Answer: 6.6 ml
Explanation:
If you had a 0.200 L solution containing 0.0140 M of Fe3+(aq), and you wished to add enough 1.27 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.
You have 0.014X0,2 = 0,0028 moles of Fe ion.
You need 0,0028 X 3 = 00084 moles of OH
The NaOH is 1.27 moles/liter
0.084/1.27 = 6.6 ml
