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AS Chemistry titration

I have to plan an experiment to find the solubility of calcium hydroxide in water (in units of gdm^-3) at room temperature.

Information given by my teacher
- Calcium hydroxide is a base. It is sparingly soluble in water at room temperature.
- Calcium hydroxide reacts wih hydrochloric acid
- Ca(OH)2(aq) + 2HCl(aq) --> CaCl2(aq) + 2H2O(l)
- The maximum amount of calcium hydroxide needed to produce 1dm3 of saturated solution at room temperature is 1.5g.

BTW, this is homework and i'm going to show my working. I need to ensure my method is reasonable as I will be doing the experiment I've planned on Monday.

Method to make the Ca(OH)2 solution
Here is what i've done so far.
Mass of Ca(OH)2(s) = 1.5g
Molar mass of Ca(OH)2(s) = 74g
No of moles of Ca(OH)2(s) = mass/molar mass = 1.5/74

Molarity of Ca(OH)2(aq)
= moles of solute/volume of solution(dm3)
= (1.5/74)/1 = 1.5/74 moldm^-3

Firstly I would weigh out approximately 1 gram of solid Ca(OH)2 and spatula it into a 250cm3 volumetric flask about 3/4 full. I would put the lid on and shake the volumetric flask between 5-10 until no more solid dissolves (ie there will be a small amount of undissolved solute in the flask) this ensures that the solution is saturated. Once the solution is saturated I will mark up the solution to 250cm3 with more water. Then I will filter the solution to remove the excess solid. For the plan of the experiment, I will assume that the maximum amount of solid (1.5g) is required to make a solution of volume 1000cm3, so that I can calculate a suitable value for the HCl concentration for titrations. I will put a lid on the volumetric flask, to ensure that CO2 doesn't react with the aqueous solution.

Calculation of HCl concentration
No of moles of HCl (according to equation)
= 2*1.5/74 = 3/74 moles

I intend to use 25cm3 of alkali per titration (easy to extract using a 25cm3 pipette).

Therefore this volume will contain (25/1000)*(1.5/74) moles = 75/148000 moles

At equivalence point.
no of moles of alkali = no of moles of moles of acid = 2*no of moles of alkali

No of moles of acid at equivalence point = 2*75/148000 = 150/148000

No of moles of acid = (volume*concentration)/1000

I want approximately 20cm3 of acid to react with the 25cm3 alkali.
150/148000 = (20*C)/1000
C = (150/148000)*(1000/20) = 0.0507 moldm^-3

So I will use a rounded down value of 0.05 moldm^-3

Details of titration
1) Once the Ca(OH)2(aq) solution has been made, pour some of the solution into a 250cm3 beaker.
2)Clean the 25cm3 pipette with the Ca(OH)2(aq) solution. Use the 25cm3 pipette and pipette filler to extract 25cm3 of the solution until the solution reaches the mark on the apparatus. Put the solution into a 100cm3 conical flask. Add 4 drops of the methyl orange indicator. Swirl the solution to ensure the indicator is mixed with the solution. The solution will turn yellow with the indicator.
3) Clean the burette with the 0.1M HCl. Then add the 0.1M HCl to the 25cm3 burette. Place a funnel in the top to ensure none is spilled. Remove the funnel once the acid has been added.
4) Make a note of the initial reading on the burette. Read this value off the burette scale, ensuring eye-level is perpendicular to the vertical burette. Read the value just under the meniscus. Turn the burette tap on to allow the HCl to neutralise the Ca(OH)2(aq). Once the titration is near it's end-point (i.e. when the volume of acid is near 20cm3, turn the tap so that the acid comes out in drops. When the solution turns from yellow to orange, close the tap. Record the final burette reading in the table. This will be the rough titration.
5) Repeat this titration 5-10 times or until at least 2 titres are within 0.1cm3 of one another. Calculate the titre using the equation;
Titre = Final reading - Initial reading
6) Calculate the average titre in cm3 by using the titre values, ignoring the rough titre value.

Readings taken from my titrations
Table of results

A description of how you will use your results to calculate the solubility of calcium hydroxide in units of gdm^-3 at room temperature.

Okay, so we measure 5-10 titrations, calculate average titre. This will be equal to the volume of acid required to neutralise the alkali.

