Question (b) If the total molar concentration of all species in the solution is

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(b) If the total molar concentration of all species in the solution is the same 2+ soj , and H20? 2. A solution contains 45 mg/L Ba2+. A consultant has suggested that the Ba+ could be removed from solution by converting it to solid barium sulfate BaSO4(s)]. To accomplish this, she proposes adding enough Na2SO4 to the solution to convert all the Ba+ to BaSO4(s), plus an extra 5 mg/L SO as a safety factor. What dose (mg/L) of Na2SO4 should be added, and what would the concentration of solids be, in ppm, if essentially all the Ba2+ precipitated?

Explanation / Answer

2 )

the conentration of Ba+2 = 45 mg / L

Atomic weight of Ba+2 = 137.3 g / mole

Moalrity = mass / molecular weight X volume

so concentration = 45 / 137.3 millimolar = 0.327 millimolar

the reaction of Ba+2 with Na2SO4 will be

Ba+2 + Na2SO4 --> BaSO4 + 2Na+

As per stoichiometry for each mole of Ba+2 to be precipitated we need to add one mole of Na2SO4.

so for 0.327 millimolar we need to add 0.327 millmoles of Na2SO4 which will give 1 mole of BaSO4

Molecular weight of Na2SO4 = 142

So 1 mole = 142 grams

0.327 millimoles = 0.327 X 142 milligrams = 46.43 mg of Na2SO4

for safety we are adding an additonal 5mg / L

so total amount of Na2SO4 = 51.43 g / L of Na2SO4

Molecular weight of BaSO4 = 233 grams

so if 0.327 millimoles of BaSO4 precipitate will be obtained, it means we will get 0.327 X 233 mg of BaSO4 / L

which = 76.19 mg / L = 76.19 ppm of BaSO4 will be obtained

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