Question

Determine the concentrations of BaBr2, Ba2+, and Br− in a solution prepared by dissolving 2.38 × 10−4 g BaBr2 in 1.00 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm).

[BaBr2]=    M

[Ba2+]=             M

[Ba2+]=             ppm

[Br−]=                 M

[Br−]= ppm

Answer 1

weight of BaBr2   = 2.38*10^-4g

gram molecular weight of BaBr2 = 297.13g/mole

molarity   = W*1000/G.M.Wt*volume in ml

                = 2.38*10^-4*1000/297.13*1

                = 0.0008M

BaBr2(aq) ---------------> Ba^2+ (aq) + 2Br^- (aq)

0.0008M                         0.0008M         2*0.0008M

[Ba^2+]    = 0.0008M

[Br^-]     = 2*0.0008M   = 0.0016M

mass of Ba^2+   = no of moles * gram atomic mass

                         = 0.0008mole*137.33g/mole

                         = 0.109864g

                        = 0.109864/1000

                         = 0.000109864Kg

                       = 0.000109864*10^6  

                         = 109.864ppm

no of moles of Br^-    = 0.0016moles

mass of Br^-     = no of moles * gram atomic mass

                      = 0.0016*80

                       = 0.128g

                        = 0.00128Kg

                        = 0.00128*10^6

                          = 1280ppm