Question

A method used by the US. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the air sample through a "bubbler" containing sodium iodide, which removes the ozone according to the following equation: \(\mathrm{O}_{3}(g)+2 \mathrm{Nal}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) \(\mathrm{O}_{2}(g)+\mathrm{I}_{2}(\mathrm{~s})+2 \mathrm{NaOH}(a q)\) (a) How many moles of sodium iodide are needed to remove \(5.95 \times 10^{-6}\) mol of \(\mathrm{O}_{3}\) ? (b) How many grams of sodium iodide are needed to remove \(1.3 \mathrm{mg}\) of \(\mathrm{O}_{3}\) ?

Short answer

(a) \(1.19 \times 10^{-5}\) moles of sodium iodide are needed to remove \(5.95 \times 10^{-6}\) moles of O3. (b) 8.126 mg of sodium iodide are needed to remove 1.3 mg of O3.

Step-by-step solution

(1. Write down the balanced chemical equation)

The balanced chemical equation for the given reaction is as follows: \(O_3(g) + 2NaI(aq) + H_2O(l) \rightarrow O_2(g) + I_2(s) + 2NaOH(aq)\)

(2. Determine the mole ratio for O3 and NaI)

The balanced chemical equation tells us that 1 mole of ozone, O3, reacts with 2 moles of sodium iodide, NaI. Thus, the mole ratio between O3 and NaI is 1:2. (a) Using the mole ratio to determine moles of NaI needed for a given amount of O3:

(3. Use the given amount of O3 to find moles of NaI)

We are given that there are \(5.95 \times 10^{-6}\) moles of O3. Using the 1:2 mole ratio between O3 and NaI, we need twice as many moles of NaI to remove this amount of O3. Moles of NaI needed = \(2 \times (5.95 \times 10^{-6})\)

(4. Calculate moles of NaI required)

Moles of NaI needed = \(2 \times (5.95 \times 10^{-6}) = 1.19 \times 10^{-5}\) moles Therefore, \(1.19 \times 10^{-5}\) moles of sodium iodide are needed to remove \(5.95 \times 10^{-6}\) moles of O3. (b) Using the mole ratio and molar masses to determine the mass of NaI needed to remove a given mass of O3:

(5. Convert mass of O3 to moles)

We are given that there are 1.3 mg of ozone, O3. To convert this mass to moles, we need to use the molar mass of O3, which is approximately 48 g/mol. Moles of O3 = \(\frac{1.3 \times 10^{-3} \text{g}}{48 \text{g/mol}}\)

(6. Calculate moles of O3)

Moles of O3 = \(\frac{1.3 \times 10^{-3} \text{g}}{48 \text{g/mol}} = 2.708 \times 10^{-5}\) moles

(7. Use the given amount of O3 to find moles of NaI)

Using the 1:2 mole ratio between O3 and NaI, we need twice as many moles of NaI to remove this amount of O3. Moles of NaI needed = \(2 \times (2.708 \times 10^{-5})\)

(8. Calculate moles of NaI required)

Moles of NaI needed = \(2 \times (2.708 \times 10^{-5}) = 5.417 \times 10^{-5}\) moles

(9. Convert moles of NaI to mass in grams)

To convert moles of NaI to mass, we will use the molar mass of NaI, which is approximately 150 g/mol. Mass of NaI needed = \(5.417 \times 10^{-5}\) moles \(\times 150 \text{g/mol}\)

(10. Calculate mass of NaI required)

Mass of NaI needed = \(5.417 \times 10^{-5} \text{moles} \times 150 \text{g/mol} = 8.126 \times 10^{-3} \text{g}\) Therefore, 8.126 mg of sodium iodide are needed to remove 1.3 mg of O3.

Key concepts

Sodium Iodide Reaction

In the reaction between ozone and sodium iodide, ozone ( \(\text{O}_3\) ) reacts with sodium iodide ( \(\text{NaI}\) ) in the presence of water to produce oxygen ( \(\text{O}_2\) ), iodine ( \(\text{I}_2\) ), and sodium hydroxide ( \(\text{NaOH}\) ). This reaction is important because it allows for the determination of ozone concentration in air samples.
This is done by measuring the amount of reactants converted in the reaction.
Sodium iodide serves as a reagent that helps in the conversion of ozone to iodine and oxygen.

• Oxygen is released in gaseous form.
• Iodine is produced as a solid.
• Sodium hydroxide is formed as an aqueous solution.

Balanced Chemical Equation

A balanced chemical equation clearly represents the conservation of mass. Each side of the equation has the same number of atoms for each element involved in the reaction.
For the reaction \(\text{O}_3 (g) + 2 \text{NaI} (aq) + \text{H}_2\text{O} (l) \rightarrow \text{O}_2 (g) + \text{I}_2 (s) + 2 \text{NaOH} (aq)\),
it ensures that the ozone molecules convert according to the dictated stoichiometry.
When balancing equations, it is crucial because:

• The reactants and products must reflect a one-to-one balance.
• The equation confirms the stoichiometry, letting you know how much of a substance is produced or consumed.
• Helps in fundamentally understanding reactivity.

Mole Ratio

The mole ratio is an essential concept that helps in converting between reactants and products in a chemical reaction.
In the provided equation (\(\text{O}_3\) reacts with \(\text{NaI}\) in a 1:2 ratio)
for every mole of ozone, you will need two moles of sodium iodide. This ratio tells you how substances react in relation to each other.
The benefits of understanding mole ratios include:

• Allows for accurate calculations in a reaction.
• Aids in determining how much of a reactant is needed or how much product is formed.
• Helps in scaling reactions up or down without losing required proportions.

Molar Mass Conversion

Molar Mass Conversion is a method used to convert between the mass of a substance and the amount in moles. It makes use of the substance's molar mass, which is defined as the mass of one mole of a chemical element or compound.
To perform a conversion, use the formula: \(\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}\).
For sodium iodide, which has a molar mass of approximately 150 g/mol, knowing this allows you to convert mass in grams to moles accurately.

• This conversion is crucial in chemical calculations to ensure the exact quantitative reaction.
• Helps in determining the exact amount of reactants needed to produce desired product output.
• Is essential in laboratory settings for preparing accurate solutions and reactions.