Trial 1
Volume of HCL is 2.51 mL
Molarity of the HCL is 0.10 M
Mols HCL = L x M
= 0.0251 L x 0.10 M
= 0.00251 mol/L
The equation for the titration is
Ca()H)2 + 2HCL --> CaCl2 + 2H2O
I will now take 0.5 mols HCl to give mols of OH-
0.00251 mol/L x 0.5 = 0.001255 mol/L
mols OH- / volume of the sample = OH- in mols/L
Don't I already have mols OH- in mols/L(0.001255 mol/L)?
Trial 1
Volume of HCL is 2.51 mL
Molarity of the HCL is 0.10 M
Mols HCL = L x M
= 0.0251 L x 0.10 M
= 0.00251 mol/L You made a math error here. 2.51 mL = 0.0025 L and that x 0.1 = 0.000251 mols.
The equation for the titration is
Ca()H)2 + 2HCL --> CaCl2 + 2H2O
I will now take 0.5 mols HCl to give mols of OH-
since you changed the post from the one earlier I can't read what I wrote. You can check it if you wish to see if I made an error. Regardless of what I wrote before, here is what you do. 0.000251 = mols HCl. There are two HCl mols per 1 mol Ca(OH)2; therefore, 1/2 x 0.000251 = mols Ca(OH)2 and that divided by 10 mL (0.010 L) = [Ca(OH)2] in that portion of the sample you titrated. I get 0.01255 M for [Ca(OH)2].
Since Ca(OH)2 ==> Ca^+2 + 2OH^-, then 0.01255 = (Ca^+2) and twice that is (OH^-). Then, substitute into the Ksp expression to solve for Ksp. I get about 7.9 x 10^-6 but you need to confirm that since I estimated. I checked a source on the internet and it had 5.5 x 10^-5 so you're in the ball park. Check my arithmetic. Check my thinking.
mols OH- / volume of the sample = OH- in mols/L
Don't I already have mols OH- in mols/L(0.001255 mol/L)?
Oh really sorry it was my own mistake:P
After I repeat this same technique for the next four trials, another question asks me to conduct a search to determine the actual value for the solubility product of calcium hydroxide(which you did, I really appreciate that). Compare your experimental value to this value by determining the percentage difference.
How to do this question when I have five different answers?
http://www.ktf-split.hr/periodni/en/abc/kpt.html
http://www.csudh.edu/oliver/chemdata/data-ksp.htm
http://en.wikipedia.org/wiki/Solubility_equilibrium
http://www.thelabrat.com/protocols/solubilityproductconstant.shtml
You can go to www.google.com and type in solubility product constant calcium hydroxide and get a number of hits. I copied and pasted four above but there are others. You can tell that there is not great agreement so you need to take care about where the information comes from. I suggest you look in a quant book. The latest quant book I have doesn't list it. I have a freshman college text copyrighted 1996 that gives 7.9 x 10^-6.
To find percentage difference do this:
[(exp value - accepted value)/accepted value]*100 = percent error.
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[(exp value - accepted value)/accepted value]*100 = percent error.
For exp value, do I have to average my five trials?