Moles%20and%20Solutions

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Moles and Solutions 1.1.7
2008 SPECIFICATIONS
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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.
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• Procedure
• Fill your burette with 0.100?mol?dm3 sulfuric
  acid, H2SO4 B2.
• Pipette 25.0?cm3 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.
• 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 ?    
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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.
•  
•  
•  

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THE MOLARITY
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A
SOLUTION UNITS concentration mol
dm-3 volume dm3 BUT
IF... concentration mol dm-3 volume cm3
MOLES CONCENTRATION x VOLUME
MOLES CONCENTRATION (mol dm-3) x VOLUME
(dm3)
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• 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
• Sulfuric acid is known as a dibasic acid. Explain
  what this means.
• Explain why a titration reading of 23.58?cm3
  would generally be unacceptable.
• Under what conditions would the acid salt,
  NaHSO4, be produced?
• Why is phenolphthalein such a good indicator to
  use in this case?
• Explain why more than three significant figures
  for the final answer would not be appropriate.
•  

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All values in cm3 rough 1st 2nd 3rd 4th
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 cm3 cm3 cm3 cm3 cm3
Mean value to 1 decimal place Concordancy is
0.10 cm3 Mean value ( that the mean of
concordant values )
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All values in cm3 rough 1st 2nd 3rd 4th
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 cm3 cm3 cm3 cm3 cm3
Mean value to 1 decimal place Concordancy is
0.10 cm3 Mean value ( that the mean of
concordant values )
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All values in cm3 rough 1st 2nd 3rd 4th
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 cm3 21.1 cm3 21.1 cm3 21.1 cm3 21.1 cm3
Mean value to 1 decimal place Concordancy is
0.10 cm3 Mean value ( that the mean of
concordant values )
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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 cm3 cm3 cm3 cm3 cm3
Mean value to 1 decimal place Concordancy is
0.10 cm3 Mean value ( that the mean of
concordant values )
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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 cm3 21.2 cm3 21.2 cm3 21.2 cm3 21.2 cm3
Mean value to 1 decimal place Concordancy is
0.10 cm3 Mean value ( that the mean of
concordant values )
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THE MOLARITY
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A
SOLUTION UNITS concentration mol
dm-3 volume dm3 BUT
IF... concentration mol dm-3 volume cm3
MOLES CONCENTRATION x VOLUME
MOLES CONCENTRATION (mol dm-3) x VOLUME
(dm3)
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MOLE OF SOLUTE IN A SOLUTION
MOLES CONCENTRATION x VOLUME
1 Calculate the moles of sodium hydroxide in
25cm3 of 2M NaOH
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MOLE OF SOLUTE IN A SOLUTION
MOLES CONCENTRATION x VOLUME
1 Calculate the moles of sodium hydroxide in
25cm3 of 2M NaOH moles conc x volume in
cm3 1000
2 mol dm-3 x 25cm3 0.05 moles
1000
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MOLE OF SOLUTE IN A SOLUTION
MOLES CONCENTRATION x VOLUME
1 Calculate the moles of sodium hydroxide in
25cm3 of 2M NaOH moles conc x volume in
cm3 1000
2 mol dm-3 x 25cm3 0.05 moles
1000 2 What volume in cm3 of
0.1M H2SO4 contains 0.002 moles ?
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MOLE OF SOLUTE IN A SOLUTION
MOLES CONCENTRATION x VOLUME
1 Calculate the moles of sodium hydroxide in
25cm3 of 2M NaOH moles conc x volume in
cm3 1000
2 mol dm-3 x 25cm3 0.05 moles
1000 2 What volume in cm3 of
0.1M H2SO4 contains 0.002 moles ? volume
1000 x moles (re-arrangement of
above) (in cm3) conc
1000 x 0.002 20 cm3 0.1 mol dm-3

