What is the concentration of bromide ion if 25.0 ml of a 0.50 M AlBr3 solution combines with 40.0 ml of a 0.35 M NaBr solution?

Question

I converted AlBr3 and NaBr to moles however, I'm not so sure what to do after that. I know the final answer for the concentration of bromide ion is 0.79 mol/L. Can someone please help ?

AlBr3 = 0.0125 mol

NaBr = 0.014 mol

Answer 1

Total final volume = 25.0 ml + 40.0 ml = 65.0 ml

Final [AlBr₃]: (25.0 ml)(0.50 M) = (65.0 ml)(x M) and x = 0.192 M AlBr₃

Final [NaBr]: (40.0 ml)(0.35 M) = (65.0 ml)(x M) and x = 0.215 M NaBr

AlBr₃ ==> Al³⁺ + 3Br^-

NaBr ==> Na⁺ + Br^-

[Br^-] from AlBr₃ = 0.192 M x 3 = 0.576 M Br^-

[Br^-] from NaBr = 0.215 M x 1 = 0.215 M Br^-

Total [Br^-] = 0.576 M + 0.215 M = 0.791 M = 0.79 M (2 significant figures)

Answer 2

The key to answering this question is to be able to answer: how many moles of Br- will be in the solution if 0.0125 moles of AlBr3 dissociates (remember the 3 on the bromine), and how many moles of Br- will be in solution if 0.014 moles of NaBr dissociates?