Question

What concentration of Br– results when 757 mL of 0.767 M KBr is mixed with 741 mL of 0.481 M FeBr2?

Answer 1

Ans. Both KBr and FeBr2 dissociate into their respective ions when dissolved inn water.

I. KBr (s) ------> K⁺(aq) + Br^-(aq)

            Stoichiometry: 1 mol KBr gives 1 mol Br-

II. FeBr₂(s) ------> Fe²⁺(aq) + 2 Br^-(aq)

            Stoichiometry: 1 mol KBr gives 2 mol Br-

So,

Total moles of Br^- = Moles of KBr + (2 x moles of FeBr₂)

Now,

A. Moles of KBr = Molarity of KBr solution x Its volume (in Liters)

                                    = 0.767 M x 0.757 L                                                 ; [1 L = 1000 mL]

                                    = (0.767 mol/ L) x 0.757 L                                       ; [M = mol/ L]

                                    = 0.580619 mol

B. Moles of FeBr₂ = 2 x [ Molarity of FeBr₂ solution x Its volume (in Liters) ]

                                    = 2 x (0.481 M x 0.741 L)                                                    

                                    = 0.712842 mol

Thus,

total moles of Br^- = 0.580619 mol (from KBr) + 0.712842 mol (from FeBr2)

                                    = 1.293461 mol

Final volume of mixed solution = 757 mL (of KBr soln.) + 741 mL (of FeBr₂ soln.)

                                                = 1498 mL

                                                = 1.498 L

Final Br^- molarity in mixed solution =

Total moles of Br^- / Total volume (in L) of mixed solution

= 1.293461 mol / 1.498 L

= 0.863 mol/ L

= 0.863 M