What is the concentration of bromide present in a 0.659 M. solution of CaBr2

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SUMMARY

The concentration of bromide in a 0.659 M solution of CaBr2 is determined by recognizing that each formula unit of CaBr2 produces two bromide ions. Therefore, the bromide concentration is 2 * 0.659 M, resulting in a bromide concentration of 1.318 M. This calculation is essential for understanding the stoichiometry involved in titration reactions, particularly in the context of determining the mass percentage of arsenic in a pesticide sample through titration with Ag+ ions.

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  • Understanding of molarity and concentration calculations
  • Knowledge of stoichiometry in chemical reactions
  • Familiarity with titration techniques and equivalence points
  • Basic principles of chemical analysis and mass percentage calculations
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  • Learn about stoichiometric relationships in chemical reactions
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Chemistry students, particularly those preparing for AP Chemistry, educators teaching analytical chemistry, and professionals involved in environmental testing and pesticide analysis.

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Homework Statement



The arsenic in a 1.42 g sample of pesticide was converted to AsO43- by suitable chemical treatment. It was then titrated using Ag+ to form Ag3AsO4 as a precipitate. If it took 33.7 mL of 0.113 M Ag+ to reach the equivalence point in this titration, what is the mass percentage of arsenic in the pesticide?

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im not sure

The Attempt at a Solution



I just need a way to get the ball rolling, this is a questions for my AP Chemistry summer review and i haven't taken chemistry since sophomore year (im going to be a senior when school starts in two days) and i forgot the equations i need to solve this kind of stuff.
 
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