How many mL of a #"0.250 mol L"^(-1)# #"BaCl"_2# solution is required to precipitate all the sulfate ions from #"10.0 mL"# of a #"10.0 % (w/v)"# solution of #"Na"_2"SO"_4# ?

The answer is 28.2 ml.

1 Answer
Sep 1, 2017

Answer:

#"28.2 mL"#

Explanation:

The idea here is that the barium cation delivered to the solution by the barium chloride solution will react with the sulfate anions to produce barium sulfate, #"BaSO"_4#, an insoluble solid that will precipitate out of the solution.

The net ionic equation that describes this double-replacement reaction looks like this

#"Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-) -> "BaSO"_ (4(s)) darr#

Now, your sodium sulfate solution contains #"10.0 g"# of sodium sulfate, the solute, for every #"100 mL"# of solution #-># this is equivalent to saying that the solution is #"10.0% w/v"# sodium sulfate.

This means that the sample contains

#10.0 color(red)(cancel(color(black)("mL solution"))) * ("10.0 g Na"_2"SO"_4)/(100color(red)(cancel(color(black)("mL solution")))) = "1.00 g Na"_2"SO"_4#

Use the molar mass of sodium sulfate to calculate the number of moles of solute used to make the solution

#1.00 color(red)(cancel(color(black)("g"))) * ("1 mole Na"_2"SO"_4)/(142.04color(red)(cancel(color(black)("g")))) = "0.00704 moles Na"_2"SO"_4#

Sodium sulfate is a soluble salt, which implies that it dissociates in aqueous solution to produce sodium cations and sulfate anions.

#"Na"_ 2"SO"_ (4(aq)) -> 2"Na"_ ((aq))^(+) + "SO"_ (4(aq))^(2-)#

Notice that every mole of sodium sulfate that dissociates produces #1# mole of sulfate anions. This means that the solution will contain #0.00704# moles of sulfate anions.

Since the sulfate anions react in a #1:1# mole ratio with the barium cations to produce the precipitate, you can say that your barium chloride solution must contain #0.00704# moles of barium cations.

Barium chloride is a soluble salt that dissociates to produce barium cations in a #1:1# mole ratio, so you can say that in order for the solution to contain #0.00704# moles of barium cations, you must dissolve #0.00704# moles of barium chloride to make the solution.

Now, a #"0.250 mol L"^(-1)# barium chloride solution will contain #0.250# moles of barium chloride for every #"1 L" = 10^3# #"mL"# of solution.

This means that your barium chloride solution must have a volume of

#0.00704 color(red)(cancel(color(black)("moles BaCl"_2))) * (10^3color(white)(.)"mL solution")/(0.250color(red)(cancel(color(black)("moles BaCl"_2)))) = color(darkgreen)(ul(color(black)("28.2 mL")))#

in order to deliver #0.00704# moles of barium cations to the reaction.

The answer is rounded to three sig figs.