[renia45]

Hey everyone.
Here is the question that I am stuck on:
A 10.0mL solution of 0.100 M Na2SO4 is added to a 20.0 mL water sample containing an unknown concentration of BaCl2 that was concentrated by 100 fold after being taken from a waste water treatment plant. After filtration of the solution, 0.5600g of the insoluble BaSO4 (s) is obtained. Find the balanced chemical equation and calculate the concentration of Ba 2+ from the sample taken at the waste water treatment plant.

So I have come up with the following chemical equation:
BaCl2 + Na2SO4 ---> 2NaCl + BaSO4

I also found the moles of Na2SO4: n=cv therefore: n=(0.100M)(0.0100L) which is 1.00E-3
Now I am not sure of what to do.
Any suggestions would be appreciated.
Thank you

[DrCMS]

First you need to check the amount of sulphate added was enough to precipitate all the barium.

Calculate how many moles of precipitate were collected and compare that with the amount of sodium sulphate you've correctly calculated.

If it was then you can calculate the concentration of barium in the 20ml water sample.

With the sample prep info you can then work out the concentration in the waste water.

[renia45]

Thank you for your reply.
Ok so I originally calculated Na2SO4 to have 1.00E-3 moles and then BaSO4 has 2.399E-3 moles which is the precipitate. So does there need to be more moles of Na2SO4? If there was more Na2SO4 then would that mean that a precipitate would form?

[Borek]

http://www.chembuddy.com/?left=balancing-stoichiometry&right=limiting-reagents

[renia45]

Thanks for the link
I was just wondering why I can't just figure out the number of moles of Na2SO4 using n=cv and since it is a 1:1 ratio between Na2SO4 and BaCl2 then the number of moles are the same which can be used to find the concentration of BaCl2 using c=n/v. And this concentration would also be the same as that of Ba2+.
But then I wouldn't be using the mass given for BaSO4. So is this mass given to determine whether there is enough Na2SO4 to make BaS04? (like the info provided by the link)

[DrCMS]

Are you sure it was 10ml of 0.1M Na₂SO₄?  or 0.5600g BaSO₄?  it looks like there is a mistake somewhere.

[renia45]

Hmm.
I will use the numbers from another similar question in the exercise since I just really need to understand the process rather than achieving the numerical answer at this point.
Ok so if there is:
20.0 mL of 0.200M Na2SO4 added to a 10.0 mL water sample containing a concentration of BaCl2 that was concentrated by 100 fold. After filtration of the solution, 0.7800g of the insoluble BaSO4 is achieved. Determine the concentration of Ba2+.
Hopefully this will work out better.
Thanks everyone

[renia45]

Ok so using the new info, here is what I figured out:
Na2SO4 = 4.00E-3 moles
BaSO4 = 3.342E-3 moles
So, from this info it is apparent that enought Na2SO4 was present to yield the precipitate BaSO4.
As a result, I assume that I need to calculate the moles of BaCl2 next using mole to mole ratios. So my problem is do I use Na2SO4 in this ratio or BaSO4? Once I have the moles of BaCl2 I can figure out the concentration using: c=n/v and then divide it by 100.
I think I got it except for the part of figuring out moles of BaCl2 however I could be completely wrong. If anyone can help me out I would really appreciate it.
Thanks so much everyone

[DrCMS]

Ok so using the new info, here is what I figured out:
Na2SO4 = 4.00E-3 moles
BaSO4 = 3.342E-3 moles
So, from this info it is apparent that enought Na2SO4 was present to yield the precipitate BaSO4.

Yes so sulphate is in excess.

As a result, I assume that I need to calculate the moles of BaCl2 next using mole to mole ratios. So my problem is do I use Na2SO4 in this ratio or BaSO4?

With sulphate in excess what happened to all the BaCl₂?

[renia45]

So all the BaCl2 was used up then. But what does this give me?

[DrCMS]

Where did all the Ba in the BaCl₂ end up?

[renia45]

in the BaSO4. So I would usethe moles of BaSO4 to find that of BaCl2 using mole to mole ratios? In other words moles BaSO4= moles of BaCl2 then since its 1:1?
Except, how is it possible to say that Na2SO4 is in excess when I was comparing the moles of BaSO4(product) to Na2SO4(reactant)? I thought you needed to compare the two reactants to find the limiting reactant?

[DrCMS]

in the BaSO4. So I would usethe moles of BaSO4 to find that of BaCl2 using mole to mole ratios? In other words moles BaSO4= moles of BaCl2 then since its 1:1?

Yes

Except, how is it possible to say that Na2SO4 is in excess when I was comparing the moles of BaSO4(product) to Na2SO4(reactant)? I thought you needed to compare the two reactants to find the limiting reactant?

You've already answered that

moles BaSO4= moles of BaCl2

[renia45]

Ok haha this is finally all starting to flow together lol. Ok so all in all:
moles of BaSO4 = moles of BaCl2
concentration of BaCl2: c = moles of BaSO4 (essentially)/ 0.020 L
Then I need to divide this concentration by 100 right and then I have my answer?

Lastly, since I did not use the information provided about Na2SO4 at all to get my numerical answer, then was it provided so that I could see that it was in excess? And how come I couldn't use the moles of Na2SO4 to calculate that of BaCl2 using mole to mole ratios (which would not achieve the same answer as it would if I used BaSO4, which is the correct one)... so ultimately how do I know with which compounds I can use mole to mole ratios?

Thank you for bearing with me

[DrCMS]

Ok haha this is finally all starting to flow together lol. Ok so all in all:
moles of BaSO4 = moles of BaCl2
concentration of BaCl2: c = moles of BaSO4 (essentially)/ 0.020 L
Then I need to divide this concentration by 100 right and then I have my answer?

Yes

Lastly, since I did not use the information provided about Na2SO4 at all to get my numerical answer, then was it provided so that I could see that it was in excess?

Yes

And how come I couldn't use the moles of Na2SO4 to calculate that of BaCl2 using mole to mole ratios (which would not achieve the same answer as it would if I used BaSO4, which is the correct one)... so ultimately how do I know with which compounds I can use mole to mole ratios?

You have to use the limiting reagent.