The arsenic in a 1.22-g sample of a pesticide was converted to AsO₄^3- by suitable chemical treatment. It was then titrated using Ag⁺ to form Ag₃AsO₄ as a precipitate. (c) If it took 25.0 mL of 0.102 M Ag⁺to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

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Step 1: First, we need to determine the stoichiometry of the reaction. The balanced chemical equation is 3Ag+ + AsO4^3- -> Ag3AsO4. This tells us that 3 moles of Ag+ react with 1 mole of AsO4^3- to form 1 mole of Ag3AsO4.

Step 2: Next, we calculate the number of moles of Ag+ used in the titration. This can be done using the formula: moles = Molarity * Volume (in liters). In this case, the molarity of Ag+ is 0.102 M and the volume used is 25.0 mL or 0.025 L.

Step 3: Once we have the moles of Ag+, we can find the moles of AsO4^3- using the stoichiometry of the reaction. Since 3 moles of Ag+ react with 1 mole of AsO4^3-, the moles of AsO4^3- is one third of the moles of Ag+.

Step 4: Now, we can find the mass of arsenic in the sample. Since AsO4^3- contains one atom of arsenic, the moles of AsO4^3- is equal to the moles of arsenic. We can convert moles of arsenic to grams using the molar mass of arsenic (74.92 g/mol).

Step 5: Finally, we can calculate the mass percentage of arsenic in the pesticide. The mass percentage is calculated by dividing the mass of arsenic by the total mass of the sample and multiplying by 100.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Titration and Equivalence Point

Titration is a quantitative analytical method used to determine the concentration of a solute in a solution. The equivalence point is reached when the amount of titrant added is stoichiometrically equivalent to the amount of substance being analyzed. In this case, the Ag+ ions react with AsO43- ions to form Ag3AsO4, allowing for the calculation of arsenic content based on the volume and concentration of Ag+ used.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced chemical equations. It allows chemists to determine the relationships between the quantities of substances consumed and produced. In this problem, stoichiometry is essential to relate the moles of Ag+ used in the titration to the moles of arsenic present in the pesticide sample.

Mass Percentage

Mass percentage is a way of expressing the concentration of a component in a mixture as a percentage of the total mass. It is calculated by dividing the mass of the component by the total mass of the mixture and multiplying by 100. In this scenario, after determining the mass of arsenic from the titration data, the mass percentage can be calculated to understand how much arsenic is present in the pesticide sample relative to its total mass.

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Textbook Question

The arsenic in a 1.22-g sample of a pesticide was converted to AsO₄^3- by suitable chemical treatment. It was then titrated using Ag⁺ to form Ag₃AsO₄ as a precipitate. (b) Name Ag₃AsO₄ by analogy to the corresponding compound containing phosphorus in place of arsenic.

Textbook Question

Federal regulations set an upper limit of 50 parts per million (ppm) of NH₃ in the air in a work environment [that is, 50 molecules of NH₃(g) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing 1.00⨉10² mL of 0.0105 M HCl. The NH₃ reacts with HCl according to: NH₃₍aq) + HCl(aq) → NH₄Cl(aq). After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The remaining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. (a) How many grams of NH₃ were drawn into the acid solution?

Textbook Question

The arsenic in a 1.22-g sample of a pesticide was converted to AsO₄^3- by suitable chemical treatment. It was then titrated using Ag⁺ to form Ag₃AsO₄ as a precipitate. (a) What is the oxidation state of As in AsO₄^3-?