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Ch.4 - Reactions in Aqueous Solution
All textbooksBrown 14th EditionCh.4 - Reactions in Aqueous SolutionProblem 113a
Chapter 4, Problem 113a

The arsenic in a 1.22-g sample of a pesticide was converted to AsO43- 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 AsO43-?

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1
Identify the chemical formula of the arsenate ion, AsO4^{3-}, and note that it contains arsenic (As) and oxygen (O).
Recall that the oxidation state of oxygen is typically -2 in most compounds.
Set up an equation to find the oxidation state of arsenic (As). Let the oxidation state of As be x. The formula for the arsenate ion is AsO4^{3-}, which means the overall charge is -3.
Write the equation for the sum of oxidation states: x + 4(-2) = -3, where x is the oxidation state of As and -2 is the oxidation state of each oxygen atom.
Solve the equation for x to determine the oxidation state of arsenic in AsO4^{3-}.

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Textbook Question

The arsenic in a 1.22-g sample of a pesticide was converted to AsO43- 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.

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Textbook Question

The arsenic in a 1.22-g sample of a pesticide was converted to AsO43- 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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Open Question
The U.S. standard for arsenate in drinking water requires that public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. If this arsenic is present as arsenate, AsO43-, what mass of sodium arsenate would be present in a 1.00-L sample of drinking water that just meets the standard? Parts per billion is defined on a mass basis as ppb = (g solute / g solution) × 109.