Question

A 20.00 mL unknown solution containing SO₄^2- was treated with Ba(NO₃)₂ to precipitate 0.2613 g of BaSO₄. What was the molarity of SO₄^2- in the unknown?

Answer 1

A 20.00 mL unknown solution containing SO₄^2- was treated with Ba(NO₃)₂ to precipitate 0.2613 g of BaSO₄. What was the molarity of SO₄^2- in the unknown?

First, define molarity of SO4-2

Molarity of SO4-2 = mol of SO4-2 / volume of solution in L

volume = 20 mL = 20*10^-3 liters

mol of SO4-2 = from stoichiometric ratio

Ba+2 + SO4-2 = BaSO4(s)

find moles of BaSO4(s)

1 mol of BaSO4 = 233.38 g/mol

x mol of BAOS4 = 0.2613 g of BaSO4

x = mass / MW = (0.2613)/(233.38) = 0.0011196 mol of BaSO4

now,

1 mol of SO4-2 is present per 1 mol of BASO4

so

0.0011196 mol of SO4- must be present in 0.0011196 mol of BaSO4

reclaculate Molarity

[SO4-2] = 0.0011196 / (20*10^-3) = 0.05598 M