumetric solutions are made by dissolving a known amount of solute in a solvent (usually water) and then adding enough solvent to get the correct volume of solution. WRONG Dissolve 1g of solute in 250cm of de-ionised water

1g 3 250cm3 WATER RIGHT Dissolve 1g of solute in water and then add enough water to make 250cm3 of solution WATER 1g 250cm3 STANDARD SOLUTION ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY 4.240g of Na2CO3 was placed in a clean beaker and dissolved in de-ionised water The solution was transferred quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water. STANDARD SOLUTION ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY

4.240g of Na2CO3 was placed in a clean beaker and dissolved in de-ionised water The solution was transferred quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm-3 ? STANDARD SOLUTION ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY 4.240g of Na2CO3 was placed in a clean beaker and dissolved in de-ionised water The solution was transferred quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm-3 ? mass of Na2CO3 in a 250cm3 solution = 4.240g molar mass of Na2CO3 = 106g mol -1 no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol

STANDARD SOLUTION ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY 4.240g of Na2CO3 was placed in a clean beaker and dissolved in de-ionised water The solution was transferred quantitatively to a 250 cm3 graduated flask and made up to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm-3 ? mass of Na2CO3 in a 250cm3 solution = 4.240g molar mass of Na2CO3 = 106g mol -1 no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol Concentration is normally expressed as moles per dm3 of solution Therefore, as it is in 250cm3, the value is scaled up by a factor of 4 thats 1000/250 STANDARD SOLUTION ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY 4.240g of Na2CO3 was placed in a clean beaker and dissolved in de-ionised water The solution was transferred quantitatively to a 250 cm3 graduated flask and made up

to the mark with de-ionised (or distilled) water. What is the concentration of the solution in mol dm-3 ? mass of Na2CO3 in a 250cm3 solution = 4.240g molar mass of Na2CO3 = 106g mol -1 no. of moles in a 250cm3 solution = 4.240g / 106g mol -1 = 0.04 mol Concentration is normally expressed as moles per dm3 of solution Therefore, as it is in 250cm3, the value is scaled up by a factor of 4 no. of moles in 1000cm3 (1dm3) = 4 x 0.04 = 0.16 mol ANS. 0.16 mol dm-3 THE MOLARITY CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.200 mol dm-3 volume pipetted out into the conical flask = 25.00 cm3 25cm3

250cm3 250cm3 THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.200 mol dm-3 volume pipetted out into the conical flask = 25.00 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of 0.200 mol dm-3 THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.200 mol dm-3 volume pipetted out into the conical flask = 25.00 cm3 25cm3 250cm3 250cm3

The original solution has a concentration of 0.200 mol dm-3 This means that there are 0.200 mols of solute in every 1 dm3 (1000 cm3) of solution THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.200 mol dm-3 volume pipetted out into the conical flask = 25.00 cm3 25cm3 250cm3 250cm3 The original solution has a concentration of 0.200 mol dm-3 This means that there are 0.200 mols of solute in every 1 dm3 (1000 cm3) of solution Take out 25.00 cm3 and you will take a fraction 25/1000 ( 1/40 ) of the number of moles THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.200 mol dm-3 volume pipetted out into the conical flask = 25.00 cm3 25cm3 250cm3 250cm3

The original solution has a concentration of 0.200 mol dm-3 This means that there are 0.200 mols of solute in every 1 dm3 (1000 cm3) of solution Take out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles moles in 1dm3 (1000cm3) moles in 1cm3 moles in 25cm3 = = = 0.200 0.200/1000 25/1000 x 0.200 = 5.0 x 10-3 mol THE MOLE CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION concentration of solution in the graduated flask = 0.200 mol dm-3 volume pipetted out into the conical flask = 25.00 cm3 25cm3 250cm3 250cm3

The original solution has a concentration of 0.200 mol dm-3 This means that there are 0.200 mols of solute in every 1 dm3 (1000 cm3) of solution Take out 25.00 cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles moles in 1dm3 (1000cm3) moles in 1cm3 moles in 25cm3 = = = 0.200 0.200/1000 25 x 0.200/1000 = 5.0 x 10-3 mol Titrations 1.1.13 VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equation NaOH

