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

Question

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?

Answer 1

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.