Analytical Chemistry Q: calculating molarity

Question

Excess (NH4)2SO4= was added to a 55.0 mL solution containing BaCl2 (MW=208.23 g/mol). The resulting BaSO4BaSO4 (MW=233.43 g/mol) precipitate had a mass of 0.2319 g after it was filtered and dried. What is the molarity of BaCl2 in the solution?

Answer 1

To find the molarity of the BaCl₂, we need to know the moles of BaCl₂ and the volume of solution. Recall that molarity is defined as moles of solute per liter of solution.

Step 1: Find the moles of BaCl₂:

(NH₄)₂SO₄(aq) + BaCl₂(aq) ==> 2NH₄Cl(aq) + BaSO4(s) .. balanced precipitation reaction

From this reaction, you can see that 1 mole BaCl₂ produces 1 mole BaSO₄

moles BaSO₄ = 0.2319 g BaSO₄ x 1 mole / 233.43 g = 0.00099345 moles

moles BaCl₂ = 0.00099345

Step 2: Calculate molarity of BaCl₂

0.00099345 mols BaCl₂ / 55.0 ml x 1000 ml / L = 0.01806 moles / L = 0.0181 M (3 sig. figs.)