+ HCl > NaCl + H2O VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equation 2. Get a molar relationship between the reactants NaOH + HCl > NaCl + H2O

moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equation 2. Get a molar relationship between the reactants 3. Calculate the number of moles of each substance M is the concentration in mol dm-3 NaOH + HCl > NaCl + H2O moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl

HCl 0.100 x NaOH M 20/1000 x 25/1000 (i) (ii) VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equation 2. Get a molar relationship between the reactants 3. Calculate the number of moles of each substance M is the concentration in mol dm-3 NaOH

+ HCl > NaCl + H2O moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl HCl 0.100 x NaOH M 20/1000 4. Look at the molar relationship and insert (i) and (ii) x

25/1000 (i) (ii) moles of NaOH = moles of HCl M x 20/1000 = 0.100 x 25/1000 VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equation 2. Get a molar relationship between the reactants 3. Calculate the number of moles of each substance M is the concentration in mol dm-3 NaOH re-arrange the numbers to obtain M HCl > NaCl + H2O

moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl HCl 0.100 x NaOH M 20/1000 4. Look at the molar relationship and insert (i) and (ii) 5. Cancel the 1000s + x 25/1000 (i) (ii) moles of NaOH = moles of HCl M x 20/1000 = 0.100 x 25/1000

M x 20 M = = 0.100 x 25 0.100 x 25 20 VOLUMETRIC CALCULATIONS Calculate the concentration of a solution of sodium hydroxide if 20cm3 is neutralised by 25cm3 of hydrochloric acid of concentration 0.100 mol dm-3. 1. Write out a BALANCED equation 2. Get a molar relationship between the reactants 3. Calculate the number of moles of each substance M is the concentration in mol dm-3 NaOH re-arrange the numbers to obtain M 6. Calculate the concentration of the NaOH HCl > NaCl

+ H2O moles of NaOH = moles of HCl you need ONE NaOH for every ONE HCl HCl 0.100 x NaOH M 20/1000 4. Look at the molar relationship and insert (i) and (ii) 5. Cancel the 1000s + x 25/1000 (i) (ii)

moles of NaOH = moles of HCl M x 20/1000 = 0.100 x 25/1000 M x 20 M = = 0.100 x 25 0.100 x 25 20 = 0.125 mol dm-3 VOLUMETRIC CALCULATIONS Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you dont understand what an equation tells you, it is easy to make a mistake. 2NaOH + H2SO4 > Na2SO4 + 2H2O

you need 2 moles of NaOH to react with every 1 mole of H2SO4 i.e moles of NaOH = 2 x moles of H2SO4 or moles of H2SO4 = moles of NaOH 2 REMEMBER... IT IS NOT A MATHEMATICAL EQUATION 2 moles of NaOH DO NOT EQUAL 1 mole of H2SO4 More examples follow VOLUMETRIC CALCULATIONS Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you dont understand what an equation tells you, it is easy to make a mistake. 2NaOH +

H2SO4 > Na2SO4 + 2H2O you need 2 moles of NaOH to react with every 1 mole of H2SO4 i.e moles of NaOH = 2 x moles of H2SO4 or moles of H2SO4 = moles of NaOH 2 2HCl +

Na2CO3 > 2NaCl + CO2 + H2O you need 2moles of HCl to react with every 1 mole of Na2CO3 i.e moles of HCl = 2 x moles of Na2CO3 or moles of Na2CO3 = moles of HCl 2 VOLUMETRIC CALCULATIONS Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you dont understand what an equation tells you, it is easy to make a mistake. MnO4 + 8H+ + 5Fe2+ > Mn2+ + 4H2O + 5Fe3+

you need 5 moles of Fe2+ to react with every 1 mole of MnO4 i.e moles of Fe2+ = 5 x moles of MnO4 or moles of MnO4 = moles of Fe2+ 5 VOLUMETRIC CALCULATIONS Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you dont understand what an equation tells you, it is easy to make a mistake. MnO4 + 8H+ + 5Fe2+ > Mn2+ + 4H2O + 5Fe3+ you need 5 moles of Fe2+ to react with every 1 mole of MnO4 i.e moles of Fe2+

