Help w/ Molarity Question Related to a CuSO4 Solution

[ptownbro]

We have a question on my daughter's homework I'm trying to help with and there is a part of it we have not been able to find support for to get to the answer. See below. We believe we've been able to answer parts "a" and "b" correctly, however, we're having trouble answering part "c" and were looking form some help!

Questions: 35.0 mL of CuSO₄ solution contains .0180 moles of solute.

a) What is the morality?
(0.018 moles / 35 mL) * (1000 mL / 1 L) = 5.143 moles/L = 5.143 M

b) What mass of CuSO₄ is present?
(35 mL) * ( 1 L / 1000 mL) * (1 mol / 22.4 L) * (159.609 g / 1 mol) = .2493g

c) If 5.0 mL of this solution is diluted to 75 ml, what is the new concentration of CuSO₄? Of Cu⁺²? Of SO₄^-2?

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We have two questions related to this:

1) Did we solve the sub-part 1 of this question correctly?
We've interpreted this to mean that: Instead of a 35.0 mL solution we now have 5.0 mL solution and we'll be finding the concentration (morality) based on this 5.0 mL alone. Then, some kind of solvent (not given) was combined with some unknown solute (not given) to dilute the 5.0 mL to 75.0 mL. So, what we did was apply the ratio from part "a" as below:

(75 mL / 35 mL) * 5.143 M = 11.0207

I assumed that we 1) the quantity and make up of the solute is the same from part "a" and 2) therefore we could use the morality from part "a" and it's relative ratio to solve the new morality

2) What the heck do we do with ions in sub-parts 2 and 3? We have no clue how these even relate this a molarity or a stoichiometry type question.

 

[Bystander]

a) What is the morality?
(0.018 moles / 35 mL) * (1000 mL / 1 L) = 5.143 moles/L = 5.143 M

Watch your decimal, and try it again.

b) What mass of CuSO4 is present?
(35 mL) * ( 1 L / 1000 mL) * (1 mol / 22.4 L) * (159.609 g / 1 mol) = .2493g

How many moles? (Given in part a) What's the molecular mass? Multiply. 22.4 L is volume of an ideal gas --- are there any ideal gases in this problem?

 

[symbolipoint]

Decimal mistake in part (a). Correct molarity is 0.5143 M for the cupric sulfate.

The question part (c) dilution uses 0.005 liters of 0.5143 moles per liter copper sulfate and becomes according to description, a volume of 0.075 liters. You can now compute the new "moles per liter" or molarity.

 

[Borek]

What the heck do we do with ions in sub-parts 2 and 3?

One molecule of Na₂CO₃ dissociates into 2Na⁺ and CO₃^2-. That means 1 mole/liter solution of Na₂CO₃ contains 2 moles/liter Na⁺ and 1 mole/liter CO₃^2-. Do you see it now?

 

[ptownbro]

Ah yes. For part a, the decimal point was off.

For part b, there is no mention of ideal gas but most of his questions are at STP (teacher is not constituent in his wording). Are you saying /implying we should have used the moles from part aa instead of the 22.4L as the basis of our clac? That makes sense if that's what you meant.

 

[Bystander]

That makes sense if that's what you meant.

You got it.

 

[ptownbro]

One molecule of Na₂CO₃ dissociates into 2Na⁺ and CO₃^2-. That means 1 mole/liter solution of Na₂CO₃ contains 2 moles/liter Na⁺ and 1 mole/liter CO₃^2-. Do you see it now?

Yes and no. =) I now get how Na₂CO₃ can break up into those ions, however, I would of thought that type of disassociation would apply to the solute not the solution. In other words, (conceptually) I thought the solute would dissolve (which I assumed is disassociation) and break up into those two ions. The solution on the other hand is not broken apart.

Forgive me if I am using the terms wrong... I work in finance so chemistry is not my thing. =)

 

[Borek]

And you are right - solution doesn't break apart, it just gets diluted.

Copper sulfate was there already dissociated (it dissociated the moment it was dissolved) and it doesn't dissociate further. When you add more water all that happens is that concentrations go down.

However, you seemed to be confused about the ions in the solution (part c of the problem), hence my explanation.

 

[ptownbro]

Thanks everyone for your help.