Okay, so volume of acid is known, concentration of acid is known.
volume of alkali, is known, concentration is effectively unknown.

Ca(OH)2(aq) + 2HCl(aq) --> CaCl2(aq) + 2H2O(l)

No of moles of HCl
= (volume*concentration)/1000
= (volume*0.05)/1000
= 0.00005V (where V is the average titre with units cm3)

No of moles of Ca(OH)2
= 0.00005V/2 = 0.000025V moles in 25cm3

Change moles of calcium hydroxide into mass (multiply by RMM)
= 0.000025V*74 grams in 25cm^3

You now have mass per 25cm3
Multiply by 40 to get mass per 1000cm3

Mass of Ca(OH)2 in gdm^-3
= 0.000025V*74*40 gdm^-3
= 0.074V gdm^-3

Alternative method
0.000025V = (concentration*volume)/1000
0.000025V = (C*25)/1000
C = (1000*0.000025V)/25 = 0.001V mol/dm3

To convert to g/moldm3
Convert 0.001V to grams
Mass = 0.002V*rfm of alkali = 0.001V*74 = 0.074V

Concentration (ie solubility) of Ca(OH)2
= 0.074V gdm^-3


I'm sure there must be an error in that. Sorry if it's a bit long. Can someone please go through it for me? I will rep you for this! Guaranteed!
Reply 1
Looks right to me ... remember to make clear that V is in cm^3 though (which is why you have put in the 1000 conversion factor).

Remember though that 1.5g is a maximum ... so it might be less required to form the saturated solution.

Rather than adding water to 1.5g as you suggest, I would dissolve the Ca(OH)2 slowly in 1dm3 of water until no more will dissolve (ie you have a small amount of undissolved solute in the flask) ... this ensures you actually have a saturated solution while still knowing how much is dissolved (you can weigh the rest of the Ca(OH)2 to work out how much you actually added).
oxymoron
Looks right to me ... remember to make clear that V is in cm^3 though (which is why you have put in the 1000 conversion factor).

Remember though that 1.5g is a maximum ... so it might be less required to form the saturated solution.

Rather than adding water to 1.5g as you suggest, I would dissolve the Ca(OH)2 slowly in 1dm3 of water until no more will dissolve (ie you have a small amount of undissolved solute in the flask) ... this ensures you actually have a saturated solution while still knowing how much is dissolved (you can weigh the rest of the Ca(OH)2 to work out how much you actually added).

What you have said sounds correct, but without the value for the mass of Ca(OH)2 I can't really do many calculations. Shall I state your method but just use the value of 1.5g (I could say, add 1.5g assuming the maximum amount of solid is required to form 1dm3 saturated solution?
too complicated...

1. Chuck loads (excess( of calcium hydroxide in any old amount of water in a big beaker and stir thoroughly until you are sure that no more will dissolve.

2. Filter the mixture into a conical flask

3. Extract 25cm3 using a graduated pipette and titrate this against 0.1M HCL using methyl orange as the indicator (repeat this step as many times as necessary to get concordant results)

4. Calculate:

Find moles calcium hydroxide from 2moles acid = 1 mole of base

change moles of calcium hydroxide into mass (multiply by RMM)

You now have mass per 25cm3

multiply by 40 to get mass per 1000cm3
charco
too complicated...

1. Chuck loads (excess( of calcium hydroxide in any old amount of water in a big beaker and stir thoroughly until you are sure that no more will dissolve.

2. Filter the mixture into a conical flask

3. Extract 25cm3 using a graduated pipette and titrate this against 0.1M HCL using methyl orange as the indicator (repeat this step as many times as necessary to get concordant results)

4. Calculate:

Find moles calcium hydroxide from 2moles acid = 1 mole of base

change moles of calcium hydroxide into mass (multiply by RMM)

You now have mass per 25cm3

multiply by 40 to get mass per 1000cm3

Your method doesn't really work for what i'm doing.
For example, my teacher says you have to do a calculation to show the concentration of hydrochloric acid you will need for a suitable titration value. You have just suggested one.
Step 1) I have to actually state the mass of alkali and volume of water I intend to use.
I'm going to go with my method, because my teacher was going along those lines in class.
Reply 5
charco
too complicated...

