91Ó°ÊÓ

Primary-standard-grade arsenic(III) oxide \(\left(\mathrm{As}_{4} \mathrm{O}_{6}\right)\) is a useful (but carcinogenic) reagent for standardizing oxidants including \(\mathrm{MnO}_{4}^{-}\)and \(\mathrm{I}_{3}^{-}\). To standardize \(\mathrm{MnO}_{4}^{-}, \mathrm{As}_{4} \mathrm{O}_{6}\) is dissolved in base and then titrated with \(\mathrm{MnO}_{4}^{-}\)in acid. A small amount of iodide (I \(^{-}\)) or iodate \(\left(\mathrm{IO}_{3}^{-}\right)\)catalyzes the reaction between \(\mathrm{H}_{3} \mathrm{AsO}_{3}\) and \(\mathrm{MnO}_{4}^{-}\). $$ \begin{gathered} \mathrm{As}_{4} \mathrm{O}_{6}+8 \mathrm{OH}^{-} \rightleftharpoons 4 \mathrm{HAsO}_{3}^{2-}+2 \mathrm{H}_{2} \mathrm{O} \\ \mathrm{HAsO}_{3}^{2-}+2 \mathrm{H}^{+} \rightleftharpoons \mathrm{H}_{3} \mathrm{AsO}_{3} \\ 5 \mathrm{H}_{3} \mathrm{AsO}_{3}+2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+} \rightarrow 5 \mathrm{H}_{3} \mathrm{AsO}_{4}+2 \mathrm{Mn}^{2+}+3 \mathrm{H}_{2} \mathrm{O} \end{gathered} $$ (a) \(\mathrm{A}\) 3.214-g aliquot of \(\mathrm{KMnO}_{4}\) (FM 158.034) was dissolved in \(1.000 \mathrm{~L}\) of water, heated to cause any reactions with impurities to occur, cooled, and filtered. What is the theoretical molarity of this solution if no \(\mathrm{MnO}_{4}^{-}\)was consumed by impurities? (b) What mass of \(\mathrm{As}_{4} \mathrm{O}_{6}\) (FM \(395.68\) ) would be just sufficient to react with \(25.00 \mathrm{~mL}\) of the \(\mathrm{KMnO}_{4}\) solution in part (a)? (c) It was found that \(0.1468 \mathrm{~g}\) of \(\mathrm{As}_{4} \mathrm{O}_{6}\) required \(29.98 \mathrm{~mL}\) of \(\mathrm{KMnO}_{4}\) solution for the faint color of unreacted \(\mathrm{MnO}_{4}^{-}\)to appear. In a blank titration, \(0.03 \mathrm{~mL}\) of \(\mathrm{MnO}_{4}^{-}\)was required to produce enough color to be seen. Calculate the molarity of the permanganate solution.

Step by step solution

01

Calculate Molarity of KMnO4 (Part a)

The molarity (M) of \KMnO_4\ solution is determined by dividing the moles of solute by the volume of the solution in liters.1. Calculate moles of \KMnO_4\: \[\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.214 \, \text{g}}{158.034 \, \text{g/mol}} = 0.02034 \, \text{mol}\]2. Calculate Molarity:\[\text{Molarity} = \frac{\text{moles}}{\text{volume of solution in L}} = \frac{0.02034 \, \text{mol}}{1.000 \, \text{L}} = 0.02034 \, \text{M}\]

02

Determine Mass of As4O6 Needed (Part b)

To find the mass of \As_4O_6\ needed, we use the stoichiometry of the reaction with \KMnO_4\.1. Use the balanced equation to find moles of \As_4O_6\ required to react with \KMnO_4\:The ratios are given by:\[5\mathbf{H}_{3}\mathbf{AsO}_3 + 2\mathbf{MnO}_4^- \to 5\mathbf{H}_{3}\mathbf{AsO}_4 + 2\mathbf{Mn}^{2+}\]Each mole of \As_4O_6\ produces 4 moles of \HAsO_3^{2-}\.2. Calculate moles of \KMnO_4\ in 25.00 mL:\[\text{moles of } \mathbf{KMnO}_4 = 0.02034 \, \text{mol/L} \times 0.02500 \, \text{L} = 0.0005085 \, \text{mol}\]3. Use stoichiometry from balanced equations:\[\text{moles of } \mathbf{As}_4\mathbf{O}_6 = \frac{5}{2} \times 0.0005085 \, \text{mol} = 0.0012712 \, \text{mol} \]4. Convert moles to mass:\[\text{mass} = \text{moles} \times \text{molar mass} = 0.0012712 \times 395.68 \, \text{g/mol} = 0.5026 \, \text{g}\]

