Question

Triiodide ions are generated in solution by the following (unbalanced) reaction in acidic solution: $$\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{I}_{3}^{-}(a q)$$ Triodide ion concentration is determined by titration with a sodium thiosulfate \(\left(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\right)\) solution. The products are iodide ion and tetrathionate ion \(\left(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\right)\) a. Balance the equation for the reaction of \(\mathrm{IO}_{3}^{-}\) with \(\mathrm{I}^{-}\) ions. b. A sample of 0.6013 g of potassium iodate was dissolved in water. Hydrochloric acid and solid potassium iodide were then added. What is the minimum mass of solid KI and the minimum volume of 3.00 M HCl required to convert all of the 1 \(\mathrm{O}_{3}^{-}\) ions to \(\mathrm{I}_{3}^{-}\) ions? c. Write and balance the equation for the reaction of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) with \(\mathrm{I}_{3}-\) in acidic solution. d. A 25.00 -mL sample of a 0.0100\(M\) solution of \(\mathrm{KIO}_{3}\) is reacted with an excess of \(\mathrm{KL}\) . It requires 32.04 \(\mathrm{mL}\) of \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution to titrate the \(\mathrm{I}_{3}^{-}\) ions present. What is the molarity of the \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}\) solution? e. How would you prepare 500.0 \(\mathrm{mL}\) of the KIO\(_{3}\)solution in part d using solid \(\mathrm{KIO}_{3} ?\)

Short answer

The balanced equation for the reaction of IO₃⁻ with I⁻ ions is: \( \mathrm{IO}_{3}^{-} + 2\mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-}\). The minimum mass of solid KI required is 1.202 g, and the minimum volume of 3.00 M HCl required is 2.41 mL. The balanced equation for the reaction of S₂O₃²⁻ with I₃⁻ in acidic solution is: \(2\mathrm{S}_{2} \mathrm{O}_{3}^{2-} + \mathrm{I}_{3}^{-} \longrightarrow 3\mathrm{I}^{-} + \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\). The molarity of the Na₂S₂O₃ solution is 0.0156 M. To prepare 500.0 mL of the KIO₃ solution, weigh 0.830 g of solid KIO₃ and dissolve it in water to make up a total volume of 500.0 mL.

Step-by-step solution

Part a: Balancing the equation

To balance the given equation, we need to ensure the same number of atoms of each element appear on both sides. The equation is: \[\mathrm{IO}_{3}^{-} + \mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-}\] There is 1 iodine atom on the left side in IO₃⁻ and 3 iodine atoms on the right side in I₃⁻. Therefore, we need 2 extra iodine atoms in the left side. We can achieve this by multiplying I⁻ by 2. The balanced equation is: \[ \mathrm{IO}_{3}^{-} + 2\mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-}\]

Part b: Mass of solid KI and volume of 3.00 M HCl required

To find the minimum mass of solid KI and the minimum volume of 3.00 M HCl required, we will first determine the moles of IO₃⁻ ions, and then convert it to moles of I⁻ ions. Finally, we will use the molar mass of KI to find the mass of KI and use the molarity of HCl to find the volume of HCl. 1. Moles of KIO₃: Given mass of KIO₃ = 0.6013 g Molar mass of KIO₃ = 39.10 (K)+ 126.9 (IO₃) = 166.0 g/mol Moles of KIO₃ = mass / molar mass = 0.6013 g / 166.0 g/mol = 3.62×10⁻³ mol 2. Moles of I⁻ ions: From the balanced equation in part a, 1 mol of IO₃⁻ reacts with 2 mol of I⁻ ions. Therefore, moles of I⁻ ions = 2 x moles of IO₃⁻ = 2 x 3.62×10⁻³ mol = 7.24×10⁻³ mol 3. Mass of KI: Molar mass of KI = 39.10 (K) + 126.9 (I) = 166.0 g/mol Mass of KI = moles of I⁻ ions x molar mass of KI = 7.24×10⁻³ mol x 166.0 g/mol = 1.202 g 4. Volume of 3.00 M HCl required: Molarity of HCl = 3.00 mol/L Since the moles of I⁻ ions are the same as KI, volume of HCl = moles of I⁻ ions / molarity of HCl = 7.24×10⁻³ mol / 3.00 mol/L = 2.41×10⁻³ L or 2.41 mL Therefore, the minimum mass of solid KI required is 1.202 g, and the minimum volume of 3.00 M HCl required is 2.41 mL.

Part c: Balancing the equation for the reaction of S₂O₃²⁻ with I₃⁻

The general equation for the reaction of S₂O₃²⁻ with I₃⁻ in an acidic solution is: \[\mathrm{S}_{2} \mathrm{O}_{3}^{2-} + \mathrm{I}_{3}^{-} \longrightarrow \mathrm{I}^{-} + \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\] To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides. Balancing the sulfur and oxygen is already done, so we only need to balance the iodine atoms. There are 3 iodine atoms in I₃⁻ and only 1 in I⁻, we need 2 additional iodine atoms on the right side. Thus, we can achieve balance by multiplying the I⁻ by 3: \[2\mathrm{S}_{2} \mathrm{O}_{3}^{2-} + \mathrm{I}_{3}^{-} \longrightarrow 3\mathrm{I}^{-} + \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\]

