Question

BACKGROUND: Synthesis of Potassium Iron (III) Oxalate Hydrate Salt

The iron(II) ions from Fe(NH₄)₂(SO₄)₂•6H₂O will be precipitated as iron(II) oxalate.

Fe^2+(aq) + C2O4^2-(aq) --> FeC2O4 (s)

The supernatant liquid, containing the ammonium and sulfate ions, as well as excess oxalate ions and oxalic acid will be decanted and discarded. The solid will then be re-dissolved and the iron(II) ions will be oxidized to iron(III) ions by reaction with hydrogen peroxide.

2Fe^2+(aq) + H2O2(aq) --> 2Fe^3+(aq) +2 OH^ - (aq)

The iron(III) ions will then be precipitated to form the hydrated potassium iron(III) oxalate complex salt (Green Salt). Since the resulting salt is soluble in water, ethanol is used to precipitate the Green Salt. It is less polar than water, so the Green Salt is not soluble in ethanol.

Analysis of Oxalate in the Complex Iron Salt

The Green Salt is treated with H₂SO₄(6 M) to release the oxalate ions from the salt as oxalic acid. The oxalate ions (in the form of oxalic acid) are then titrated with potassium permanganate (a strong oxidizing agent) to determine the concentration of the oxalate. (The endpoint will be detected when a faint pink color (MnO₄⁻ion) remains for at least 15 seconds.)

MnO4^- (aq) + C2O4^2-(aq) --> Mn^2+ (aq) + CO2 (g)

         purple                                colorless

The iron(III) ions in solution will be colored and interfere with the detection of the faint pink endpoint. To counteract this, concentrated phosphoric acid will be added to the mixture, so that colorless Fe(PO₄)₂³⁻complex ion will be formed instead of colored Fe(III) complexes.

Analysis of Oxalate in the Complex Salt Balance the redox reaction that occurred during the titration: Mn2+(aq) + CO2(g) MnO,-(aq) + С,042-(aq) (a1) O% Trial 2 T's Tri Blank "Rough Run"Trial 1 Trigl 3 Tria 4 (optional) Mass of Green Salt used Initial Buret Volume Final Buret Volume 4.3 13.e 13.23.2 m 4-5 m 13-6.mll 12.43mLI 12-43,儿| la 'fml 13.0mL | 0-10 w.e Net Corrected Volume Moles of C20 mol of C,O, g Green Salt Average mol of C20, /g Green Salt Suspected Outlier (Kuspet Standard Deviation (s,) Rejected or Accepted? (circle one) Gesp Genit Final Average mol of C,0,2 g Green Salt

* Moles of C2O4^2- (Trial 1 only) ?

* Moles of C2O4^2- / g Green Salt (Trial 1 only) ?

Answer 1

According to the balanced titration equation

2 MnO₄^- ( aq ) + 16 H⁺ (aq)+ 5 C₂O₄^2- ( aq )  $\rightarrow$ 2 Mn⁺² ( aq ) + 8H₂O (l) + 10 CO₂ (g)

2 moles of MnO₄^- completely reacts with 5 moles of C₂O₄^2-

Therefore one mole of MnO₄^- reacts with  $\frac{5}{2}$  moles of C₂O₄^2-

Therefore one mole of MnO₄^- reacts with 2.5 moles of C₂O₄^2-

number of moles of C₂O₄^2- = 2.5 $\times$ number of moles of  MnO₄^-

number of moles of MnO₄^- = ( molarity of KMnO₄ ) $\times$ ( volume of KMnO₄ used )

given molarity of KMnO₄ = 0.01976 M

volume of KMnO₄ used = Net corrected burette volume = 12.93 mL

Therefore number of moles of MnO₄^- = ( 0.01976 M) $\times$ ( 12.93 mL )

= 0.2555 milli moles

= 0.0002555 moles

Therefore number of moles C₂O₄^2- = 2.5 $\times$ number of moles of  MnO₄^-

= 2.5 $\times$ 0.0002555 moles

= 0.00063875 moles

number of moles C₂O₄^2- / of green salt =  $\frac{number of moles C_{2}O_{4}^{2-}}{mass of green salt}$

=  

0.00063875moles 0,1090q

= 0.00586 moles / g