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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?
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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?
0.100 mol dm-3 How many moles will be in 1
dm3 ? 0.100 mol How many moles will be in
250cm3 ? 0.100/4 0.025 mol
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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?
0.100 mol dm-3 How many moles will be in 1
dm3 ? 0.100 mol How many moles will be in
250cm3 ? 0.100/4 0.025 mol What is
the formula of anhydrous sodium carbonate?
Na2CO3 What is the relative formula mass?
106 What is the molar mass? 106g mol -1
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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?
0.100 mol dm-3 How many moles will be in 1
dm3 ? 0.100 mol How many moles will be in
250cm3 ? 0.100/4 0.025 mol What is
the formula of anhydrous sodium carbonate?
Na2CO3 What is the relative formula mass?
106 What is the molar mass? 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)
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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?
0.100 mol dm-3 How many moles will be in 1
dm3 ? 0.100 mol How many moles will be in
250cm3 ? 0.100/4 0.025 mol What is
the formula of anhydrous sodium carbonate?
Na2CO3 What is the relative formula mass?
106 What is the molar mass? 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.
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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?
0.100 mol dm-3 How many moles will be in 1
dm3 ? 0.100 mol How many moles will be in
250cm3 ? 0.100/4 0.025 mol What is
the formula of anhydrous sodium carbonate?
Na2CO3 What is the relative formula mass?
106 What is the molar mass? 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.
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odds or evens . . . Up to 30 !! Now do all 31,
32, 33, 34, 35
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odds or evens . . .
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odds or evens . . .
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odds or evens . . . Up to 30 !! Now do all 31,
32, 33, 34, 35
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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
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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 250cm3 of de-ionised water
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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 250cm3 of de-ionised
water RIGHT Dissolve 1g of solute in
water and then add enough water to make 250cm3 of
solution
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STANDARD SOLUTION
ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY
4.240g of Na2CO3 was placed in a clean The
solution was transferred beaker and dissolved in
de-ionised water quantitatively to a 250
cm3 graduated flask and made up to the
mark with de-ionised (or distilled)
water.
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STANDARD SOLUTION
ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY
4.240g of Na2CO3 was placed in a clean The
solution was transferred beaker and dissolved in
de-ionised water 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 ?
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STANDARD SOLUTION
ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY
4.240g of Na2CO3 was placed in a clean The
solution was transferred beaker and dissolved in
de-ionised water 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
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STANDARD SOLUTION
ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY
4.240g of Na2CO3 was placed in a clean The
solution was transferred beaker and dissolved in
de-ionised water 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
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STANDARD SOLUTION
ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY
4.240g of Na2CO3 was placed in a clean The
solution was transferred beaker and dissolved in
de-ionised water 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
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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
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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
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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
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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
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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)
0.200 moles in 1cm3 0.200/1000 moles in
25cm3 25/1000 x 0.200
5.0 x 10-3 mol
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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)
0.200 moles in 1cm3 0.200/1000 moles in
25cm3 25 x 0.200/1000
5.0 x 10-3 mol
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Titrations 1.1.13
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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.
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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 gt NaCl H2O
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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 gt NaCl H2O 2. Get a
molar relationship between the reactants
moles of NaOH moles of HCl you need
ONE NaOH for every ONE HCl
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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 gt NaCl H2O 2. Get a
molar relationship between the reactants
moles of NaOH moles of HCl you need
