1.53 X 10-2 moldm-3 3.40 X 10-2 moldm-3 1.50 X 10-1 moldm-3
0.0153 moldm-3 0.0340 moldm-3 0.150 moldm-3
We need 2 moles of HCl to every one of X(OH)2, so we can work with concentration as:
0.0306 moldm-3 0.0680 moldm-3 0.300 moldm-3
0.3/10 = 0.03 0.3/5 = 0.06 0.3/1 = 0.3
Therefore concentration HCl = 0.3 moldm-3 because it is convenient.
GROUP I
LiOH NaOH KOH
5.2 moldm-3 10.5 moldm-3 17.1 moldm-3
5.2/0.3 = 17.3 10.5/0.3 = 35 17.1/0.3 = 57
too difficult to dilute by 17.3 etc, so:
5.2/20 = 0.26 10.5/40 = 0.26 17.1/50 = 0.34
Group II (diluting acid)
Group I (diluting alkali)
I have not included Mg in these calculations because its concentration is 0.0004 moldm-3. I think it is impractical to use it based on mathematics. I am discounting KOH from the experiment because 17 moldm-3 KOH is lethal. I am going to use the following hydroxides in my experiment:
Ca(OH)2 LiOH NaOH Ba(OH)2
METHOD
Dilution
Apparatus
50 cm3; 25 cm3 pipettes
250 cm3; 500 cm3; 1000 cm3 standard flasks
5.2 moldm-3 LiOH
10.5 moldm-3 NaOH
0.3 moldm-3 HCl
Distilled water
To dilute substance A for titration with substance B
- Wash the correct size pipette with water, then wash with a little of A and then fill to the line with A.
- Add the A to the relevant clean standard flask
- Using a clean measuring cylinder measure out the appropriate volume of distilled water and add to the standard flask.
Titration
To react together a solution of HCl and LiOH to find the concentration of the LiOH.
Apparatus
100 cm3 burette
0.3 moldm-3 HCl
LiOH
25 cm3 pipette
Clean Conical flask
Phenolphthalein (Indicator)
Clamp & Clamp stand
-
Wash the burette with water, then wash with a little of the 0.3 moldm-3 HCl that it will contain and finally fill it with the acid, remembering to fill the jet.
-
Wash a 25cm3 pipette with water, then wash with a little of the LiOH and then fill to the line with the LiOH, and transfer the LiOH to a clean conical flask.
- Add phenolphthalein to the LiOH (just enough to see the colour clearly)
- Set up the apparatus as shown:
- Add the HCl a little at a time, until the indicator just changes colour. Note the reading on the burette.
- Wash out the conical flask, top up the burette if necessary and repeat the experiment several times to check your results.
CALCULATING RESULTS
The equation for calculating concentration is:
Concentration (moldm-3) = Number of moles in solution
Volume (dm-3)
Before we can use this equation we must know four of the following:
- The volume of acid
- The volume of alkaline hydroxide
- The concentration of acid
- The balanced equation for the reaction showing the number of moles of the two reactants
- The concentration of alkaline hydroxide
- For the titration of LiOH with HCl.
Example Results
1) Volume HCl is 22cm3
2) Volume LiOH is 25cm3
3) Concentration of HCl is 0.3 moldm-3
4) The balanced reactants for the neutralisation reaction is
LiOH + HCl
1 mol 1 mol
LiOH and HCl react on a 1:1 ratio, therefore the number of moles of HCl used equals the number of moles of LiOH used.
Number of moles in solution of HCl
No moles in solution = Concentration (moldm-3) X Volume (dm3)
No moles in solution = 0.3 moldm-3 X 22 cm3
No moles in solution = 6.6
Number of moles in solution of LiOH
Equals the number of moles of HCl
No moles in solution = 6.6
Now that we know these four things we can calculate the fifth:
5) Concentration of LiOH
Concentration (moldm-3) = Number of moles in solution
Volume (dm-3)
Concentration (moldm-3) = 6.6
25dm-3
Concentration = 0.264 moldm-3
To obtain the true concentration this value must by multiplied out (in this instance by 20)
0.264 X 20 = 5.28
-
For the titration of Ca(OH)2 with HCl.
Example Results
1) Volume HCl is 50cm3
2) Volume Ca(OH)2 is 25cm3
3) Concentration of HCl is 0.03 moldm-3
4) The balanced reactants for the neutralisation reaction is
Ca(OH)2 + 2HCl
1 mol 2 mol
The Ca(OH)2 and HCl react on a 1:2 ratio, therefore the number of moles of Ca(OH)2 used is half the number of moles of HCl used.
Number of moles in solution of HCl
No moles in solution = Concentration (moldm-3) X Volume (dm3)
No moles in solution = 0.03 moldm-3 X 50 cm3
No moles in solution = 1.5
Number of moles in solution of Ca(OH)2
Equals half the number of moles of HCl
No moles in solution = 1.5
2
No moles in solution = 0.75
Now that we know these four things we can calculate the fifth:
5) Concentration of Ca(OH)2
Concentration (moldm-3) = Number of moles in solution
Volume (dm-3)
Concentration (moldm-3) = 0.75
25dm-3
Concentration = 0.03 moldm-3
RISK ASSESSMENT
- A Sensible amount of care must be taken when handling breakable apparatus.
- If cuts occur use standard first aid.
- Sodium hydroxide solution (greater than 0.5M) is corrosive and poisonous.
- The same can be applied to lithium hydroxide.
- These precautions must be taken for handling all alkali hydroxides in the experiment. They vary in risk posed, but unifying the care taken saves confusion and injury.
- Contact with eyes and skin must be avoided, so a lab coat and goggles must be worn at all times.
- If spillage occurs, goggles and gloves must be worn. Mop up with plenty of water.
- Standard first aid.
- Hydrochloric acid, bench, 2M, dilute
- It is an irritant, so to avoid contact with eyes and skin, goggles and a lab coat must be worn at all times.
- If spillage occurs mop up with plenty of water
- Standard first aid
- Phenolphthalein is harmful when solid and flammable in solution
- Do not expose to flames
- Wear a lab coat and goggles at all times
- If spillage occurs mop up with plenty of water
- Standard first aid
Bibliography
- Nuffield Advanced Science – Book Of Data
- Cambridge Advanced Sciences Chemistry 1
- The Usborne Illustrated Dictionary of Science
