Question

What concentration of Br– results when 781 mL of 0.681 M KBr is mixed with 789 mL of 0.649 M FeBr2?

Answer 1

In this case, let's calculate the moles:

moles of HBr = 0.681 mol/L * 0.781 L = 0.531861 moles

moles of FeBr₂ = 0.649 * 0.789 = 0.512061 moles * 2 = 1.024122 moles

The extra 2 comes from the fact that every moles of FeBr2 releases two Br^- ions to the solution. Now the total moles are:

moles Br^- = 1.024122 + 0.531861 = 1.555983 moles

And finally the concentration with a total volume of (0.789+0.781 = 1.57 L):

[Br^-] = 1.555983 moles / 1.57 L = 0.9911 M

Hope this helps