ONE NaOH for every ONE HCl 3. Calculate the
number of moles of each substance HCl
0.100 x 25/1000 (i) M is the
concentration in mol dm-3 NaOH M x
20/1000 (ii)
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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 gt NaCl H2O 2. Get a
molar relationship between the reactants
moles of NaOH moles of HCl you need
ONE NaOH for every ONE HCl 3. Calculate the
number of moles of each substance HCl
0.100 x 25/1000 (i) M is the
concentration in mol dm-3 NaOH M x
20/1000 (ii) 4. Look at the molar
relationship and insert (i) and (ii) moles
of NaOH moles of HCl M x
20/1000 0.100 x 25/1000
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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 gt NaCl H2O 2. Get a
molar relationship between the reactants
moles of NaOH moles of HCl you need
ONE NaOH for every ONE HCl 3. Calculate the
number of moles of each substance HCl
0.100 x 25/1000 (i) M is the
concentration in mol dm-3 NaOH M x
20/1000 (ii) 4. Look at the molar
relationship and insert (i) and (ii) moles
of NaOH moles of HCl M x
20/1000 0.100 x 25/1000 5. Cancel
the 1000s M x 20
0.100 x 25 re-arrange the numbers to
obtain M M 0.100 x 25
20
50
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 gt NaCl H2O 2. Get a
molar relationship between the reactants
moles of NaOH moles of HCl you need
ONE NaOH for every ONE HCl 3. Calculate the
number of moles of each substance HCl
0.100 x 25/1000 (i) M is the
concentration in mol dm-3 NaOH M x
20/1000 (ii) 4. Look at the molar
relationship and insert (i) and (ii) moles
of NaOH moles of HCl M x
20/1000 0.100 x 25/1000 5. Cancel
the 1000s M x 20
0.100 x 25 re-arrange the numbers to
obtain M M 0.100 x 25
20 6. Calculate the concentration of
the NaOH 0.125 mol dm-3
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VOLUMETRIC CALCULATIONS
Care must be taken when dealing with reactions
that do not have a 11 molar ratio. If you dont
understand what an equation tells you, it is easy
to make a mistake. 2NaOH H2SO4 gt
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
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VOLUMETRIC CALCULATIONS
Care must be taken when dealing with reactions
that do not have a 11 molar ratio. If you dont
understand what an equation tells you, it is easy
to make a mistake. 2NaOH H2SO4 gt
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 gt 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
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VOLUMETRIC CALCULATIONS
Care must be taken when dealing with reactions
that do not have a 11 molar ratio. If you dont
understand what an equation tells you, it is easy
to make a mistake. MnO4 8H
5Fe2 gt 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
54
VOLUMETRIC CALCULATIONS
Care must be taken when dealing with reactions
that do not have a 11 molar ratio. If you dont
understand what an equation tells you, it is easy
to make a mistake. MnO4 8H
5Fe2 gt 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 gt 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
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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 gt 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 (ii) where V is the volume of
alkali in cm3
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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 gt 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 (ii) where V is the volume of
alkali in cm3 substitute numbers moles
of NaOH 2 x moles of H2SO4 0.100
x V/1000 2 x 0.120 x
20/1000 cancel the 1000s 0.100 x V
2 x 0.120 x 20 re-arrange
Volume of NaOH (V) 2 x 0.120 x
20 48.00 cm3 0.100
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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.  
62
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) 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
63
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
0.168 mol dm3
0.136 mol dm3
0.118 mol dm3
1.0 mol dm3
64
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)
0.12 mol
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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
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• A5 is the LIMEWATER
• C5 is the solution in the Volumetric Flask
• Its the diluted HYDROCHLORIC ACID (B5)

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All values in cm3 rough 1st 2nd 3rd 4th
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 cm3 cm3 cm3 cm3 cm3
Mean value to 1 decimal place Concordancy is
0.10 cm3 Mean value ( that the mean of
concordant values )
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• 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) 2H2O(l)
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• 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.
So whats the concentration of this dilute HCl
solution ? Molarity moles / volume 0.007
/0.250 ( thats 250/1000 ) 0.028 moles /dm3
3.50 cm3 HCl 2.00 Molar Moles 3.50/1000 x
2.00 or 0.0035 x 2.00 0.007 moles
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• 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 2H2O(l)
Moles of Ca(OH)2 0.00112 / 2
0.00056moles in 25 cm3 of lime water
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• 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 Ca(OH)2 40.1 16.0 x 2 1.0 x
2 74.1 grms
mass of Ca(OH)2 74.1 x 0.0224 1.65984 1.66
grms
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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
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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 ? Moles ?
100.0 cm3 HNO3 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