Problem 120 - Chemistry The Science in Context

Question

Ozone, \(\mathrm{O}_{3},\) reacts with iodide ion \(\left(\mathrm{I}^{-}\right)\) in basic solution to form \(\mathrm{O}_{2}\) and \(\mathrm{I}_{2}\) by the unbalanced chemical equation: $$ \mathrm{O}_{3}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{O}_{2}(g)+\mathrm{I}_{2}(a q) $$ a. Balance the equation. b. A saturated solution of ozone in \(125 \mathrm{mL}\) of water at \(0^{\circ} \mathrm{C}\) is treated with \(10 \mathrm{mL}\) of \(2.0 \mathrm{MKI}\). After the reaction is complete, the solution is titrated with \(0.100 M \mathrm{H}^{+} .\) If \(54.7 \mathrm{mL}\) of acid is needed, what is the concentration of \(\mathrm{O}_{3}\) in a saturated solution?

Step-by-Step Solution

Answer

Based on the calculations made through these five steps, it is concluded that the concentration of ozone (\(\mathrm{O}_{3}\)) in the saturated solution is 0.0552 M.

Step 1: Balancing the chemical equation

To balance the equation, we need to make sure that the number of atoms of each element on both sides of the equation is equal. The balanced chemical equation is: $$ 2\mathrm{O}_{3}(a q)+10\mathrm{I}^{-}(a q) \rightarrow 3\mathrm{O}_{2}(g)+5\mathrm{I}_{2}(a q) $$

Step 2: Finding the initial moles of iodide ions

We are given a 10 mL solution of \(2.0 \mathrm M \mathrm{KI}\). To find the initial moles of iodide ions, use the formula: moles of \(\mathrm{I}^{-}\)= Volume in Liters \(\times\) Molarity $$ \text{moles of }\mathrm{I}^{-}=0.010 \text{ L} \times 2.0 \mathrm M = 0.020 \text{ moles} $$

Step 3: Finding the moles of iodine after the reaction

After the titration with \(0.100 \mathrm M\) \(\mathrm{H^{+}}\), 54.7 mL of the acid is needed. The balanced equation for iodine reacting with the acid is: $$ 2\mathrm{H}^{+}(a q) + \mathrm{I}_{2}(a q) \rightarrow 2\mathrm{HI}(a q) $$ Using the titration information, we can calculate moles of left over \(\mathrm{I}_{2}\) by: $$ \text{moles of } \mathrm{I}_{2} = \frac{1\text{ mol } \mathrm{I}_{2}}{2 \text{ mol } \mathrm{H^{+}}} \times 0.0547 \text{ L} \times 0.100 \mathrm M $$ $$ \text{moles of } \mathrm{I}_{2} = 0.002735 \text{ moles} $$

Step 4: Finding the moles of O3 in the reaction

Using the balanced chemical equation, we can find the moles of \(\mathrm{O}_{3}\) that reacted with the iodide ions to form iodine: $$ \text{moles of }\mathrm{O}_{3}\text{ reacted} = \frac{2\text{ mol } \mathrm{O}_{3}}{5\text{ mol } \mathrm{I}_{2}} \times (\text{Initial moles of }\mathrm{I}^{-} - \text{Leftover moles of}\mathrm{I}_{2} $$ $$ \text{moles of }\mathrm{O}_{3}\text{ reacted} = \frac{2}{5} \times (0.020 - 0.002735) = 0.006906 \text{ moles} $$

Step 5: Finding the concentration of O3 in a saturated solution

Now that we know the moles of \(\mathrm{O}_{3}\) reacted, we can find the concentration of the saturated solution using the total volume of the solution (125 mL) and the moles of \(\mathrm{O}_{3}\): $$ \text{Concentration of } \mathrm{O}_{3} = \frac{\text{moles of }\mathrm{O}_{3}}{\text{ volume in L}} $$ $$ \text{Concentration of } \mathrm{O}_{3} = \frac{0.006906}{0.125} \mathrm M = 0.0552 \mathrm M $$ The concentration of \(\mathrm{O}_{3}\) in the saturated solution is 0.0552 M.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry. When a chemical reaction occurs, the number of atoms for each element must remain the same before and after the reaction. This means you have to adjust the coefficients in front of each compound to maintain a balance between reactants and products. In the given reaction:

• From the unbalanced equation: \[ \mathrm{O}_{3}(a q)+\mathrm{I}^{-}(a q) \rightarrow \mathrm{O}_{2}(g)+\mathrm{I}_{2}(a q) \]
• Balance the oxygen by starting with the heaviest and most complex molecule, which is \( \mathrm{O}_{3} \) on the reactant side.
• You have 3 oxygen atoms on the reactant side and 2 on the product side. To balance, you need 3 \( \mathrm{O}_{2} \) molecules to give you 6 total oxygen atoms.
• To balance the iodide ions, 10 \( \mathrm{I}^- \) ions are needed to produce 5 molecules of \( \mathrm{I}_{2} \).

The final balanced equation is \[ 2\mathrm{O}_{3}(a q)+10\mathrm{I}^{-}(a q) \rightarrow 3\mathrm{O}_{2}(g)+5\mathrm{I}_{2}(a q) \]. Always ensure that both sides of the equation contain the same number of each type of atom to confirm that it is balanced.

Molarity Calculations

Molarity is a measure of the concentration of a solute in a solution, expressed as moles of solute per liter of solution. Calculating molarity requires understanding both the amount of solute and the volume of solution it is dissolved in. For the exercise, we needed to find how much iodide ion was initially present:

• Using the formula, \[\text{moles of }\mathrm{I}^{-}= \text{Volume in Liters} \times \text{Molarity}\]
• The initial moles of \( \mathrm{I}^{-} \) is computed using 10 mL (or 0.010 L) of a 2.0 M \( \mathrm{KI} \): thus the initial moles is 0.020 moles.

Understanding molarity calculations allows you to connect the volume and concentration of solutions to the quantity of solute, making it easier to predict the behavior of solutions during experiments.

Redox Reactions

Redox reactions involve the transfer of electrons between chemical species, resulting in a change in their oxidation states. This type of reaction plays a crucial role in various chemical processes, including the reaction between ozone and iodide ions.

• In our exercise, ozone acts as an oxidizing agent, gaining electrons, whereas iodide ions act as reducing agents, donating electrons.
• Recognizing the oxidizing and reducing agents is key to understanding the chemistry of the reaction.
• The balanced equation is: \[ 2\mathrm{O}_{3}(a q)+10\mathrm{I}^{-}(a q) \rightarrow 3\mathrm{O}_{2}(g)+5\mathrm{I}_{2}(a q) \]. This tells us that ozone reduces iodide ions to iodine molecules while itself getting reduced to oxygen gas.

In redox reactions, it's essential to balance both the charge and the atoms involved, ensuring that the electrons lost in oxidation equal those gained in reduction. Mastery of redox principles facilitates the analysis of reactions and the prediction of products.

Titration Analysis

Titration is an analytical technique used to determine the concentration of a solute within a solution. It involves adding a solution of known concentration to a solution of unknown concentration until a reaction is complete.In the exercise, a titration was used to analyze the leftover iodine:

• 54.7 mL of 0.100 M \(\mathrm{H}^{+}\) was required to completely react with the \(\mathrm{I}_{2}\) present.
• The reaction is: \[ 2\mathrm{H}^{+}(a q) + \mathrm{I}_{2}(a q) \rightarrow 2\mathrm{HI}(a q) \], demonstrating that each mole of \(\mathrm{I}_{2}\) requires 2 moles of \(\mathrm{H}^{+}\).

By calculating the moles of \(\mathrm{H}^{+}\), we could determine the moles of \(\mathrm{I}_{2}\) that reacted and consequently deduce how much \(\mathrm{O}_{3}\) was reacted initially. Titration provides precise quantitative analysis of unknown concentrations and is valuable in a variety of chemical investigations.