Volume of NaOH needed to precipitate copper(II) but not calcium

I have $\ce{NaOH}: C = \pu{0.1M}, V\_{\ce{NaOH}}$ that I want to progressively add in a solution (in water) of:
$$\begin{align}
\ce{Cu^{2+}}&: \pu{0.01M}; & K\_\mathrm{sp} &= 10^{-18.6} \\
\ce{Ca^{2+}}&: \pu{0.01M}, & K\_\mathrm{sp} &= 10^{-5.26}
\end{align}$$
with initial volume: $V\_\mathrm{init} = \pu{0.1L}$. I want to add $\ce{NaOH}$ as long as $\ce{Cu(OH)2}$ precipitates, without causing any precipitation of $\ce{Ca(OH)2}$.


Knowing that the equilibrium will be displaced to the left and I will have precipitation, I have done: 
$$
\begin{array}{ccc}
\ce{Cu(OH)2 &->&Cu^2+ &+ &2OH^-}\\
&&0.001&&0.1V\_{\ce{NaOH}}\\
&&0.001-x&&0.1V\_{\ce{NaOH}} - 2x\\
\end{array}
$$


Also, since I want to achieve $99.999\%$ precipitation of $\ce{Cu(OH2)}$ I have got: $$x = 99.999\cdot10^{-5}$$


Using all that in the expression of $K\_\mathrm{sp}$ I get : 
$$10^{-18.6} = \left (\frac{0.001 - 99.999 \cdot 10^{-5}}{0.1+V\_{\ce{NaOH}}} \right)\cdot\left(\frac{0.1\cdot V\_{\ce{NaOH}} - 2\cdot99.999\cdot10^{-5}}{0.1 + V\_{\ce{NaOH}}}\right)^2$$ 


However for this expression I get a value of $V\_{\ce{NaOH}}$ that is extremely small. Is there anything wrong with my reasoning and if yes, how can I determine the volume for which i get the precipitation I want?




You need to precipitate $\pu{9.9999\times 10^-4 mol}$ of $\ce{Cu^{2+}}$, hence you need to add $2(\pu{9.9999\times 10^-4 mol}) = \pu{1.99998\times 10^-3 mol}$ of $\ce{NaOH}$, this is:


$\displaystyle{V\_\ce{NaOH} = \frac{n\_\ce{NaOH}}{c\_\ce{NaOH}} = \frac{\pu{1.99998\times 10^-3 mol}}{\pu{0.1 mol L-1}} = \pu{19.9998 mL} \approx \pu{20.0 mL}}$


There are $\pu{1\times 10^-8 mol}$ of $\ce{Cu^{2+}}$ left in solution, i.e. $\ce{[Cu^{2+}]} = \pu{8.33 \times 10^{-8} mol L^-1}$. Then, $\displaystyle{\ce{[OH-]} = \sqrt{\frac
{K\_\text{sp}}{\ce{[Cu^{2+}]}}} = \pu{1.74 \times 10^{-6} mol L-1}}$.


The final concentration of $\ce{Ca^{2+}}$ is $\pu{0.8333 mol L-1}$, so $Q\_\text{ps} = \ce{[Ca^{2+}][OH-]^2} = \pu{2.52 \times 10^{-12}} < K\_\text{sp}$, i.e. $\ce{Ca(OH)2}$ won't precipitate at this point. 


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