Help with Molarity? (Fundamentals of Chemistry)

[Starlight]

When 50.0 mL of a 0.150M barium nitrate solution is mixed with 50.0 mL of a 0.350 M sodium sulfate solution, a precipitate of BaSO4 is formed. What is the concentration of sulfate ions in the resulting solution?

a) 0.0500 M

b) 0.100 M

c) 0.165 M

d) 0.350 M

how would you go about figuring out this problem?

[charmanderp]

I am a lowly Year 12 student, but I'm can still help you out :p

If the two substances were the same ie. we were adding 50ml of 0.54M HCL to 1L of 0.1M HCL, we could simply use the C1V1=C2V2 principle.

However, that is not the case. We are adding two different parties here, and therefore to calculate the concentration of one of the components in the resulting product, we need to calculate the number of moles (molecules) by which it will be in excess.

Therefore, we need to calculate the number of moles of sulfate ions (in sodium sulfate) and the number of moles of barium ions (in barium nitrate). This can be done using the principle of n=CV, where the number of moles of a substance in solution is given by concentration multiplied by volume in litres.

n(Ba(NO3)2)= 0.15x0.05 = 0.0075mol
n(Na2SO4)= 0.35x0.05 = 0.0175mol

Here the mole ratio of Ba (barium) to Ba(NO3)2 is 1:1, as is the mole ratio of SO4 (sulfate) to Na2SO4.

So we are combining 0.0075mol of barium ions with 0.0175mol of sulfate ions.

As you can see, we have more sulfate ions than barium ions. This means that the sulfate ions are in excess, and that excess is what we will use to calculate the concentration of sulfate ions in the resulting solution.

In order to find out the amount by which sulfate is in excess, we need to write a balanced equation.

The unbalanced equation is as such (without states):

Ba(NO3)2 + Na2SO4 --> BaSO4 + Na2(NO3)2

So as we can see, now, for every 1 mole of Na2SO4, we need 1 moles of BaNO3. We'll now recall the values we found earlier:

n(Ba(NO3)2)= 0.15x0.05 = 0.0075mol

n(Na2SO4)= 0.35x0.05 = 0.0175mol

Since the mole ratio is 1:1 and there is less Ba(NO3)2, this is the limiting reagent (the value which limits how much precipitate might be produced). We only need 0.0075mol of Na2SO4, yet we have 0.0175mol, which means (0.0175-0.0075)mol is in excess. Which is 0.01mol, in excess, which will just be floating around in solution, with no barium ions left to form a precipitate with.

As mentioned earlier, n=CV. If we transpose this equation we find that C(concentration) is equal to number of moles divided by volume (in litres.

So we have C=0.01(mol in excess)/0.05 (litres of barium nitrate solution)+0.05 (litres of sodium sulfate solution).

C=0.01/0.1

which is equal to option b) 0.100M

I hope I've explained it clearly I made sure I was reasonably thorough, since I can see you haven't done chemistry before, but if you want a more concise explanation or anymore assistance, don't hesitate to shoot me a message.


To anyone else who sees this, there is indeed the possibility that I am wrong, so please let me know if I am!

[Hehetymen]

Same answer I got. Nice explanation.

[Starlight]

That's great! Thanks a lot, I kind of got halfway through the problem initially but I forgot about the use of the n= cv equation