Part d: Molarity of Na₂S₂O₃ solution

Given information: Volume of KIO₃ solution = 25.00 mL Molarity of KIO₃ solution = 0.0100 M Volume of Na₂S₂O₃ solution = 32.04 mL 1. Moles of KIO₃: Moles of KIO₃ = Molarity x Volume = 0.0100 M x 0.025 L = 2.50 x 10⁻⁴ mol 2. Moles of I₃⁻ produced: From the balanced equation in part a, 1 mol of IO₃⁻ produces 1 mol of I₃⁻. Therefore, moles of I₃⁻ = 2.50 x 10⁻⁴ mol (same as KIO₃) 3. Moles of S₂O₃²⁻ reacted: From the balanced equation in part c, 1 mol of I₃⁻ reacts with 2 mol of S₂O₃²⁻ ions. Therefore, moles of S₂O₃²⁻ ions = 2 x moles of I₃⁻ ions = 2 x 2.50 x 10⁻⁴ mol = 5.00 x 10⁻⁴ mol 4. Molarity of Na₂S₂O₃ solution: Molarity = Moles / Volume = 5.00 x 10⁻⁴ mol / 0.03204 L = 0.0156 M Therefore, the molarity of the Na₂S₂O₃ solution is 0.0156 M.

Part e: Preparing 500.0 mL of KIO₃ solution

To prepare 500.0 mL of the KIO₃ solution, we will use the molarity and volume along with the molar mass of KIO₃. Given information: Volume of KIO₃ solution = 500.0 mL Molarity of KIO₃ solution = 0.0100 M 1. Moles of KIO₃: Moles of KIO₃ = Molarity x Volume = 0.0100 M x 0.500 L = 5.00 x 10⁻³ mol 2. Mass of KIO₃: Molar mass of KIO₃ = 166.0 g/mol Mass of KIO₃ = Moles x Molar mass = 5.00 x 10⁻³ mol x 166.0 g/mol = 0.830 g To prepare 500.0 mL of the KIO₃ solution, weigh 0.830 g of solid KIO₃ and dissolve it in water to make up a total volume of 500.0 mL.

Key concepts

Balancing Equations

Balancing chemical equations is like a puzzle where the pieces must fit perfectly. Each element in a chemical reaction should have the same number of atoms on both sides of the equation. This is crucial because atoms cannot be created or destroyed in a chemical reaction, according to the Law of Conservation of Mass.
In the provided exercise, the combination of iodate ions (\(\mathrm{IO}_{3}^{-}\)) with iodide ions (\(\mathrm{I}^{-}\)) forms triiodide ions (\(\mathrm{I}_{3}^{-}\)). Initially, we see one iodine atom on the left and three on the right. To balance it, add additional iodide ions, specifically two more, to the left. This adjustment ensures the number of iodine atoms is equal on both sides.

Balancing equations is an essential skill in chemistry as it helps in quantitative predictions in reactions, revealing how much of each substance will react and what quantities will form as products.

Stoichiometry

Stoichiometry connects the dots between balanced chemical equations and real-world chemical quantities, like masses and volumes. It allows the calculation of reactants and products in a chemical reaction. In the context of the exercise, knowing the molar mass and using stoichiometric coefficients, you can determine how much of each reactant is needed and the expected amount of products formed.

For instance, when solving the exercise, you first calculate the moles of potassium iodate (\(\mathrm{KIO}_3\)) using its mass and molar mass. Then, based on the balanced equation, you convert these moles to the required moles of iodide ions (\(\mathrm{I}^-\)). This step is pivotal as it helps in further calculations for mass and volume of other reagents needed.

• Calculate moles from known mass and molar mass.
• Use balanced equations to find mole ratios.
• Convert to desired unit (mass, volume, etc.).

Mastering stoichiometry unlocks the ability to scale reactions from the lab to industrial settings, ensuring efficient and economical use of resources.

Titration

Titration is an analytical technique used to determine the concentration of an unknown solution. By slowly adding a titrant of known concentration to the solution until the reaction reaches completion, you can accurately measure the unknown concentration.
In the provided exercise, a titration with sodium thiosulfate solution is used to determine the concentration of triiodide ions generated from the reaction. As sodium thiosulfate reacts with the triiodide ion, we monitor this process until all the triiodides have reacted. The volume of sodium thiosulfate used helps in calculating the concentration of the ions present.

This method is not just about exact measurement; it extends to a wide range of applications in chemistry, including quality control and substance purity checks. Titration is valued for its precision and reliability in determining concentrations of reactants in a chemical reaction.

Molarity Calculations

Molarity, expressed in moles per liter (M), is a common unit of concentration in chemistry. It tells us how many moles of solute are present in a given volume of solution. Calculating molarity is crucial when preparing solutions and conducting experiments, as seen in the exercise.
You can calculate molarity by dividing the number of moles of solute by the volume of the solution in liters. For example, if you have a measured volume of sodium thiosulfate solution used in titration, you could determine its molarity by dividing the moles of sodium thiosulfate used by the volume of solution in liters.

This simple yet powerful concept allows chemists to design and conduct experiments with precision, ensuring that reactions occur under controlled and known conditions. Understanding molarity is foundational for exploring more advanced topics in chemistry and for practical applications like solution preparation and titration.