How would you prepare 1.000 L of 0.05000 M KBrO3

How would you prepare 1.000 L of 0.05000 M KBrO₃?

1. How would you prepare 2.5 L of approximately 0.06 MI₃^– solution? Calculate the molar concentration of KMnO₄ in this solution.

The Correct Answer and Explanation is :

To prepare 1.000 liter of a 0.05000 M potassium bromate (KBrO₃) solution, follow these steps:

1. Calculate the required mass of KBrO₃:

• Molarity (M): .05000 mol/L – Volume (V): .000 L – Molar mass of KBrO₃: 67.00 g/mol
  sing the formula:
  [ \text{Mass (g)} = \text{Molarity (mol/L)} \times \text{Volume (L)} \times \text{Molar Mass (g/mol)} ]
  [ \text{Mass} = 0.05000 \, \text{mol/L} \times 1.000 \, \text{L} \times 167.00 \, \text{g/mol} ]
  [ \text{Mass} = 8.350 \, \text{g} ]

1. Dissolve the calculated mass:

• eigh 8.350 g of pure KBrO₃ using an analytical balance. – ransfer the KBrO₃ into a 1-liter volumetric flask. – dd distilled or deionized water to the flask, filling it to about halfway. – wirl the flask gently to dissolve the KBrO₃ completely.

1. Dilute to the final volume:

• nce the KBrO₃ is fully dissolved, add more distilled water to the flask until the bottom of the meniscus reaches the 1.000-liter mark. – ap the flask and invert it several times to ensure thorough mixing.
  Preparation of 2.5 L of Approximately 0.06 M Triiodide (I₃⁻) Solution:

riiodide ions (I₃⁻) are typically prepared by mixing iodine (I₂) with iodide ions (I⁻) in solution, where the equilibrium is:
[ \text{I}_2 + \text{I}^- \leftrightarrow \text{I}_3^- ]
To prepare approximately 0.06 M I₃⁻ solution:

1. Calculate the moles of I₃⁻ needed:

• Molarity (M): .06 mol/L – Volume (V): .5 L
  [ \text{Moles of I}_3^- = 0.06 \, \text{mol/L} \times 2.5 \, \text{L} = 0.15 \, \text{mol} ]

1. Determine the required masses:

• Iodine (I₂): ach mole of I₃⁻ requires 1 mole of I₂. – Potassium iodide (KI): ach mole of I₃⁻ requires 1 mole of I⁻, provided by KI.
  Molar masses:
• ₂: 253.8 g/mol – I: 166.0 g/mol
  Masses required:
• ₂:
  [ 0.15 \, \text{mol} \times 253.8 \, \text{g/mol} = 38.07 \, \text{g} ]
• I:
  [ 0.15 \, \text{mol} \times 166.0 \, \text{g/mol} = 24.90 \, \text{g} ]

1. Dissolve the reagents:

• issolve 24.90 g of KI in approximately 1.5 L of distilled water in a 2.5-liter volumetric flask. – dd 38.07 g of I₂ to the KI solution. – tir the mixture until all the iodine dissolves, forming a deep brown solution due to the formation of I₃⁻ ions.

1. Dilute to the final volume:

• nce the iodine is fully dissolved, add distilled water to the flask until the solution reaches the 2.5-liter mark. – ix thoroughly to ensure uniformity.
  Calculating the Molar Concentration of KMnO₄ in the I₃⁻ Solution:

f potassium permanganate (KMnO₄) is used to oxidize iodide ions (I⁻) to triiodide (I₃⁻), the balanced redox reaction in acidic solution is:
[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{I}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} + 2\text{I}_3^- ]
From the stoichiometry:

• mole of KMnO₄ produces 2 moles of I₃⁻.
  o produce 0.15 moles of I₃⁻:
  [ \text{Moles of KMnO}_4 = \frac{0.15 \, \text{mol}}{2} = 0.075 \, \text{mol} ]
  he total volume of the solution is 2.5 L, so the molarity of KMnO₄ used would be:
  [ \text{Molarity of KMnO}_4 = \frac{0.075 \,