Question

Aluminum can be analyze by the following rather complicated procedure. The procedure features 8-hydroxyquinoline, C₉H₇ON, which reacts with Al³⁺ to form a solid and reacts with Br₂to form Br₂C₉H₅ON. We will abbreviate the formula of 8-hydroxyquinoline as H₂L so we don't have to keep writing C₉H₇ON. Al³⁺ reacts with H₂L at pH 5 in reaction (1) to form a solid. The solid can be filtered and rinsed to give pure Al(H₂L)₃. In more acidic solution, the reaction goes in the opposite direction so Al(H₂L)₃ can be dissolved in warm hydrochloric acid in reaction (2). This gives a solution in which Al³⁺ and 8-hydroxyquinoline are in their stoichiometric ratio of 1:3.

(1)		Al3+ + 3 H2L → Al(HL)3(s) + 3 H+
(2)		Al(HL)3(s) + 3 H+ → Al3+ + 3 H2L

Potassium bromate is a primary standard. Bromate ion reacts with excess bromide ions to give bromine Br₂ in reaction (3). Bromine reacts with 8-hydroxyquinoline in reaction (4).

(3)		BrO3- + 5 Br- + 6 H+ → 3 Br2 + 3 H2O
(4)		Br2 + 2 H2L → Br2L + 2 H+ + 2 Br-

In this analysis, excess Br₂ is added to the H₂L from reaction (2) and then unreacted Br₂ is back titrated. The back titration is accomplished with the following steps. An excess of KI is added and the leftover bromine reacts with iodide ion to form triiodide ion in reaction (5). Then triiiodide is titrated with thiosulfate in reaction (6).

(5)		Br2 + 3 I- → I3- + 2 Br-
(6)		I3- + 2 S2O32- → 3 I- + S4O62-

A sample solution containing aluminum is mixed with excess 8-hydroxyquinoline (reaction 1). The precipitate is filtered, rinsed, and quantitatively transferred into another beaker with hydrochloric acid and then heated to dissolve the solid (reaction 2). 0.3363 g of KBrO₃ (FM 167.01) is added to a 100 mL volumetric flask, dissolved and diluted to volume; 25.00 mL of this solution is added to the beaker along with an excess of potassium bromide (reactions 3 and 4). 1 g of potassium iodide is added (reaction 5) and the I₃^- that forms is titrated with 13.57 mL of 0.04490 M sodium thiosulfate (reaction 6) to a starch endpoint. How many mg of Al were in the original sample solution?
mg

Answer 1

To solve this, we need to start from back. In the last step, the I3- that forms is titrated with 13.57 mL of 0.04490 M sodium thiosulfate (reaction 6) to a starch endpoint.

Volume of sodium thiosulfate = 13.57 ml = 0.01357 L

Concentration of sodium thiosulfate = 0.04490 M

Moles of sodium thiosulfate = volume*concentration = 0.01357*0.04490 = 0.000609 mol

In reaction 6, 2 moles of sodium thiosulfate reacts with one mole of I3-.

So, 0.000609 moles of sodium thiosulfate reacts with I3- = 0.000609/2 = 0.000305 mol of I3-

This means that 0.000305 mol of I3- are formed in reaction 5.

In reaction 5, 1 mole of Br2 gives 1 mole of I3-.

Moles of Br2 used in reaction 5 is 0.000305 mol.

In reaction 3, KBrO3 reacts with KBr.

Mass of KBrO3 = 0.3363 g

Molecular mass of KBrO3 = 167.01 g/mol

Moles of KBrO3 = mass/molar mass = 0.3363/167.01 = 0.00201 mol

This is added to 100 ml volumetric flask.

25.00 ml of this is taken to react with KBr.

So, moles of KBrO3 taken = 0.00201*(25/100) = 0.000503 mol

In reaction 3, 1 mole of KBrO3 gives 3 moles of Br2.

So, 0.000503 moles of KBrO3 will give = 0.000503*3 = 0.001509 mol of Br2.

Moles of Br2 formed in reaction 3 = 0.001509 mol

Moles of Br2 used in reaction 5 = 0.000305 mol

So, moles of Br2 used in reaction 4 = 0.001509 - 0.000305 = 0.001204 mol

In reaction 4, 1 mol of Br2 reacts with 2 moles of H2L.

Moles of H2L reacted = 0.001204*2 = 0.002408 mol

Al3+ and 8-hydroxyquinoline are in their stoichiometric ratio of 1:3.

So, moles of Al3+ = 0.002408/3 = 0.000803 mol

Molar mass of Al = 27 g/mol

Mass of Al3+ used = moles*molar mass = 0.000803*27 = 0.0217 g = 21.7 mg

mass of Al3+ used = 21.7 mg