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.

Question

Rosa Pennington

Chemistry

7 September, 04:59

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.

+2

Answers (1)

Know the Answer?

Pranav Chen · Answer 1

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