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 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: \[ \mathrm{IO}_{3}^{-} + 2\mathrm{I}^{-} \longrightarrow \mathrm{I}_{3}^{-}\]

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 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.

03

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: \[2\mathrm{S}_{2} \mathrm{O}_{3}^{2-} + \mathrm{I}_{3}^{-} \longrightarrow 3\mathrm{I}^{-} + \mathrm{S}_{4} \mathrm{O}_{6}^{2-}\]

04

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.

05

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.

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