If 56.8 mL of BaCl2 solution is needed to precipitate all the sulfate ion in a 772-mg sample of Na2SO4, what is the molarity of the solution?

Question

Just need a little bit of help. 

thanks!

Answer 1

BaCl2(aq) + Na2SO4(aq) ==> BaSO4(s) + 2NaCl(aq)  balanced equation

Note mole ratio of BaCl2:Na2SO4 = 1:1

mmoles Na2SO4 present = 772 mg x 1 mmole/142 mg = 5.437 mmoles Na2SO4 present

mmoles BaCl2 needed = 5.437 mmoles Na2SO4 x 1 mmole BaCl2/1 mmole Na2SO4 = 5.437 mmoles BaCl2 needed

5.437 mmoles BaCl2/56.8 ml = 0.0957 mmole/ml = 0.0957 mol/L = 0.0957 M BaCl2

 

Answer 2

The molar ratio of Ba2+ to SO42- in the reaction is 1:1.  So, set up a dimensional analysis (chain calculation) as follows:

 1) calculate the molar mass of Na2SO4 (since we want the molarity of the Na2SO4 solution).

2) Then,  (1 mole Na2SO4/142 g) (772 x 10^-3 g Na2SO4/56.8 x 10^-3L Ba2+)  =  0.0957 mol Na2SO4/L solution or0.0957 M