Chemistry-Finding concentrations of two compounds

[taxidriverhk]

Homework Statement

Given balanced chemical reaction:
Ba(NO₃)₂ (aq) + Na₂SO₄ (aq) → BaSO₄ (s) + 2NaNO₃ (aq)

Suppose 200 mL of Ba(NO₃)₂ and 200 mL of Na₂SO₄ of unknown concentrations are mixed together and forms the reaction above. Then, there are 0.0120 M of NaNO₃ produced, and 0.0004 M of excess sulfate ion. Find the initial concentration of each reactant.

Homework Equations

Molarity = number of moles/volume of solvent

The Attempt at a Solution

I led x be the concentration of Ba(NO₃)₂
and y be the concentration of Na₂SO₄
Then I used the law of conservation of mass to find the total mass of reactants and products, then wrote the equation below
5.6x + 28.4y = 0
But I am now not sure what to do next, do I need to find the volume of the solution produced or to do something else?
Please give me some hints or concepts so that I can figure it out, I will be extremely appreciated...

 

[Borek]

It can be solved in many ways, but the simplest approach is to look for things that have not changed. How many moles of NO₃^- in the final solution? How is this number related to the initial number of moles of Ba(NO₃)₂?

 

[taxidriverhk]

It can be solved in many ways, but the simplest approach is to look for things that have not changed. How many moles of NO₃^- in the final solution? How is this number related to the initial number of moles of Ba(NO₃)₂?

I think I may get some ideas from what you asked
I now try to write out the net ionic equation for the reaction below
Ba²⁺ + 2NO₃^- + 2Na⁺ + SO₄^2- → BaSO₄ + 2Na⁺ + 2NO₃^-
And I find out that NO₃^- and Na⁺ are spectator ions, that means their concentrations did not change right?
But the problem is I don't know how to find the final concentrations of each ion from the given concentration of NaNO₃ which is 0.0120M, should I divide it half and half, or is there some other ways? Please advise..

 

[Borek]

And I find out that NO₃^- and Na⁺ are spectator ions, that means their concentrations did not change right?

Concentration changes, but not because of teh reaction - you mix two solutions, so they are diluted.

But the problem is I don't know how to find the final concentrations of each ion from the given concentration of NaNO₃ which is 0.0120M, should I divide it half and half, or is there some other ways? Please advise..

If concentration of NaNO₃ is 0.0120M, concentrations of both Na⁺ and NO₃^- are also 0.0120M. This is not always the case, but here NaNO₃ dissociates into 1+1 ions.

 

[taxidriverhk]

So the final concentration of each ion is below
Ba²⁺ = 0 M
SO₄^2- = 0.0004 M
Na⁺ = 0.0120 M
NO₃^- = 0.0120 M
is it right?
If they are right, then I will be able to find the initial concentration of Ba(NO₃)₂ by first finding the number of moles of NaNO₃, and use the chemical equation to convert it to the number of moles of Ba(NO₃)₂, and finally divide it by 0.2 L to find out the initial concentration, did I think it right?
But how about the initial concentration of another compound? I got confused by the excess 0.0004M sulfate acid
Please advise, thank you.

 

[Borek]

You are right about initial concentration of Ba(NO₃)₂.

Part of the sulfate is in the precipitate (this part you can easily calculate from known initial concentration of barium), part of the sulfate is still in the solution. Sum of these was present in the initial solution of sodium sulfate.