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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 \(\mathrm{KIO}_{3} ?\)

Step by step solution

01

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 IO鈧冣伝 and 3 iodine atoms on the right side in I鈧冣伝. Therefore, we need 2 extra iodine atoms in the left side. We can achieve this by multiplying I鈦� by 2. The balanced equation is: \[ \mathrm{IO}_{3}^{-} + 2\mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-}\]

02

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 IO鈧冣伝 ions, and then convert it to moles of I鈦� 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 KIO鈧�: Given mass of KIO鈧� = 0.6013 g Molar mass of KIO鈧� = 39.10 (K)+ 126.9 (IO鈧�) = 166.0 g/mol Moles of KIO鈧� = mass / molar mass = 0.6013 g / 166.0 g/mol = 3.62脳10鈦宦� mol 2. Moles of I鈦� ions: From the balanced equation in part a, 1 mol of IO鈧冣伝 reacts with 2 mol of I鈦� ions. Therefore, moles of I鈦� ions = 2 x moles of IO鈧冣伝 = 2 x 3.62脳10鈦宦� mol = 7.24脳10鈦宦� 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鈦� ions x molar mass of KI = 7.24脳10鈦宦� 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鈦� ions are the same as KI, volume of HCl = moles of I鈦� ions / molarity of HCl = 7.24脳10鈦宦� mol / 3.00 mol/L = 2.41脳10鈦宦� 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.

03

Part c: Balancing the equation for the reaction of S鈧侽鈧兟测伝 with I鈧冣伝

The general equation for the reaction of S鈧侽鈧兟测伝 with I鈧冣伝 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 I鈧冣伝 and only 1 in I鈦�, we need 2 additional iodine atoms on the right side. Thus, we can achieve balance by multiplying the I鈦� by 3: \[2\mathrm{S}_{2} \mathrm{O}_{3}^{2-} + \mathrm{I}_{3}^{-} \longrightarrow 3\mathrm{I}^{-} + \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\]

04

Part d: Molarity of Na鈧係鈧侽鈧� solution

Given information: Volume of KIO鈧� solution = 25.00 mL Molarity of KIO鈧� solution = 0.0100 M Volume of Na鈧係鈧侽鈧� solution = 32.04 mL 1. Moles of KIO鈧�: Moles of KIO鈧� = Molarity x Volume = 0.0100 M x 0.025 L = 2.50 x 10鈦烩伌 mol 2. Moles of I鈧冣伝 produced: From the balanced equation in part a, 1 mol of IO鈧冣伝 produces 1 mol of I鈧冣伝. Therefore, moles of I鈧冣伝 = 2.50 x 10鈦烩伌 mol (same as KIO鈧�) 3. Moles of S鈧侽鈧兟测伝 reacted: From the balanced equation in part c, 1 mol of I鈧冣伝 reacts with 2 mol of S鈧侽鈧兟测伝 ions. Therefore, moles of S鈧侽鈧兟测伝 ions = 2 x moles of I鈧冣伝 ions = 2 x 2.50 x 10鈦烩伌 mol = 5.00 x 10鈦烩伌 mol 4. Molarity of Na鈧係鈧侽鈧� solution: Molarity = Moles / Volume = 5.00 x 10鈦烩伌 mol / 0.03204 L = 0.0156 M Therefore, the molarity of the Na鈧係鈧侽鈧� solution is 0.0156 M.

05

Part e: Preparing 500.0 mL of KIO鈧� solution

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.