1. Chuck loads (excess( of calcium hydroxide in any old amount of water in a big beaker and stir thoroughly until you are sure that no more will dissolve.

2. Filter the mixture into a conical flask

3. Extract 25cm3 using a graduated pipette and titrate this against 0.1M HCL using methyl orange as the indicator (repeat this step as many times as necessary to get concordant results)

4. Calculate:

Find moles calcium hydroxide from 2moles acid = 1 mole of base

change moles of calcium hydroxide into mass (multiply by RMM)

You now have mass per 25cm3

multiply by 40 to get mass per 1000cm3



Could use this method but use the 1.5g you are given as a ROUGH value to calculate the conc of acid to use (as you did).

Could still say how much you will add ... add 1.5g of solid to 1dm3 if you like ... but make it clear that it will not all necessarily dissolve (the solution will be saturated though ... and it is your job to work out the concentration of this saturated solution).
Charco's method is better.

You can do a calculation to show a suitable HCl conc by estimating a reasonable titre and working backwards.
endeavour
Charco's method is better.

You can do a calculation to show a suitable HCl conc by estimating a reasonable titre and working backwards.

so what's wrong with my method?
Method to make the Ca(OH)2 solution
Here is what i've done so far.
Mass of Ca(OH)2(s) = 1.5g
No of moles of Ca(OH)2(s) = 1.5/74 gmol^-1

Molarity of Ca(OH)2(aq)
= moles of solute/volume of solution(dm3)
= (1.5/74)/1 = 1.5/74 moldm^-3

Firstly I would weigh out an excess of solid Ca(OH)2 and put it into a weighing bottle. I would note the mass of the weighing bottle and the solid (1). I would dissolve the Ca(OH)2(s) slowly in 1dm3 of water until no more will dissolve (ie there will be a small amount of undissolved solute in the flask) this ensures that the solution is saturated while still knowing how much is dissolved. Once the solution is saturated, I would reweigh the weighing bottle and the solid (2). The mass of solid used is; (1) - (2). For the plan of the experiment, I will assume that the maximum amount of solid (1.5g) is required to make the solution so that I can calculate a suitable value for the HCl concentration for titrations.



"Firstly I would weigh out an excess of solid Ca(OH)2 and put it into a weighing bottle. I would note the mass of the weighing bottle and the solid (1)."

why? you are making a saturated solution - it really doesn't matter how much anything weighs beforehand

you are told that a MAXIMUM of 1.5 g dissolves in a litre

Why do you need to make up a litre - are you going to do 40 titrations?

Your method overcomplicates things and wastes time. To perform this practical you need only 1g of calcium hydroxide and 250cm3 water. Making volumetric solutions wastes time that could be better spent gazing out of the window or removing portions of fluff from belly button.
The actual amounts you use are immaterial because that is what you are going to calculate from the results of the titration.

If you wish to calculate the molarity of the HCl then do it from the info given i.e. 1.5 g/dm3 being equivalent to 1.5/74 moles Ca(OH)2 - this needs twice that molar HCl approximately (although even this isn't true - it only needs to be thereabouts) so anything around about 0.05 molar should be OK. You'll have a titre of about 20cm3 acid which is fine in terms of errors.

Simplicity is the key to all good science...
Reply 9
Yes I agree with charco here - if that is the aim of the practical (ie to work out the solubility of Ca(OH)2) then that is the quickest/easiest method (and also most accurate).
I just assumed when I first read the method that you had been told you *had* to make up 1dm3 of solution for some reason. But if the only information you were given is what you have told us, then just making up any old volume of saturated solution (eg about 1g in about 250cm3) is fine ... just say you will make sure it is saturated by ensuring there is undissolved solid.

Another thought ... if you have to mention how you could extend this, you could say you would investigate how the solubility changes with temperature.
oxymoron
Yes I agree with charco here - if that is the aim of the practical (ie to work out the solubility of Ca(OH)2) then that is the quickest/easiest method (and also most accurate).
I just assumed when I first read the method that you had been told you *had* to make up 1dm3 of solution for some reason. But if the only information you were given is what you have told us, then just making up any old volume of saturated solution (eg about 1g in about 250cm3) is fine ... just say you will make sure it is saturated by ensuring there is undissolved solid.

