Part D Let us assume that Fe(OH)2 (s) is completely insoluble, which signifies t

Question

Part D Let us assume that Fe(OH)2 (s) is completely insoluble, which signifies that the precipitation reaction with NaOH(a) (presented in the transition) would go to completion Fe2+ (aq) + 2NaOH(aq) Fe(OH), (s) + 2Na+ (aq) If you had a 0.500 L solution containing 0.0140 M of Fe2+ (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. Express the volume to three significant figures and include the appropriate units Hints minimal volume of NaOH(aq) solutionValue Units

Explanation / Answer

From the reaction

1 mole of Fe+2 reacts with 2 moles of NaOH

Given data ;

Molarity of Fe+2 = 0.014 M

V = 0.5 L

MOlarity of NaOH = 1.27 M

Moles of Fe+2 = Molaity*Volume in L = 0.5*0.014 = 0.007

So

0.007 moles of Fe+2 reacts with 2*0.007 = 0.014 moles of NaOH

MOles of NaOH = 0.014

0.014 = 1.27*V

V = 0.011023 L

V = 11.023 ml

So answer is 11.023 ml of NaOH

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