Question

Determine the concentrations of BaBr2, Ba2 , and Br– in a solution prepared by dissolving 1.87 × 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).

Answer 1

Molarity = no of moles / volume in liters

no of moles = weight / molar mass

= 0.000187 / 297.14

= 6.3 x 10^-7 moles

M = 6.3 x 10^-7 moles / 2.5

M = 2.52 x 10^-7 M

one mole of BaBr2 can give one mole of Ba2+ + 2 Br-

we can have Ba2+ concentration 2.52 x 10^-7 M

Br- concentration = 2 x 2.52 x 10^-7 M

= 5.04 x 10^-7 M

1 ppm = 1 part "substance / 1 x 10⁶ parts solution

moles/L of Ba2+ = (2.52 x 10^-7mol x 137.33 g/mol)/L

moles/L of Ba2+ = 3.46 x 10^-5 grams/L

but we need to cnvert in to mg/L

3.46 x 10^-5 x 1000 milligrams/L

0.0346 mg/L or 0.0346 ppm of Ba2+

similarly

= 5.04 x 10^-7 x 78.9

= 397.656 x 10^-7 grams/L

= 397.656 x 10^-7 x 1000 mg/L

= 0.0397 ppm of Br^-