# Question ae1b3

Jul 30, 2017

Here's what I got.

#### Explanation:

The idea here is that you must calculate the reaction quotient, ${Q}_{c} 3$, and compare it with the solubility product constant, ${K}_{s p}$, of barium sulfate.

If you have

${Q}_{c} > {K}_{s p}$

then barium sulfate will precipitate out of solution.

Using the chemical equation that you balanced correctly

${\text{BaCl"_ (2(aq)) + "Na"_ 2"SO"_ (4(aq)) -> "BaSO"_ (4(s)) darr + 2"NaCl}}_{\left(a q\right)}$

you can say that $1$ mole of barium chloride will react with $1$ mole of sodium sulfate to precipitate $1$ mole of barium sulfate.

However, you don't have to worry about that here because you're looking to find the concentrations of the two reactants in the resulting solution.

Now, the total volume of the resulting solution will be

$\text{10 mL + 10 mL = 20 mL}$

This essentially means that you're doubling the volume of the two solutions while keeping the number of moles of the two solutes constant, so you can say that the concentrations of the two solutions will be halved.

The two ions that are of interest here are the barium cations, ${\text{Ba}}^{2 +}$, and the sulfate anions, ${\text{SO}}_{4}^{2 -}$.

Since both of them are produced in $1 : 1$ mole ratios in their respective solutions

${\text{BaCl"_ (2(aq)) -> "Ba"_ ((aq))^(2+) + 2"Cl}}_{\left(a q\right)}^{-}$

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

you can say that the resulting solution will contain

["Ba"^(2+)] = "0.010 M"/2 = "0.0050 M"

["SO"_4^(2-)] = "0.100 M"/2 = "0.050 M"#

Barium sulfate is considered insoluble in water, so an equilibrium will be established in aqueous solution between the undissolved solid and the solvated ions.

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

Your goal here is to figure out if the concentrations of the two ions are high enough to allow for precipitation to take place.

The reaction quotient, which uses the starting concentrations of the two ions in the resulting solution, will be equal to

${Q}_{c} = \left[{\text{Ba"^(2+)] * ["SO}}_{4}^{2 -}\right]$

Plug in your values to find--for the sake of simplicity, I'll leave the reaction quotient without added units

${Q}_{c} = \text{0.0050 M" * "0.050 M} = 2.5 \cdot {10}^{- 4}$

As you can see, you have

$2.5 \cdot {10}^{- 4} > 5.81 \cdot {10}^{- 8}$

which means that barium sulfate will precipitate out of the solution.