Question

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.

Answer 1

`"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.