Solubility of calcium hydroxide determined by titration

For a lab I had to determine the solubility and $K_{sp}$ of $\ce{Ca(OH)2}$ by performing two different titrations. I am given the formula to find solubility in molarity it is as written below.

Solubility of $\ce{Ca(OH)2}$ in $\mathrm{molarity = \frac{Moles \space of \space solute}{Volume \space of \space solution}}$

I know the moles of solute, which is $\ce{Ca(OH)2}$, from calculations: 0.0001664 moles. The only question I have is what volume do I use? I tried using the volume of $\ce{Ca(OH)2}$ used, $0.01~\mathrm{L}$, which gave a result of $0.01664~\mathrm{M}$. This closer to the literature value of $0.0216~\mathrm{M}$ than my second method. Second method I used the volume of the entire solution at the end of the titration, $0.03020~\mathrm{L}$, which resulted in a solubility of $0.005510~\mathrm{M}$. My question is if either of these methods are correct? If not, could you please point me in the right direction?

Thanks for helping!!

P.S. Sorry I'm not very good with formatting and computers and I don't know how to edit my equations to look pretty or very clear

Information:

Amount of $\ce{Ca(OH)2}$ added: 10.00 mL

Amount of $\ce{HCl}$ added: $20.20~\mathrm{mL}$, concentration of $0.01647~\mathrm{M}$

Steps:

1) Find amount of HCl added to $\ce{Ca(OH)2}$ and indicator.

Volume of $\ce{HCl}$ added = $20.20~\mathrm{mL}$, molarity of $\ce{HCl}$ = $0.01647~\mathrm{M}$, moles of $\ce{HCl}$ added= 0.0003327

*Since the competing equilibrium of the strong base, calcium hydroxide, and the strong acid, $\ce{HCl}$, cancel out the only equilibrium left is water. So the amount of $\ce{HCl}$ added will be proportional to the amount of $\ce{H3O+}$ and $\ce{OH-}$.

2) Find molarity of $\ce{OH-}$ and $\ce{Ca^{2+}}$ ions

Molarity of $\ce{OH-}$ = $\mathrm{\frac{0.0003327~moles}{0.01000~L} = 0.03327~M}$

Molarity of $\ce{Ca^{2+}}$ is equal to $0.5~\ce{OH-}$ because the chemical equation of the dissociation of $\ce{Ca(OH)2 <=> Ca^{2+} + 2OH-}$

3) Find Ksp. \begin{align*} K_{sp} &= [\ce{Ca^{2+}}][\ce{OH-}] \\ &=[0.01664][0.03327]^2 \\ &=1.842\cdot10^{-5} \end{align*} 4) Find solubility of $\ce{Ca(OH)2}$ in molarity

Solubility in Molarity= moles of solute/ volume of solution solubility=0.0001664/0.03020L solubility= 0.005510M OR solubility=0.0001664/0.01000L solubility= 0.01664M

• The volume used should be the total volume of the solution. Since its not matching the actual literature value, perhaps you could edit the question to include all your steps and calculation. Maybe did something wrong there. – Binary Geek Mar 19 '15 at 4:50

I just realized what I have done wrong!!! The solubility of the $Ca(OH)_2$ is proportional to the concentration of $Ca^{2+}$ and 1/2 of the $OH^-$ ion concentration and there is no need for me to divide by the volume of any solution. This is because the amount of ions present in the solution directly reflects how much the salt has dissociated or dissolved.