Problem 88 - Chemistry and Chemical Reactivity

Question

Fireworks contain \(\mathrm{KClO}_{3}\). To analyze a sample for the amount of \(\mathrm{KClO}_{3}\) a chemist first reacts the sample with excess iron(II), $$\begin{array}{r}\mathrm{ClO}_{3}^{-}(\mathrm{aq})+6 \mathrm{Fe}^{2+}(\mathrm{aq})+6 \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq}) \longrightarrow \\\\\mathrm{Cl}^{-}(\mathrm{aq})+9 \mathrm{H}_{2} \mathrm{O}(\ell)+6 \mathrm{Fe}^{3+}(\mathrm{aq})\end{array}$$ and then titrates the resulting solution with \(\mathrm{Ce}^{4+}\) [in the form of \(\left.\left(\mathrm{NH}_{4}\right)_{2}\mathrm{Ce}\left(\mathrm{NO}_{3}\right)_{6}\right]\) $$\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Ce}^{4+}(\mathrm{aq}) \longrightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Ce}^{3+}(\mathrm{aq}) $$ to determine the quantity of iron(II) that did not react with \(\mathrm{ClO}_{3}^{-}\). (This is referred to as a "back titration." Suppose a 0.1342-g sample of a firework was treated with 50.00 mL. of \(0.0960 \mathrm{M} \mathrm{Fe}^{2+}\) The unreacted \(\mathrm{Fe}^{2+}\) ions then required \(12.99 \mathrm{mL}\) of \(0.08362 \mathrm{M} \mathrm{Ce}^{4+} .\) What is the weight percent of \(\mathrm{KClO}_{3}\) in the original sample?

Step-by-Step Solution

Answer

The weight percent of KClO₃ in the original sample is 56.53%.

Step 1: Calculate Initial Fe²⁺ Moles

First, determine the initial number of moles of Fe²⁺ used in the reaction. This is done by multiplying the volume of Fe²⁺ solution by its molarity. The volume must be converted from milliliters to liters.\[\text{Initial moles of Fe}^{2+} = 50.00 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} \times 0.0960 \text{ M}\]\[= 0.0048 \text{ moles Fe}^{2+}\]

Step 2: Calculate Moles of Ce⁴⁺ Used in Titration

Next, find the number of moles of Ce⁴⁺ used in the back titration.\[\text{Moles of Ce}^{4+} = 12.99 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} \times 0.08362 \text{ M}\]\[= 0.001086858 \text{ moles Ce}^{4+}\]

Step 3: Calculate Unreacted Fe²⁺ Moles

The moles of unreacted Fe²⁺ are equal to the moles of Ce⁴⁺ used, because Ce⁴⁺ reacts with Fe²⁺ in a 1:1 ratio during the titration.\[\text{Unreacted moles of Fe}^{2+} = 0.001086858 \text{ moles}\]

Step 4: Calculate Reacted Fe²⁺ Moles

Subtract the moles of unreacted Fe²⁺ from the initial moles to find the moles of Fe²⁺ that reacted with ClO₃⁻.\[\text{Reacted moles of Fe}^{2+} = 0.0048 \text{ moles} - 0.001086858 \text{ moles}\]\[= 0.003713142 \text{ moles}\]

Step 5: Calculate Moles of ClO₃⁻

From the balanced equation, 6 moles of Fe²⁺ react with 1 mole of ClO₃⁻. Use this stoichiometry to find the moles of ClO₃⁻.\[\text{Moles of ClO}_3^- = \frac{0.003713142 \text{ moles Fe}^{2+}}{6}\]\[= 0.000618857 \text{ moles ClO}_3^-\]

Step 6: Calculate Mass of KClO₃

Calculate the mass of KClO₃ using its molar mass (122.55 g/mol for K, 35.45 g/mol for Cl, 48.00 g/mol for three O atoms; total molar mass of 122.55 + 35.45 + 48.00 = 122.55 g/mol). Multiply moles by the molar mass.\[\text{Mass of KClO}_3 = 0.000618857 \text{ moles} \times 122.55 \text{ g/mol}\]\[= 0.075878 \text{ g}\]

Step 7: Calculate Weight Percent of KClO₃

Find the weight percent of KClO₃ in the original sample by dividing the mass of KClO₃ by the mass of the sample and multiplying by 100.\[\text{Weight percent of KClO}_3 = \left(\frac{0.075878 \text{ g}}{0.1342 \text{ g}}\right) \times 100\]\[= 56.53\%\]

Key Concepts

Back Titration

Back titration is a type of analytical chemistry method used to determine the concentration of a substance that is difficult to analyze directly. It involves two stages. First, an excess of a standard solution (titrant) is added to the unknown sample. Then, the amount of unreacted titrant is quantified by a subsequent titration. This process is particularly useful when the reaction between the analyte and titrant is slow or when the endpoint is difficult to observe.

In the exercise example, excess iron(II) ions are initially added to the sample. After allowing it to react completely, the unreacted iron(II) is measured using a titration with cerium(IV) ions. This is a classic scenario where back titration is advantageous because it ensures the complete reaction of all available analyte without the need for exact initial calibration of the titrant added.

Back titration allows chemists to accurately determine analyte concentrations indirectly. This method is especially handy when direct titration is impractical.

Potassium Chlorate Analysis

Potassium chlorate (\(\mathrm{KClO}_{3}\)) is an important compound often found in fireworks and other pyrotechnics. Its analysis typically involves redox reactions, as it acts as a powerful oxidizing agent. Understanding the role of potassium chlorate in such mixtures is critical for both safety and quality control purposes.

In the given exercise, potassium chlorate is reduced by reacting it with iron(II) ions in an acidic solution. This process reduces the chlorate to chloride ions while oxidizing the iron(II) to iron(III). By subsequently determining how much iron(II) was converted, the amount of potassium chlorate in the sample can be back-calculated.

This specific method involves a clear understanding of redox chemistry, as every mole of potassium chlorate interacts with six moles of iron(II) ions. Knowing how to interpret these stoichiometric relationships is crucial for accurate calculations.

Such analysis not only ensures that fireworks perform as intended but also helps in regulating safety standards by verifying that the chemical compositions are within expected parameters.

Chemical Stoichiometry

Stoichiometry is the chemistry of measuring the quantitative relationships in chemical reactions. It is the backbone of understanding how substances interact in exact proportions. In the context of potassium chlorate analysis, stoichiometry helps to determine the number of moles involved in each step of the reaction and the titration process.

In the exercise at hand, stoichiometry allows the chemist to know exactly how many moles of iron(II) will react with one mole of chlorate ions, as given by the balanced chemical equation: \[\mathrm{ClO}_{3}^{-} + 6 \mathrm{Fe}^{2+} + 6 \mathrm{H}_{3} \mathrm{O}^{+} \rightarrow \mathrm{Cl}^{-} + 9 \mathrm{H}_{2} \mathrm{O} + 6 \mathrm{Fe}^{3+}\]

This stoichiometric relationship is used to reliably calculate the amount of potassium chlorate based on the reacted iron(II) ions. The calculations hinge on understanding and correctly applying these stoichiometric ratios to find the reaction's endpoint and determine the composition of the original mixture.

Proper understanding of stoichiometry is not only essential for this exercise but is pivotal for chemists in predicting the outcomes of reactions, ensuring reactants are used efficiently, and determining yields of products.