Question

\(25 \mathrm{~mL}\) of household bleach solution was mixed with \(30 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{KI}\) and \(10 \mathrm{~mL}\) of \(4 \mathrm{~N}\) acetic acid. In the titration of the liberated iodine, \(48 \mathrm{~mL}\) of \(0.25 \mathrm{~N} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) was used to reach the end point. The molarity of the household bleach solution is (A) \(0.48 \mathrm{M}\) (B) \(0.96 \mathrm{M}\) (C) \(0.24 \mathrm{M}\) (D) \(0.024 \mathrm{M}\)

Short answer

The molarity of the household bleach solution is (C) 0.24 M.

Step-by-step solution

Understand the Reaction

Realize that when household bleach (which contains sodium hypochlorite, NaOCl) reacts with potassium iodide (KI), iodine (I2) is liberated. Then, the iodine reacts with sodium thiosulfate (Na2S2O3) in the titration:

Write the Balanced Equations

The overall reaction can be represented by two steps. First, NaOCl reacts with KI to produce I2:\[NaOCl + 2KI \rightarrow NaCl + I2 + K2O\]Second, I2 reacts with Na2S2O3 in the titration:\[I2 + 2Na2S2O3 \rightarrow 2NaI + Na2S4O6\]

Calculate Moles of Na2S2O3

Use the volume and normality of Na2S2O3 to calculate the moles of Na2S2O3 used in the titration:\[0.048 \text{ L} \times 0.25 \text{ N} = 0.012 \text{ mol} \text{ of Na2S2O3}\]

Determine Moles of I2

From the balanced equation of the reaction between I2 and Na2S2O3, 1 mole of I2 reacts with 2 moles of Na2S2O3. Therefore, using stoichiometry:\[0.012 \text{ mol Na2S2O3} \times \frac{1}{2} = 0.006 \text{ mol I2}\]

Determine Moles of NaOCl

From the balanced equation of the reaction between NaOCl and KI, each mole of NaOCl produces 1 mole of I2. Therefore, the moles of NaOCl are equal to the moles of I2 liberated:\[0.006 \text{ mol I2} = 0.006 \text{ mol NaOCl}\]

Calculate the Molarity of the NaOCl Solution

Use the volume of bleach solution and the moles of NaOCl to calculate its molarity:\[Molarity = \frac{moles}{volume}\]\[Molarity = \frac{0.006 \text{ mol}}{0.025 \text{ L}} = 0.24 \text{ M}\]

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships of reactants and products in a chemical reaction. Breathroughs in understanding about this topic allows chemists to predict the amounts of substances consumed and produced in a given reaction. In the context of our household bleach example, stoichiometry is used to understand the relationship between the reactants (sodium hypochlorite and potassium iodide) and the iodine produced.

By applying stoichiometric calculations, we were able to determine the number of moles of iodine produced, and from there, infer the amount of the original bleach present in our sample. The key here is to always ensure your reaction equation is balanced before performing any calculations, as this ensures the conservation of mass and provides accurate stoichiometric ratios.

• Identify the reactants and products.
• Write the balanced chemical equation.
• Use the balanced equation to set up mole ratios.
• Apply these ratios to find the amount of a reactant or product.

These steps make it possible to solve for quantities that were not immediately provided in the original problem.

Molarity Calculation

Molarity, often represented by the symbol M, is a measure of the concentration of a solute in a solution. It's defined as the number of moles of solute dissolved in one liter of solution. Molarity calculations are a staple in any chemistry practice because they enable chemists to quantify the concentration of substances.

In our exercise, we calculated the molarity of the household bleach solution. To perform a molarity calculation:

• Determine the number of moles of the solute.
• Measure the volume of the solution in liters.
• Divide the number of moles by the volume to obtain the molarity.

For instance, with the reaction between bleach and iodide, we used the titration results to find the moles of sodium hypochlorite and then divided this value by the volume of the bleach solution, yielding the molarity. This process is critical in experiments and industrial processes where specific solution concentrations are required.

Redox Reactions

Redox reactions are a type of chemical reaction where electrons are transferred between two species. This stands for reduction-oxidation reactions; reduction refers to a gain of electrons whereas oxidation refers to a loss of electrons. Redox reactions are fundamental to numerous processes in chemistry, including titrations involving oxidation and reduction.

Our exercise included a classic redox titration: iodine being reduced to iodide ions by sodium thiosulfate. Understanding redox reactions requires comprehension of oxidation states and recognition of the reducing and oxidizing agents.

• Identify which species are oxidized and reduced.
• Write half-reactions for both processes.
• Combine the half-reactions, ensuring that the number of electrons is balanced.

Mastery of redox reactions opens up avenues to solving complex titration problems like ours and highlights the inner workings of electron transfer in chemical equations.