03

Account for Titration Results (Part c)

1. Correct for the blank titration:The blank titration consumes \0.03 \, \text{mL}\ of the solution, so we subtract this from the actual titration volume:\[\text{volume for reaction} = 29.98 \, \text{mL} - 0.03 \, \text{mL} = 29.95 \, \text{mL}\]2. Calculate moles of \As_4O_6\ used:\[\text{moles of } \mathbf{As}_4\mathbf{O}_6 = \frac{0.1468 \, \text{g}}{395.68 \, \text{g/mol}} = 0.000371 \, \text{mol}\]3. Based on the stoichiometry the moles of \KMnO_4\ react as:\[\text{moles of } \mathbf{KMnO}_4 = \frac{2}{5} \times 0.000371 = 0.0001484 \, \text{mol}\]4. Calculate the molarity of \KMnO_4 \:Convert \29.95 \, \text{mL}\ to liters:\[29.95 \, \text{mL} = 0.02995 \, \text{L}\]Now, find the molarity:\[\text{Molarity} = \frac{0.0001484 \, \text{mol}}{0.02995 \, \text{L}} = 0.004953 \, \text{M}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arsenic(III) Oxide

Arsenic(III) oxide, known chemically as \(\mathrm{As}_4\mathrm{O}_6\), is a white, toxic powder primarily used in chemical research and synthesis. One of its significant applications is as a primary standard in titrations involving oxidants like permanganate ions \((\mathrm{MnO}_4^-)\). It acts as a reducing agent that can help standardize oxidizing agents' concentration in solution.

In the context of permanganate titration, \(\mathrm{As}_4\mathrm{O}_6\) undergoes a redox reaction. This means it transfers electrons to the oxidant during the reaction, changing its oxidation state. The reaction leads to the formation of arsenic acid \((\mathrm{H}_3\mathrm{AsO}_4)\), highlighting its utility in redox chemistry. However, remember that arsenic compounds are carcinogenic, requiring proper handling and disposal methods to ensure safety.

When analyzing its usage, especially in educational labs, it's crucial to be aware of the stoichiometric complexities it brings into these reactions. The stoichiometry tells us how many moles of \(\mathrm{As}_4\mathrm{O}_6\) react with permanganate in what ratios, providing a clear view of the chemical equation's balancing requirements.

Molarity Calculation

Molarity (\(M\)) is a key concept in chemistry that measures the concentration of a solution. It is defined as the number of moles of solute per liter of solution. Calculating molarity is foundational when preparing solutions for titration, allowing for precise concentration determination.

To calculate molarity, you first need the mass of the solute and its molar mass. By dividing the mass of the solite by its molar mass, you find the number of moles. Following this, divide the moles by the volume of the solution in liters.

Here's a quick breakdown using the example of \(\mathrm{KMnO}_4\):

• Start with the given mass: 3.214 g of \(\mathrm{KMnO}_4\).
• Find moles using the formula: \(\text{moles} = \frac{3.214\, \text{g}}{158.034\, \text{g/mol}}\).
• Convert the solution's volume to liters if necessary, then calculate molarity: \(\text{Molarity} = \frac{\text{moles}}{\text{volume in L}}\).

This calculation is vital for preparing solutions with accurate concentrations, especially in quantitative analysis, which impacts reaction outcomes and data interpretation.

Redox Reaction Stoichiometry

Redox reactions are a cornerstone in chemistry, involving the transfer of electrons between molecules. Stoichiometry in these reactions refers to the quantitative relationships between reactants and products in a chemical equation.

During a redox titration involving \(\mathrm{As}_4\mathrm{O}_6\) and \(\mathrm{MnO}_4^-\), understanding the stoichiometry is crucial. You need to recognize how many moles of each reactant are required and how they relate to each other in the balanced equation.

Breaking down a typical reaction: the process begins with \(\mathrm{As}_4\mathrm{O}_6\) undergoing conversion to \(\mathrm{H}_3\mathrm{AsO}_3\), then the interaction with \(\mathrm{MnO}_4^-\) converts \(\mathrm{H}_3\mathrm{AsO}_3\) to \(\mathrm{H}_3\mathrm{AsO}_4\). The completion of this sequence depends on the accurate balance of reactants, which can be determined through stoichiometry.

In practical titration exercises, ensure calculations for the moles involved respect the stoichiometric coefficients of the balanced equation. It ensures not just completion of the reaction but also the accuracy of data collected from the experiment, forming a core component of analytical chemistry.