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Chapter 4: Problem 113

The arsenic in a \(1.22-\) g sample of a pesticide was converted to \(\mathrm{AsO}_{4}^{3-}\) 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 As in AsO \(_{4}^{3-2}(\mathbf{b})\) 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?

Short Answer

Expert verified
The oxidation state of As in AsO4^(3-) is +5. The compound Ag3AsO4 can be named as silver arsenate by analogy to the corresponding compound containing phosphorus (silver phosphate). The mass percentage of arsenic in the pesticide is 5.22%.

Step by step solution

01

Determine oxidation state of As in AsO4^(3-)#

To determine the oxidation state of arsenic in AsO4^(3-), we can use the general rule that the sum of oxidation states of all the elements in a compound is equal to the charge of the compound. In AsO4^(3-), we have As and 4 oxygen atoms, so we can write: x + 4(-2) = -3 Where x represents the oxidation state of As and -2 is the known oxidation state of O in this compound. Solving for x, x = -3 + 8 x = +5 Thus, the oxidation state of As in AsO4^(3-) is +5.
02

Naming Ag3AsO4 by analogy to the corresponding compound containing phosphorus#

In order to find the name of Ag3AsO4 by analogy to the corresponding compound containing phosphorus, we must first identify the analogous phosphorus compound. The phosphorus compound will have the same structure but with arsenic (As) replaced by phosphorus (P). The phosphorus compound is Ag3PO4, which is known as silver phosphate. By analogy, Ag3AsO4 can be named as silver arsenate.
03

Calculate moles of Ag+ at the equivalence point#

From the given titration data, it takes 25.0 mL of 0.102 M Ag+ solution to reach the equivalence point. To compute the moles of Ag+ ions at the equivalence point, we can use the formula: n(Ag+) = M x V n(Ag+) = (0.102 mol/L) x (25.0 mL) x (1 L/1000 mL) = 0.00255 mol
04

Calculate moles of As in the pesticide sample#

Since 3 moles of Ag+ reacts with 1 mole of AsO4^(3-) to form Ag3AsO4, the mole ratio of Ag+:As is 3:1. We can now use the stoichiometry to compute the moles of As in the pesticide sample as: n(As) = n(Ag+) / 3 = 0.00255 mol / 3 = 0.00085 mol
05

Calculate mass percentage of arsenic in the pesticide#

Now that we have the moles of arsenic, we can calculate its mass in the pesticide sample: mass(As) = n(As) × molar mass(As) = 0.00085 mol × 74.92 g/mol = 0.06369 g To find the mass percentage of arsenic in the pesticide, we can use the formula: mass percentage = (mass(As) / mass of pesticide sample) × 100 % mass percentage = (0.06369 g / 1.22 g) × 100 % = 5.22 % Thus, the mass percentage of arsenic in the pesticide is 5.22%.

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Most popular questions from this chapter

Classify each of the following substances as a nonelectrolyte, weak electrolyte, or strong electrolyte in water: (a) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) , (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (ethanol), \((\mathbf{c}) \mathrm{NH}_{3},(\mathbf{d}) \mathrm{KClO}_{3}\), \((\mathbf{e}) \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\).

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