Ask Question
1 May, 00:51

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.

+3
Answers (1)
  1. 1 May, 01:08
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers