What concentration of Br − Br− results when 601 mL 601 mL of 0.559 M KBr...

Question

Question

What concentration of Br − Br− results when 601 mL 601 mL of 0.559 M KBr 0.559 M KBr is mixed with 687 mL 687 mL of 0.267 M FeBr 2 ? 0.267 M FeBr2?

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Answer 1

Answer #1

millimole of Br^— from KBr = Molarity x volume ( mL) = 0.559 x 601

millimole of Br^— from FeBr₂ = 2 x 0.267 x 687 ( since FeBr₂ gives 2 Br^— ion)

total millimole of Br^— =0.559 x 601 + 2x 0.267 x 687 = 702.817 millimole

total volume in mL = 687 + 601 =1288 mL

Molarity of Br^— = millimole / volume ( mL)

= 702.817 / 1288 = 0.545 M

kindly rate plz

Answer 2

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