Problem 113 - Chemistry The Central Science In Si Units

Question

The arsenic in a \(1.22-\mathrm{g}\) 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 \(\mathrm{AsO}_{4}^{3-} ?(\mathbf{b})\) 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 \(0.102 \mathrm{MAg}^{+}\) to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

Step-by-Step Solution

Answer

(a) +5; (b) Silver arsenate; (c) 5.22%.

Step 1: Determine the Oxidation State of As in \( \mathrm{AsO}_{4}^{3-} \)

The compound \( \mathrm{AsO}_{4}^{3-} \) contains arsenic and oxygen. The general rule is that oxygen has an oxidation state of \(-2\). Let \( x \) be the oxidation state of arsenic. The sum of oxidation states should equal the overall charge of the ion, which is \(-3\). Thus, we have the equation: \[ x + 4(-2) = -3 \] Simplifying this gives: \[ x - 8 = -3 \] Therefore, \( x = +5 \). Thus, the oxidation state of arsenic in \( \mathrm{AsO}_{4}^{3-} \) is +5.

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

By analogy to the corresponding phosphorus compound \( \mathrm{Ag}_{3} \mathrm{PO}_{4} \), which is named silver phosphate, the compound \( \mathrm{Ag}_{3} \mathrm{AsO}_{4} \) is named silver arsenate.

Step 3: Calculate Moles of \( \mathrm{Ag}^+ \) Used in the Titration

The concentration of \( \mathrm{Ag}^+ \) is \( 0.102 \mathrm{M} \) and the volume used is \( 25.0 \mathrm{~mL} \), which must be converted to liters: \( 25.0 \, \text{mL} = 0.0250 \, \text{L} \). The number of moles of \( \mathrm{Ag}^+ \) is calculated as follows: \[ \text{moles of } \mathrm{Ag}^+ = 0.102 \times 0.0250 = 0.00255 \text{ moles} \].

Step 4: Determine Moles of \( \mathrm{As} \) Present

From the balanced reaction \( \mathrm{3Ag}^+ + \mathrm{AsO}_4^{3-} \rightarrow \mathrm{Ag}_3\mathrm{AsO}_4 \), we see 3 moles of \( \mathrm{Ag}^+ \) react with 1 mole of \( \mathrm{AsO}_4^{3-} \), which means the moles of \( \mathrm{AsO}_4^{3-} \) are one-third of the moles of \( \mathrm{Ag}^+ \). Thus, \( \text{moles of } \mathrm{AsO}_4^{3-} = \frac{0.00255}{3} = 0.00085 \text{ moles} \). Since each \( \mathrm{AsO}_4^{3-} \) ion contains one arsenic atom, the moles of arsenic are the same.

Step 5: Calculate Mass of Arsenic

The molar mass of arsenic is \( 74.92 \, \text{g/mol} \). Thus, the mass of arsenic is given by: \[ \text{mass of } \mathrm{As} = 0.00085 \, \text{moles} \times 74.92 \, \text{g/mol} = 0.0637 \, \text{g} \].

Step 6: Determine Mass Percentage of Arsenic in the Pesticide

The mass percentage of arsenic in the sample is calculated using the formula: \[ \text{mass \, percentage} = \left(\frac{\text{mass of } \mathrm{As}}{\text{mass of sample}}\right) \times 100\% \]. Substituting the values, we get: \[ \text{mass \, percentage} = \left(\frac{0.0637}{1.22}\right) \times 100\% \approx 5.22\% \].

Key Concepts

Oxidation State Determination

Oxidation states help us understand how electrons are distributed in a compound. To determine the oxidation state of an element in a compound, we consider known oxidation states of other elements involved. In the compound \( \text{AsO}_4^{3-} \), we know that oxygen typically has an oxidation state of \(-2\). We set the oxidation state of arsenic as \( x \). The sum of the oxidation states must equal the charge of the ion, which is \(-3\). Therefore, we write the equation: \[ x + 4(-2) = -3 \] Solving this equation: \[ x - 8 = -3 \] Solving for \( x \), we find: \[ x = +5 \] Thus, the oxidation state of arsenic in \( \text{AsO}_4^{3-} \) is +5. This understanding is crucial for predicting the behavior of elements in redox reactions, which are fundamental in titration chemistry.

Precipitation Reactions

Precipitation reactions occur when two solutions react and form an insoluble solid, or precipitate. In this case, when \( \text{Ag}^+ \) ions are added to the solution containing \( \text{AsO}_4^{3-} \), they react to form \( \text{Ag}_3\text{AsO}_4 \), a solid precipitate. This occurs due to the low solubility of silver arsenate in water, making it fall out of solution.Precipitation reactions are pivotal in analytical chemistry. They allow us to remove ions from solution and sometimes weigh the precipitate to determine the quantity of the ion present. For instance, in a titration, when we reach the equivalence point, the amount of titrant added is chemically equivalent to the substance being titrated. In this case, the reaction of \( 3 \text{Ag}^+ \) ions with \( 1 \text{AsO}_4^{3-} \) forms \( \text{Ag}_3\text{AsO}_4 \) as a precipitate.

Stoichiometry

Stoichiometry is the science of measuring the quantities of reactants and products in chemical reactions. This concept ensures that we have the correct amounts of each substance required for a reaction. In the context of titration, stoichiometry helps determine how much of a titrant is needed.To find out how much silver ion is used, we convert the volume of \( \text{Ag}^+ \) solution from milliliters to liters and use the molarity to calculate moles:

• Volume of \( \text{Ag}^+ \) = 25.0 mL = 0.0250 L
• Molarity of \( \text{Ag}^+ \) = 0.102 M
• Moles of \( \text{Ag}^+ \) = 0.102 \times 0.0250 = 0.00255 \text{ moles}

The balanced chemical equation \( 3\text{Ag}^+ + \text{AsO}_4^{3-} \to \text{Ag}_3\text{AsO}_4 \) shows that 3 moles of silver react with 1 mole of arsenate. Thus, the moles of arsenate equals one-third of the moles of silver:\[ \text{moles of } \text{AsO}_4^{3-} = \frac{0.00255}{3} = 0.00085 \text{ moles} \]Since each \( \text{AsO}_4^{3-} \) contains one arsenic, we have the same number of moles of arsenic.

Mass Percentage Calculation

Calculating mass percentage involves determining what portion of a sample comprises a specific element or compound. To find the mass percentage of arsenic in the pesticide, we must first compute the mass of arsenic present. Using stoichiometry, we know the moles of arsenic from previous steps:

• Moles of arsenic = 0.00085 moles
• Molar mass of arsenic = 74.92 g/mol

Calculate the mass:\[ \text{mass of } \text{As} = 0.00085 \times 74.92 = 0.0637 \text{ g} \]With this mass, we determine the mass percentage in the original sample of 1.22 g:\[ \text{mass percentage} = \left(\frac{0.0637}{1.22}\right) \times 100\% \approx 5.22\% \]This result shows that arsenic comprises approximately 5.22% of the pesticide by mass. Such calculations are vital for quality control and ensuring product safety.