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:
$$\begin{align}
\ce{Cu^{2+}}&amp;: \pu{0.01M}; &amp; K_\mathrm{sp} &amp;= 10^{-18.6} \\
\ce{Ca^{2+}}&amp;: \pu{0.01M}, &amp; K_\mathrm{sp} &amp;= 10^{-5.26}
\end{align}$$
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: 
$$
\begin{array}{ccc}
\ce{Cu(OH)2 &amp;-&gt;&amp;Cu^2+ &amp;+ &amp;2OH^-}\\
&amp;&amp;0.001&amp;&amp;0.1V_{\ce{NaOH}}\\
&amp;&amp;0.001-x&amp;&amp;0.1V_{\ce{NaOH}} - 2x\\
\end{array}
$$

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 $Q_\text{ps} = \ce{[Ca^{2+}][OH-]^2} = \pu{2.52 \times 10^{-12}} &lt; K_\text{sp}$, i.e. $\ce{Ca(OH)2}$ won't precipitate at this point.