Solubility product when you add an acid to a solid?
Are not you complicating the problem?

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 $\pu{pH = 1.52}$.