Problem 114 - Chemistry: The Central Science

Question

The arsenic in a 1.22-g sample of a pesticide was converted to \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) by suitable chemical treatment. It was then titrated using \(\mathrm{Ag}^{+}\) to form \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) as a precipitate. (a) What is the oxidation state of \(\mathrm{As}\) in \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) ? (b) Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took \(25.0 \mathrm{~mL}\) of \(0.102 \mathrm{M} \mathrm{Ag}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

Step-by-Step Solution

Answer

The oxidation state of As in \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) is +5. The compound \(\mathrm{Ag}_{3}\mathrm{AsO}_{4}\) is named silver arsenate. The mass percentage of arsenic in the pesticide is 5.22%.

Step 1: Find the oxidation state of As in \(\mathrm{AsO}_{4}{ }^{3-}\)

Let's consider the arsenate ion, \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\). To find the oxidation state of As, we set up an equation based on the oxidation states of each atom in the ion: As + 4(O) = -3 Since the oxidation state of oxygen is typically -2, we can substitute this value into the equation: As + 4(-2) = -3 Now, solve for the oxidation state of As: As - 8 = -3 As = +5 The oxidation state of As in \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) is +5.

Step 2: Name \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\)

To name the compound \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\), we must identify the corresponding compound containing phosphorus in place of arsenic. The corresponding phosphorus compound is \(\mathrm{Ag}_{3} \mathrm{PO}_{4}\), which is called silver phosphate. Replacing the phosphorus with arsenic, we can name the compound as silver arsenate.

Step 3: Calculate the moles of \(\mathrm{Ag}^{+}\) used in the titration

We are given that \(25.0 \mathrm{~mL}\) of \(0.102 \mathrm{M} \mathrm{Ag}^{+}\) was used to titrate the \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) in the pesticide. We can convert the volume and concentration to moles by using the following equation: moles of \(\mathrm{Ag}^{+}\) = volume of \(\mathrm{Ag}^{+}\) × concentration of \(\mathrm{Ag}^{+}\) moles of \(\mathrm{Ag}^{+}\) = (25.0 mL × 0.102 mol/L) × (1 L / 1000 mL) moles of \(\mathrm{Ag}^{+}\) = 0.00255 mol

Step 4: Calculate the moles of \(\mathrm{AsO}_{4}{ }^{3-}\) in the pesticide

Since the reaction between \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) and \(\mathrm{Ag}^{+}\) forms \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) precipitate, the stoichiometry of the reaction can be determined from the balanced equation: 3 \(\mathrm{Ag}^{+}\) + \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) -> \(\mathrm{Ag}_{3} \mathrm{AsO}_{4}\) From the stoichiometry, we can see that 3 moles of \(\mathrm{Ag}^{+}\) reacts with 1 mole of \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\). So, we can find the moles of \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) in the pesticide: moles of \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) = moles of \(\mathrm{Ag}^{+}\) ÷ 3 moles of \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) = 0.00255 mol ÷ 3 moles of \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) = 0.00085 mol

Step 5: Obtain the mass of As in the pesticide

Now that we have the moles of \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) in the pesticide, we can obtain the mass of As in the pesticide by multiplying the moles of \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) by the molar mass of As: mass of As = moles of \(\mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\) × molar mass of As mass of As = 0.00085 mol × (74.92 g/mol) mass of As = 0.0637 g

Step 6: Calculate the mass percentage of As in the pesticide

We can find the mass percentage of As in the pesticide by dividing the mass of As by the mass of the pesticide and multiplying by 100: mass percentage of As = (mass of As / mass of the pesticide) × 100 mass percentage of As = (0.0637 g / 1.22 g) × 100 mass percentage of As = 5.22% The mass percentage of arsenic in the pesticide is 5.22%.

Key Concepts

Oxidation State

In chemistry, the oxidation state is a fundamental concept used to understand electron transfer and bonding in molecules. It represents an imaginary charge that an atom would have if all bonds between different atoms were 100% ionic. Calculating the oxidation state helps in determining how many electrons are lost or gained by an element in a compound. For arsenic in the arsenate ion - Consider the chemical formula: \( \mathrm{AsO}_{4}{\underline{\phantom{xx}}}^{3-}\). - Oxygen typically has an oxidation state of -2. - With four oxygen atoms, their total contribution is \(4 \times (-2) = -8\).- The entire compound has a charge of -3, implying:\[\mathrm{As} + (-8) = -3\]- Solving the equation gives us the oxidation state of arsenic as +5.

Titration

Titration is a common laboratory method used to determine the concentration of a substance in a solution. In this process, a solution of known concentration, called the titrant, is gradually added to a measure of the substance being analyzed, known as the analyte, until a reaction is completed. This point is known as the equivalence point.For arsenic determination:

• Arsenic ion is converted to arsenate ion \( \mathrm{AsO}_{4}^{3-}\).
• \( \mathrm{Ag}^{+} \) is used as the titrant to form the precipitate \( \mathrm{Ag}_{3}\mathrm{AsO}_{4}\).
• The volume of \(25.0~\mathrm{mL}\) of \(0.102~\mathrm{M}\) \(\mathrm{Ag}^{+}\) used helps calculate arsenic content.

Calculating the equivalence involves simple stoichiometry, ensuring the titration reaction correctly reflects the stoichiometric ratios in the balanced chemical equation.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is a key concept ensuring that the law of conservation of mass is satisfied—meaning atoms are neither created nor destroyed. The heart of stoichiometry is balanced equations from which we derive the mole ratios necessary in calculations.In this chemical reaction:- The balanced equation is: \[3~\mathrm{Ag}^{+} + \mathrm{AsO}_{4}^{3-} \rightarrow \mathrm{Ag}_{3}~\mathrm{AsO}_{4}\]- For every mole of \(\mathrm{AsO}_{4}^{3-}\), three moles of \(\mathrm{Ag}^{+}\) are required.- Knowing the moles of \(\mathrm{Ag}^{+}\) used (0.00255 mol), you can determine the moles of arsenate ion: \[\text{Moles of } \mathrm{AsO}_{4}^{3-} = \frac{0.00255}{3} \approx 0.00085~\text{mol}\]

Precipitation Reaction

A precipitation reaction involves the formation of a solid, known as a precipitate, when two solutions are combined. This type of reaction is an important aspect of chemistry, especially with respect to purification and analysis of compounds.In the exercise, the reaction between \(\mathrm{Ag}^{+}\) and \(\mathrm{AsO}_{4}^{3-}\) results in the formation of the insoluble compound \(\mathrm{Ag}_{3}~\mathrm{AsO}_{4}\), indicating it precipitates out of the solution.Characteristics of the precipitation reaction:

• Component ions combine to form a less soluble, solid compound.
• Agitates or forms a cloudy solution indicating solid formation.
• Important for determining the endpoint in titrations or when separating components in mixtures.

Understanding these characteristics helps when applying precipitation reactions in chemical analysis and purification techniques.