(a) A standard iodine solution was standardized against a 0.4123 g primary stand

Question

(a) A standard iodine solution was standardized against a 0.4123 g primary standard As4O6 by dissolving the As4O6 in a small amount of NaOH solution, adjusting the pH to 8, and titrating, requiring 40.28 mL iodine solution. What is the concentration of the iodine solution?

As4O6 (s) + 6H2O -> 4H3AsO3

H3AsO3 +I3^- + H2O -> H3AsO4 + 3I^- + 2H^+

(b) The purity of a hydrazine (N2H4) sample is determined by titration with triiodide. A sample of the oily liquid weighting 1.4286 g is dissolved in water and diluted to 1 L in a volumetric flask. A 50.00 mL aliquot is taken with a pipette and titrated with the standard iodube solution in (a), requiring 42.41 mL. What is the percent purity by weight of the hydrazine?

N2H4 + 2I3^- -> N2 + 6I^- +4H^+

Explanation / Answer

Number of moles of As4O6 = Mass / Molecular mass

                                                = 0.4123 g / 395.68 g mol-1 = 0.001042 moles.

As4O6(s) + 6 H2O ---> 4 H3AsO3

4 H3AsO3 + 4 I3- + 4 H2O ---------> 4 H3AsO4 + 12 I- + 8 H+

According to stoichiometry of reaction, 1 mol As4O6 reacts with 4 moles I3- moles I- . So, number of moles of I3- must be equal to (4 x number of moles of As4O6) for the reaction to be complete.

So, number of moles of I3- = 4 x 0.001042 moles = 0.004168 moles.

Molarity of the solution, [I3-] = Number of moles / Vol. of solution in L   

                                    = 0.004168 moles / 0.04028 L = 0.103 M

Part B: 1 mol hydrazine completely reacts with 2 moles of I3-. So, number of moles of hydrazine must be equal to (2 x number of moles of I3-) for the reaction to be complete.

Using, M1V1 = M2V2     --- equation 1

Where, M1= molarity of solution 1, V1= volume of solution 1 (standard I3- solution)

            M2= molarity of solution 2, V2= volume of solution 2 (Unknown N2H4 solution.)

Or, 0.103 M x 42.41 mL = 2­ x M2 x 50.00 mL

Or, M2 = 0.087365

Thus, Molarity of N2H4 = 0.087365 M

So, the number of moles present in 1L solution = 0.087365 moles

Mass in 1L solution = moles x molecular mass

= 0.043682 x 32.046 g mol-1 = 1.399833372 g

% purity = (Actual titrimetric mass / initial mass of sample) x 100

            = (1.399833372 g / 1.4286 g) x 100 = 97.98 %

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