Determine the concentrations of babr2, ba2 , and br– in a solution prepared by dissolving 2.37 × 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).

Formulae:

Molarity, M = # moles of solute / Liters of solution
# moles = grams / molar mass

Reaction:



1) BaBr2

molar mass = 137.33 + 2*79.90 = 297.13 g/mol

# moles = grams / molar mass = 2.37*10^-4 g / 297.13 g/mol = 7.98*10^-7 moles

Liters of solution: in this case the volumen is that of the water because the volume of the solute is negligible.

M = 7.98 * 10^-7 mol / 1 L = 7.98 * 10^-7 M

2) Ba+2

The chemical reaction tells that the molar ratio of Ba+2 ions to BaBr2 is 1:1, then that same proportion remains for the molarity and you have

M =  7.98 * 10^-7 M

3) Br -

The chemical reaction shows that the molar ratio of Br- ions to BaBr2 is 2:1, then the Molarity will be also 2:1

M = 2*7.98 * 10^-7 M = 1.60*10^-6 M

ppm

To find ppm concetration of the ions first determine their weight divide by the mass of solution and multiply by 10^6

Ba+2 mass = # of moles * molar mass = 7.98*10^-7 mol * 137.33 g/mol = 0.00011 g

ppm = [0.00011g/1000 g of solution] * 1000000 = 0.11 ppm

(note that I didn't add the 0.00011 g to the mass of the solution because 0.00011 is negligible in front of 1000 g)

Br - mass = 2 * 7.98*10^-7 mol * 79.90 g/mol = 0.00013 g

ppm = 0.00013 * 1000000 / 1000 = 0.13 ppm

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