Determine the concentrations

Question

Determine the concentrations of BaBr2, Ba^2+, and Br− in a solution prepared by dissolving  × 10^−4 g BaBr2 in 2.00 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm).

[BaBr2] =_____ M

[Ba^2+]=_____ M [Ba^2+]=_____ ppm

[Br−]=_____ M [Br−]=_____ ppm

Answer 1

BaBr₂ ==> Ba²⁺ + 2Br^- (this is the dissolution process when placed in water)

molar mass BaBr2 = 297 g/mol

[BaBr₂] = 1x10^-4 g x 1 mol/297 / 2.00 L = 1.68x10^-7 M

[Ba²⁺] = 1.68x10^-7 M

[Br^-] = 2 x 1.68x10^-7 = 3.36x10^-7 M

To find ppm of each species:

[Ba²⁺]: 1.68x10^-7 mol/L x 137 g/mol x 1000 mg/g = 0.0231 mg/L = 0.0231 ppm

[Br^-]: 3.36x10^-7 mol/L x 79.9 g/mol x 1000 mg/g = 0.0268 mg/L = 0.0268 ppm