Question

Determine the concentrations of BaBr2, Ba2 , and Br– in a solution prepared by dissolving 1.76 × 10–4 g BaBr2 in 2.50 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 = _______ ppm

[Br-]=________M = ________ppm

(Please show steps!)

Answer 1

no of moles of BaBr2   = 1.76*10^-4/297   = 5.92*10^-7 moles

molarity = W/G.M.Wt* volume of solution in L

              = 1.76*10^-4/297*2.5

               = 2.37*10^-7 M

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

2.37*10^-7 M             2,37*10^-7 M 2*2.37*10^-7

[BaBr2]   = 2.37*10^-7 M    

[Ba^2+]   = 2.37*10^-7 M      = 5.92*10^-7 moles * 137/2500   = 3.24*10^-8 * 10^6 = 0.0324ppm

[Br^-]      = 4.74*10^-7 M        = 5.92*10^-7 * 2*80/2500          = 3.78*10^-8 *10^6    = 0.0378ppm