= 5 x moles of MnO4 or moles of MnO4 = moles of Fe2+ 5 2MnO4 + 5H2O2 + 6H+ > 2Mn2+ + 5O2 + 8H2O you need 5 moles of H2O2 to react with every 2 moles of MnO4 i.e moles of H2O2 = 5 x moles of MnO4 2 or moles of MnO4 =

2 x moles of H2O2 5 VOLUMETRIC CALCULATIONS Calculate the volume of sodium hydroxide (concentration 0.100 mol dm-3) required to neutralise 20cm3 of sulphuric acid of concentration 0.120 mol dm-3. 2NaOH + H2SO4 > Na2SO4 + 2H2O you need 2 moles of NaOH to react with every 1 mole of H2SO4 therefore moles of NaOH = 2 x moles of H2SO4 moles of H2SO4 = 0.120 x 20/1000 (i)

moles of NaOH = 0.100 x V/1000 where V is the volume of alkali in cm3 (ii) VOLUMETRIC CALCULATIONS Calculate the volume of sodium hydroxide (concentration 0.100 mol dm-3) required to neutralise 20cm3 of sulphuric acid of concentration 0.120 mol dm-3. 2NaOH + H2SO4 > Na2SO4 + 2H2O you need 2 moles of NaOH to react with every 1 mole of H2SO4 therefore moles of NaOH = 2 x moles of H2SO4 moles of H2SO4 = 20/1000

(i) moles of NaOH = 0.100 x V/1000 where V is the volume of alkali in cm3 (ii) substitute numbers x moles of NaOH = 2 x moles of H2SO4 0.100 x V/1000 = 2 x 0.120 x 20/1000 0.100 x V = 2 x 0.120 x 20

Volume of NaOH (V) = 2 x 0.120 x 20 = 48.00 cm3 0.100 cancel the 1000s re-arrange 0.120 Exercise 1.1.13 Titrations Simple volumetric calculations In this series of calculations you should start by writing the equation for the reaction taking place then generate the molarity/volume ratio. In some cases you will need to calculate the molarity of the solutions before you start the main part of the question. 1 25.0 cm3 of sodium hydroxide ( NaOH ) reacts with 21.0 cm3 of 0.2 mol dm3 hydrochloric acid (HCl) 2 25.0 cm3 of sodium hydroxide ( NaOH ) reacts with 17.0 cm3 of 0.1 mol dm3 sulphuric acid (H2SO4) 3 20.0 cm3 of hydrochloric acid (HCl) reacts with 23.6 cm3 of 0.1 mol dm3 NaOH 4 20.0 cm3 of hydrochloric acid (HCl) reacts with 20.0 cm3 of a solution of NaOH containing 40 g dm3 of NaOH 5 25.0 cm3 of nitric acid ( HNO3)reacts with 15.0 cm3 of a solution of 0.2 mol dm3 Ammonium hydroxide (NH4OH) 6 25.0 cm3 of a solution of barium chloride ( BaCl2) reacts with 20.0 cm3 of a solution of 0.05

mol dm3 sulphuric acid (H2SO4) 7 25.0 cm3 of a solution of sodium chloride (NaCl) reacts with 10.0 cm3 of a 0.02 mol dm3 silver nitrate ( AgNO3) 8 10.0 cm3 of a solution of Aluminium chloride (AlCl3) reacts with 30.0 cm3 of 0.01 mol dm-3 silver nitrate ( AgNO3) 1 25.0 cm3 of sodium hydroxide ( NaOH ) reacts with 21.0 cm3 of 0.2 mol dm3 hydrochloric acid NaOH + HCl NaCl +H2O 0.168 mol dm3 (HCl) 2 NaOH + H SO Na SO +2 H2O 2 25.0 cm3 of sodium hydroxide ( 2 NaOH )4reacts with 2 17.0 cm 4 3 of 0.1 mol dm3 sulphuric acid 0.136 mol dm (H2SO4) NaOH + HCl NaCl +H2O 3 20.0 cm3 of hydrochloric acid (HCl) reacts with 23.6 cm3 of 0.1 mol dm3 NaOH 0.118 mol dm

