Question

Standardization of Sodium Thiosulfate Titrant Weight of KIO3 = 1.5062 g; Concentration of KIO3 (M) = 0.0141 M

T1: Initial = 0.70 mL; Final = 29.60 mL; Volume of Titrant = 28.90 mL

T2: Initial = 2.10 mL; Final = 30.70 mL; Vol. of Titrant = 28.60 mL

T3: Initial = 2.20 mL; Final = 30.20 mL; Vol. of Titrant = 28.00 mL

Vitaminc C Analysis

Trial 1: Initial = 1.60 mL; Final = 15.70 mL; Vol. of Titrant = 14.10 mL

Trial 2: Initial = 2.10 mL; Final = 16.10 mL; Vol. of Titrant = 14.00 mL

Trial 3: Initial = 8.10 mL; Final = 22.40 mL; Vol of Titrant = 14.30 mL

Other info: Mass of Na2S2O3 = 6.5201 g; Mass of Na2CO3 = 0.02860g; Mass of KIO3 = 1.5062g

Calculate the total moles of triiodide added to the flask containing Vitamin C.

Answer 1

KIO₃ produces triiodide ion, I₃^- as per the reaction below.

IO₃^- (aq) + 8 I^- (aq) + 6 H⁺ (aq) -----------> 3 I₃^- (aq) + 3 H₂O (l) .........(1)

Triiodide, I₃^- reacts with ascorbic acid in Vitamin C as below.

C₆H₈O₆ (aq) + I₃^- (aq) ----------> C₆H₆O₆ (aq) + 3 I^- (aq) + 2 H⁺ (aq) ..........(2)

The excess triiodide reacts with thiosulfate ion, S₂O₃^2- as below.

2 S₂O₃^2- (aq) + I₃^- (aq) --------> S₄O₆^2- (aq) + 3 I^- (aq) .......(3)

Mass of KIO₃ taken = 1.5062 g

The atomic masses are

K: 39.098 g/mol

I: 126.904 g/mol

O: 15.999 g/mol

Molar mass of KIO₃ = (1*39.098 + 1*126.904 + 3*15.999) g/mol

= 213.999 g/mol

Mol(s) IO₃^- corresponding to 1.5062 g = (1.5062 g)/(213.999 g/mol)

= 0.007038 mol.

As per stoichiometric equation (1),

1 mol IO₃^- = 3 mols I₃^-.

Therefore,

Mol(s) I₃^- added to the flask = (0.007038 mol)*(3 mols)/(1 mol)

= 0.021114 mol

≈ 0.02111 mol (ans, correct to 4 sig figs).