A solution of sodium thiosulfate was prepared by weighing out a certain mass of crystalline sodium thiosulfate (Na2S2O3·5H2O) on a clock glass, dissolving it in deionized water and making the solution up carefully to 500 cm³ in a volumetric flask - Leaving Cert Chemistry - Question 1 - 2007

To dissolve the crystalline thiosulfate, a clean clock glass was rinsed with deionized water to remove impurities. A controlled amount of sodium thiosulfate was weighed and then placed in a beaker. Deionized water was added to the beaker, and the mixture was stirred until all the solid dissolved completely. The solution was then poured through a funnel into a 500 cm³ volumetric flask. Deionized water was added to the flask until the bottom of the meniscus reached the 500 cm³ mark at eye level. This ensured that the solution was accurately prepared.

The solution of sodium thiosulfate is completely soluble in water; however, iodine can form a complex with thiosulfate ions. To bring iodine into solution, potassium iodide (KI) must be added, which will dissociate to release iodide ions (I⁻), allowing the iodine to remain in the aqueous phase.

A color change was observed in the conical flask because the freshly prepared starch acts as an indicator. Initially, the iodine solution is brown-yellow in color. As sodium thiosulfate is added, it reduces iodine to iodide ions, and the color fades. Near the endpoint, the addition of starch forms a blue-black complex with iodine, indicating the presence of free iodine in the solution, resulting in a dramatic color change.

To find the molarity (M) of the sodium thiosulfate solution, we can use the formula:

$M = \frac{n}{V}$

where $n$ is the number of moles of solute, and $V$ is the volume of solution in liters. Given that the average titre is 20.0 cm³ (0.020 L) and reacts with 0.05 M iodine:

1. Calculate moles of Iodine reacted: $n = M \times V = 0.05 \times 0.020 = 0.001$ moles of iodine.

2. According to the reaction stoichiometry, it takes 2 moles of thiosulfate for every mole of iodine: $n (S_2O_3^{2-}) = 2 \times 0.001 = 0.002$ moles of thiosulfate.

3. The solution volume is 0.500 L: $M (S_2O_3^{2-}) = \frac{0.002}{0.500} = 0.004$ M.

To find the concentration in grams per liter, we first calculate the molar mass of Na2S2O3·5H2O:

• Na: 2 x 23.0 = 46.0 g/mol
• S: 2 x 32.1 = 64.2 g/mol
• O: 3 x 16.0 = 48.0 g/mol
• H: 10 x 1.0 = 10.0 g/mol

Total = 46.0 + 64.2 + 48.0 + 10.0 = 168.2 g/mol.

Concentration in grams/L: $C = M \times Molar Mass = 0.004 \times 168.2 = 0.6728$ g/L.