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?
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
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