3 NaOH + HCl NaCl +H2O 4 20.0 cm3 of hydrochloric acid (HCl) reacts with 20.0 cm3 of a solution of NaOH containing 40 g dm3 of NaOH 1.0 mol dm 3 3 5 25.0 cm3 of nitric acid ( HNO3)reacts with 15.0 cm3 of a solution of 0.2 mol dm3 Ammonium hydroxide (NH4OH) HNO3 +NH4OH NH4NO3 +H2O 0.12 mol 6 25.0 cm3 of a solution of barium chloride ( BaCl2) reacts with 20.0 cm3 of a solution of 0.05 mol dm3 sulphuric acid (H2SO4) BaCl2 + H2SO4 BaSO4 +2HCl 7 25.0 cm3 of a solution of sodium chloride (NaCl) reacts with 10.0 cm3 of a 0.02 mol dm3 silver nitrate ( AgNO3) 8 10.0 cm3 of a solution of Aluminium chloride (AlCl3) reacts with 30.0 cm3 of 0.01 mol dm-3

silver nitrate ( AgNO3) 9 25.0 cm3 of HxA reacts with 25.0 cm3 of 0.2 mol dm3 NaOH to give Na2A 10 25.0 cm3 of H3PO4 reacts with 100.0 cm3 of 0.1 mol dm3 NaOH to give NaH2 PO4 11 25 cm3 of a solution of 0.1 mol dm3 NaOH reacts with 50 cm3 of a solution of hydrochloric acid. What is the molarity of the acid? 12 25 cm3 of a solution of 0.2 mol dm3 KOH reacts with 30 cm3 of a solution of nitric acid. What is the concentration of the acid in moles dm3? 13 I n a titration 25 cm3 of ammonia solution react with 33.30 cm3 of 0.1 mol dm3 HCl. What isthe concentration of the ammonia solution in g dm3? 14 I n the reaction between iron(I I ) ammonium sulphate and potassium manganate(VI I ) solution. 25 cm3 of the Fe2+solution reacted with 24.8 cm3 of 0.020 mol dm3 KMnO4 solution. What is the molarity of the iron(I I ) ammonium sulphate solution? ( you dont need the equation ) 15 10 cm3 of a solution of NaCl react with 15 cm3 of 0.02 mol dm3 silver nitrate solution. What is the concentration of the NaCl solution in g dm3?

16 25 cm3 of a solution of an acid HxA containing 0.1 mol dm3 of the acid in each 1000 cm3 of solution reacts with 75 cm3 of a solution of 0.1 mol dm3 NaOH. What is the value of x? 17 25 cm3 of a solution of sodium carbonate react with 10 cm3 of a 0.1 mol dm3 HCl. What is the concentration of the sodium carbonate? 18 What volume of 0.1 mol dm3 HCl will be needed to react with 25 cm3 of 0.2 mol dm3 NaOH? 19 What volume of 0.05 mol dm3 H2SO4 will be needed to react with 25 cm3 of 0.2 mol dm3 NaOH? 20 What volume of 0.02 mol dm3 KMnO4 will be needed to react with 25 cm3 of 0.1 mol dm3 FeSO4 solution? The last five questions will require you to use the skills you have learnt in this se ction, together with those f rom other sections. 21 What weight of silver chloride will be produced if 25 cm3 of 0.1 mol dm3 silver nitrate is added to excess sodium chloride solution? 22 What weight of calcium carbonate will dissolve in 100 cm3 of 0.2 mol dm3 HCl?

23 What volume of carbon dioxide will be produced if 100 cm3 of 0.2 mol dm3 HNO3 is added to excess sodium carbonate solution? 24 What weight of magnesium will dissolve in 10 cm3 of 1 mol dm3 HCl and what volume of hydrogen will be produced? 25 What volume of ammonia gas will be produced in the f ollowing reaction . . if 50 cm3 of 0.5 mol dm3 sodium hydroxide is boiled with 50 cm3 of 0.4 mol dm3 ammonium chloride solution? (Care: one of these is in excess.) NaOH(aq) +NH4Cl(aq) NaCl(aq) +H2O(l) +NH3(g) 1.1.13 Titrations Calculations Answers 1 0.168 mol dm3 2 0.136 mol dm3 3 0.118 mol dm3 4

