Question

In an iodometric estimation, the following reactions occur \(2 \mathrm{Cu}^{2+}+4 \mathrm{I}^{-} \longrightarrow \mathrm{Cu}_{2} \mathrm{I}_{2}+\mathrm{I}_{2} ; \mathrm{I}_{2}+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow 2 \mathrm{NaI}+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) \(0.12\) mole of \(\mathrm{CuSO}_{4}\) was added to excess of \(\mathrm{KI}\) solution and the liberated iodine required \(120 \mathrm{~mL}\) of hypo. The molarity of hypo solution was: (a) 2 (b) \(0.20\) (c) \(0.1\) (d) \(1.0\)

Short answer

2

Step-by-step solution

Understand the Chemical Reactions

The first reaction depicts copper(II) sulfate reacting with potassium iodide to form copper(I) iodide and iodine. The second reaction describes iodine reacting with sodium thiosulfate (hypo) to form sodium iodide and sodium tetrathionate. Both reactions indicate the stoichiometry of the substances involved which will be necessary for calculations.

Calculate Moles of Iodine Produced

From the stoichiometry of the first reaction, 1 mole of CuSO4 would produce 1 mole of I2. Since we have 0.12 moles of CuSO4, it will produce 0.12 moles of I2.

Calculate Moles of Hypo Solution Used

From the stoichiometry of the second reaction, 1 mole of I2 reacts with 2 moles of Na2S2O3. Hence, 0.12 moles of I2 would react with 0.24 moles of Na2S2O3.

Calculate Molarity of Hypo Solution

Molarity is defined as moles of solute divided by volume of solution in liters. The volume of the hypo solution is 120 mL, which is 0.12 liters. Using the moles of Na2S2O3 calculated, the molarity (M) is given by M = moles/volume = 0.24 moles / 0.12 liters.

Solve for the Molarity

M = 0.24 / 0.12 = 2 M. Therefore, the molarity of the hypo solution is 2.

Key concepts

Chemical Stoichiometry

Chemical stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the law of conservation of mass, meaning the mass of the reactants must equal the mass of the products. In stoichiometry, the balanced chemical equation provides the ratio in which substances react and form products.

For instance, looking at the reaction \(2 \mathrm{Cu}^{2+} + 4 \mathrm{I}^{-} \longrightarrow \mathrm{Cu}_{2} \mathrm{I}_{2} + \mathrm{I}_{2}\), the stoichiometric coefficients tell us that two copper ions react with four iodide ions to produce one molecule of copper(I) iodide and one molecule of iodine. Using this stoichiometric relationship, we can predict the amount of each substance involved given any starting quantity. In the exercise example, the stoichiometry is used to deduce how much iodine is produced from a known quantity of copper sulfate.

Molarity Calculation

The concept of molarity refers to the concentration of a solution and is expressed as the number of moles of solute (the substance being dissolved) per liter of solution. It is a crucial factor in many aspects of chemistry, especially when dealing with reactions in solution.

The formula to calculate molarity (M) is \( M = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \). In the textbook exercise, the molarity is found by first calculating the moles of iodine produced and then the moles of sodium thiosulfate (hypo) that reacts with this iodine. Knowing the volume of the sodium thiosulfate solution allows for the calculation of molarity. Remembering to convert volume from milliliters to liters is essential, as molarity is always expressed in moles per liter.

Redox Reaction

A redox reaction, short for reduction-oxidation reaction, involves the transfer of electrons from one substance to another. These reactions are composed of two half-reactions: oxidation, where a substance loses electrons, and reduction, where a substance gains electrons.

In the iodometric estimation from the exercise, the reaction \(2 \mathrm{Cu}^{2+} + 4 \mathrm{I}^{-} \longrightarrow \mathrm{Cu}_{2} \mathrm{I}_{2} + \mathrm{I}_{2}\) shows the iodide ion \(\mathrm{I}^{-}\) being oxidized to iodine \(\mathrm{I}_{2}\), while the copper ion \(\mathrm{Cu}^{2+}\) is reduced to copper(I) iodide \(\mathrm{Cu}_{2} \mathrm{I}_{2}\). Understanding the flow of electrons in such reactions is essential for predicting the outcome and calculating the molarity of substances involved, as seen in the subsequent reaction where iodine is further reacted with sodium thiosulfate.