If you had a 0.800 l solution containing 0.0240 m of fe3+(aq), and you wished to add enough 1.38 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.

0.041 L = 41.3 mL

Explanation:

This problem we will solve by considering the stoichiometry of the reaction and the definition of molarity.

Number of moles in .800 L solution:

0.800 L x 0.0240 M = 0.800 L x .0240 mol/L = 0.0192  mol Fe³⁺

to form the precipitate Fe(OH)₃ we will need 3 times .0192

mol NaOH required = 0.057

given the concentration of 1.38 mol M NaOH we can calculate how many milliliters of NaOH will contain 0.057 mol:

1.L/1.38 mol NaOH x 0.057 mol NaOH = 0.041 L

0.041 L x 1000 mL/1L = 41.3 mL