Let us assume that cu(oh)2(s) is completely insoluble, which signifies that the precipitation reaction with naoh(aq) (presented in the transition) would go to completion. cu2+(aq)+2naoh(aq) → cu(oh)2(s)+2na+(aq) if you had a 0.800 l solution containing 0.0190 m of cu2+(aq), and you wished to add enough 1.35 m naoh(aq) to precipitate all of the metal, what is the minimum amount of the naoh(aq) solution you would need to add? assume that the naoh(aq) solution is the only source of oh−(aq) for the precipitation.