Question

If 500 mg of Fe³⁺ and 50 mg of OH^- are added to 1L of pure water, what will be the final equilibrium concentration of Fe³⁺ and OH^-. Assume water does not dissociate.

Fe(OH)₃ => Fe³⁺+3OH^-, pKs =38.57, MW_Fe = 55.85 g/mol, MW_OH = 17.01 g/mol

Answer 1

Fe3+ moles mass of Fe3+ in grams / Mol weight of Fe3+

           = ( 0.5 /55.85) = 0.0089526

Oh- mole = mass / Mola mass of OH-

                     = ( 0.05 / 17.01) = 0.0029395

OH- moles required per Fe3+ moles = ( 3) Fe3+ moles ( balanced equations shows 1Fe3+ reacts with 3OH-)

          = 3 x 0.0089526 = 0.02686

but we had only 0.0029395 OH- moles , hence OH- is limitig reagent

we have reaction Fe3+ (aq) + 3OH- (aq) <--> Fe(OH)3

Fe3+ moles reacted = ( 1/3) OH- moles = ( 1/3) 0.0029395 = 0.00098

Moles of Fe3+ moles left = initial Fe3+ moles - Fe3+ moles reacted

               = 0.0089526-0.00098 = 0.008

[Fe3+] = moles / voluem = 0.008/1 = 0.008

we have   pk = 38.57 , Ks = 10^ -38.57 = 2.69 x 10^ -39

Ks means dissociation of complex given by

Fe(OH)3 (s) <--> Fe3+ (aq) + 3OH- (aq) ,

Ks = [Fe3+] [OH-]^3                     ( solid Fe(OH03 not included in Ks equation)

2.69 x 10^ -39 = ( 0.008) [OH-]^3

[OH-] = 6.96 x 10^ -13 M

[Fe3+] = 0.008 M

Thus equilibrium concentratins are Fe3+ and OH- are obatined