1.0 mol dm3 5 0.12 mol dm3 6 0.040 mol dm3 7 0.0080 mol dm3 8 0.010 mol dm3 9 0.10 mol dm3 120 cm3 H2 10 0.40 mol dm3 11 0.050 mol dm3 12

0.167 mol dm3 13 2.26 g dm3 14 0.099 mol dm3 15 1.755 g dm3 16 3.0 17 0.02 mol dm3 18 50 cm3 19 50 cm3

20 25 cm3 21 0.359 g 22 1.0 g 23 240 cm3 24 0.12 g Mg 25 480 cm3 + A5 is the LIMEWATER C5 is the solution in the Volumetric Flask Its the diluted HYDROCHLORIC ACID (B5) All values

in cm3 rough 1st 2nd 3rd initial 0.00 20.20 0.00 0.05 final 22.15 41.30 22.20 21.20 titre

22.15 21.10 22.20 21.15 Mean value Mean value to 1 decimal place Concordancy is 0.10 cm3 Mean value ( that the mean of concordant values ) 4th cm3 Calculate the amount, in mol, of HCl in the volumetric flask. Calculate the amount, in mol, of HCl used in titration. Calculate the amount, in mol, of Ca(OH)2 reacting. Calculate the amount, in mol, of Ca(OH)2 in 1 dm3. Calculate mass of Ca(OH)2 in 1 dm3. But first of all write the equation for the reaction 2HCl(aq) + Ca(OH)2 (aq) CaCl2 (aq) + 2H O(l) 2 Calculate the amount, in mol, of HCl in the volumetric flask.

Use a graduated pipette to add 3.50 cm3 of 2.00 mol dm3 hydrochloric acid to a 250 cm3 volumetric flask. 3.50 cm3 HCl 2.00 Molar Moles = 3.50/1000 x 2.00 or 0.0035 x 2.00 = 0.007 moles So whats the concentration of this dilute HCl solution ? Molarity = moles / volume = 0.007 /0.250 ( thats 250/1000 ) = 0.028 moles /dm3 Calculate the amount, in mol, of HCl in the volumetric flask. Calculate the amount, in mol, of HCl used in titration. Calculate the amount, in mol, of Ca(OH)2 reacting Molarity of HCl is 0.028 moles /dm3 Lets assume its was 40.0 cm3 HCl needed to neutralise Moles of HCl Moles = 0.028 x 40.0/1000 = 0.00112 moles 2HCl(aq) + Ca(OH)2 (aq) CaCl2+ Moles of 2HCa(OH)2

2O(l) = 0.00112 / 2 = 0.00056moles in 25 cm3 of lime water Calculate the amount, in mol, of HCl in the volumetric flask. Calculate the amount, in mol, of HCl used in titration. Calculate the amount, in mol, of Ca(OH)2 in 1 dm3. Calculate mass of Ca(OH)2 in 1 dm3 to 3 significant figures. Moles of Ca(OH)2 in 25 cm = 0.00056moles So moles in 1000 cm3 must be = 0.00056moles x 40 = 0.0224 moles Molar mass of 40.1 16.0 x 2 1.0 x 2 74.1 grms Ca(OH)2 mass of Ca(OH)2 = 74.1 x 0.0224 = 1.65984 1.66 grms Question 23 What volume of carbon dioxide will be produced if 100 cm3 of 0.2 mol dm3 HNO3 is added to excess sodium carbonate solution?

Na2CO3 + 2HNO3 2NaNO3 + CO2 + H2O Question 23 What volume of carbon dioxide will be produced if 100 cm3 of 0.2 mol dm3 HNO3 is added to excess sodium carbonate solution? Na2CO3 + 2HNO3 2NaNO3 + CO2 + H2O Volume of CO2 ? 100.0 cm3 HNO3 Moles ? 0.20 Molar Moles = 0.100 x 0.20 = 0.020 Ratio : CO2 : HNO3 1:2 Moles CO2 = 0.020/2 = 0.01 mole Volume = 0.01 x 24.0 dm3 = 0.24 dm3