Another thought ... if you have to mention how you could extend this, you could say you would investigate how the solubility changes with temperature.


I have a funny feeling that calcium hydroxide solubility changes with temperature differently to most substances - I think that it gets less soluble as you increase the temp or at least there is very little variation - I don't say this as a fact, I've just got a hunch that I've heard it somewhere :confused:
Reply 11
charco
I have a funny feeling that calcium hydroxide solubility changes with temperature differently to most substances - I think that it gets less soluble as you increase the temp or at least there is very little variation - I don't say this as a fact, I've just got a hunch that I've heard it somewhere :confused:


Yes Calcium hydroxide is more soluble in cold water than warm water (it is an exception to the rule which is why it would be interesting to investigate).

EDIT: Oxygen also behaves in this way which is why you see bubbles when you heat up water as the oxygen is forced to come out of solution and escape as a gas.
oxymoron
Yes Calcium hydroxide is more soluble in cold water than warm water (it is an exception to the rule which is why it would be interesting to investigate).

EDIT: Oxygen also behaves in this way which is why you see bubbles when you heat up water as the oxygen is forced to come out of solution and escape as a gas.


Yes, but whereas it makes sense for a gas to behave in this way as there is a considerable increase in entropy going from solution to gas state, this shouldn't be the case for a solid dissolving (and usually isn't)

Gibbs free energy change

delta G = delta H - T delta S

So if the change in entropy (S) is positive then Gibbs free energy gets more negative as the temperature rises - i.e the process of solution --> gas gets more likely.

But the reverse argument is also true for the solubility of a solid. There is a greater degree of disorder (entropy) in solution than in solid form. Therefore dissolving involves an increase in entropy. Increase in temperature increases solubility.

But this doesn't hold for calcium hydroxide, suggesting that as the temperature rises Gibbs free energy gets more positive. The enthalpy change of the dissolution process must be fairly constant as enthalpy changes very little with temperature (it's usually considered constant for energetics purposes)

So the only possibility is that the entropy of the solution is LESS than that of the solid!!!

I find this hard to swallow but can't see an alternative. Any ideas?

:confused:
Reply 13
charco
Yes, but whereas it makes sense for a gas to behave in this way as there is a considerable increase in entropy going from solution to gas state, this shouldn't be the case for a solid dissolving (and usually isn't)

Gibbs free energy change

delta G = delta H - T delta S

So if the change in entropy (S) is positive then Gibbs free energy gets more negative as the temperature rises - i.e the process of solution --> gas gets more likely.

But the reverse argument is also true for the solubility of a solid. There is a greater degree of disorder (entropy) in solution than in solid form. Therefore dissolving involves an increase in entropy. Increase in temperature increases solubility.

But this doesn't hold for calcium hydroxide, suggesting that as the temperature rises Gibbs free energy gets more positive. The enthalpy change of the dissolution process must be fairly constant as enthalpy changes very little with temperature (it's usually considered constant for energetics purposes)

So the only possibility is that the entropy of the solution is LESS than that of the solid!!!

I find this hard to swallow but can't see an alternative. Any ideas?

:confused:


I think it must have something to do with the entropy change of the water. When the Ca(OH)2 is dissolved in the water, water molecules become more ordered, attracted arround the ions to solvate them. This results in a decrease in entropy. This effect is stronger for smaller, more charged ions.

I realise this doesn't explain everything (especially comparing with other solids) but it's the only way I can come up with at the moment to explain it.
Problem is though that this would apply to other calcium salts as well - and I don't think that it does.

It may be caused by the OH ions or the Ca2+ ions

How does the solubility of say, calcium nitrate (incidentally one of my favourite chemicals - Bengal Saltpetre) vary with temperature?

Or another soluble hydroxide such as Potassium or Strontium?
Reply 15
Yes I know ... I don't know of the top of my head how the other solids vary with temperature, but I from what I have just read (in a couple of books and online), there is no way to predict the solubility of solids and how it will vary. It relies on a comlex balance between entropy change of the compound and the solvent as well as the enthalpy change (ie lattice enthalpy etc). These vary from compound to compound but in MOST cases result in solubility increasing with temperature (but not all - there are still a substantial minority that behave in the opposite way or not changing much at all. The only example I have found though is Calcium Hydroxide!)

I don't know - interesting though.