# 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.

Sep 1, 2017

$\text{28.2 mL}$

#### 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, ${\text{BaSO}}_{4}$, an insoluble solid that will precipitate out of the solution.

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

${\text{Ba"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-) -> "BaSO}}_{4 \left(s\right)} \downarrow$

Now, your sodium sulfate solution contains $\text{10.0 g}$ of sodium sulfate, the solute, for every $\text{100 mL}$ of solution $\to$ this is equivalent to saying that the solution is $\text{10.0% w/v}$ 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.

${\text{Na"_ 2"SO"_ (4(aq)) -> 2"Na"_ ((aq))^(+) + "SO}}_{4 \left(a q\right)}^{2 -}$

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 ${\text{0.250 mol L}}^{- 1}$ barium chloride solution will contain $0.250$ moles of barium chloride for every $\text{1 L} = {10}^{3}$ $\text{mL}$ of solution.

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

$0.00704 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles BaCl"_2))) * (10^3color(white)(.)"mL solution")/(0.250color(red)(cancel(color(black)("moles BaCl"_2)))) = color(darkgreen)(ul(color(black)("28.2 mL}}}}$

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

The answer is rounded to three sig figs.