Question

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.

Answer 1

Answer: (0.150 L) x (0.0240 M Fe{3+}) x (3 mol OH{-} / 1 mol Fe{3+}) / (1.39 M NaOH) = 0.00777 L = 7.77 mL NaOH