# Solubility product when you add an acid to a solid?

I ran into some trouble when I was resolving some exercises about solubility product principle.The exercise says:

A concentrated, strong acid is added to a solid mixture of $0.015\ \mathrm{mol}$ samples of $\ce{Fe(OH)2}$ and $\ce{Cu(OH)2}$ placed in $1.0\ \mathrm{L}$ of water. At what values of pH will the dissolution of each hydroxide be complete?

So, I don't know how start when you have an acid and an insoluble salt when using the Ksp equation.

I know I can't to do this.

$\ce{Fe(OH)2 + HCl -> FeCl2 + H2O}$

$\ce{Cu(OH)2 + HCl -> CuCl2 + H2O}$

Not even do I have a common ion in order to establish a reaction. Do you know where I should begin? I tried using $K_\mathrm{sp}$, but I already have molar solubility, is $0.015\ \mathrm{M}$ for both. I can't ignore that. I don't find useful $K_\mathrm{sp}$ in this case. I know there's something that I'm not understanding. Hope you can help me to figure it out.

• Think of the problem this way. At what pH would the Fe(II) and Cu(II) solutions just start to ppt? At that balancing point if the pH is any higher the ppt will form. If the pH is lower then no ppt will form. – MaxW Nov 25 '15 at 4:39
• As @MaxW has stated, look up the solubility data for the different salts in water at different pH values. – Technetium Nov 25 '15 at 7:18
• Instead of using the data from the equation of formation of the different salts, you need to write the equation of both salt products dissolving in water then use the Ksp equation with that data. Also 1M of each of the hydroxides reacts with 2M Hcl to form the dichloride. – Technetium Nov 25 '15 at 7:27

The solution is simple. You have $\pu{0.015M}$ of $\ce{Me(OH)2}$ and according to the reaction you will need twice as much acid (based on stoichiometry): $\pu{0.030M}$. Knowing that $\pu{pH = - \log [H^+]}$, the dissolution will be complete at $\pu{pH = 1.52}$.