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-?
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.

Key Concepts
Parts Per Billion (ppb)
Molar Mass of Sodium Arsenate
Dilution and Concentration Calculations
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.
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?
Potassium superoxide, KO2, is often used in oxygen masks (such as those used by firefighters) because KO2 reacts with CO2 to release molecular oxygen. Experiments indicate that 2 mol of KO2(s) react with each mole of CO2(g). (b) Indicate the oxidation number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced?
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?