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!