Determine the concentrations

Question

Determine the concentrations of BaBr2, Ba^2+, and Br− in a solution prepared by dissolving 2.91 × 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).

I need help in calculating my ppm's :(

[BaBr2] = 4.89x 10^-7 M

[Ba^2+]= 4.89x10^-7 M [Ba^2+]=_____ ppm

[Br−]= 9.78x10^-7 M [Br−]=_____ ppm

Answer 1

I see you corrected the original value of x10^-4 g to 2.91x10^-4 g. Ok. No problem.

To get ppm, which is the same as mg/L, just convert moles/L to mg/L

For Ba²⁺:

4.89x10^-7 mol/L x 137 g Ba/mol x 1000 mg/g = 0.0670 mg/L = 0.0670 ppm Ba²⁺ (3 sig. figs.)

For Br-:

9.78x10^-7 mol/L x 79.9 g Br/mol x 1000 mg/g = 0.0781 mg/L = 0.0781 ppm Br- (3 sig. figs.)