How Much Solution Can Be Prepared with Specific Chloride Concentration?

[Saitama]

Homework Statement

Available are 1L of 0.1M NaCl and 2L of 0.2M CaCl₂ solutions. Using only these two solutions what maximum volume of a solution can be prepared having [Cl^-]=0.34M exactly. Both electrolytes are strong.
a)2.5L
b)2.4L
c)2.3L
d)None

Homework Equations

NaCl-->Na⁺+Cl^-
CaCl₂-->Ca²⁺+2Cl^-

The Attempt at a Solution

I have no idea on how to start with this one. Any ideas are greatly appreciated.

 

[I like Serena]

Hey Pranav! :)

Let's see... what can we do...?

A. We can mix the solutions together
B. We can let some of the volume vaporize, but I'm not sure if we're allowed to do that.

What molarity do you get for instance if you mix everything together?

 

[chemisttree]

You have 2L of .2 M CaCl2 but what concentration of Cl- do you have for this reagent?

 

[Borek]

Evaporation is most likely out of the question.

 

[Saitama]

Yes, Evaporation is out of the question.

We have 0.1 moles of NaCl from which we get 0.1 moles of Cl- ions. Similarly we have 0.4 moles of CaCl2, from which we can obtain 0.8 moles of Cl-. If we mix both of the solutions together, we get 0.9 moles of Cl- ions and total volume becomes 3L. Therefore, molarity of Cl- ions is 0.3M.

@chemisttree: From CaCl2 we have 0.4M concentration of Cl- ions.

 

[Infinitum]

From CaCl2 we have 0.4M concentration of Cl- ions.

Yep.

Now, conserve number of moles in the equation, before after getting the solution. Call the initial volume of chlorine ions from NaCl as V1, and CaCl2 V2, and proceed.

 

[Saitama]

Yep.

Now, conserve number of moles in the equation, before after getting the solution. Call the initial volume of chlorine ions from NaCl as V1, and CaCl2 V2, and proceed.

Before mixing the moles for chlorine ions are 0.9.
I am confused, how should i form the equation here?

 

[Infinitum]

Before mixing the moles for chlorine ions are 0.9.
I am confused, how should i form the equation here?

Yes, but that is the moles for both the solutions together. You only need to conserve moles of the volume you are using

 

[Saitama]

Yes, but that is the moles for both the solutions together. You only need to conserve moles of the volume you are using

So you mean the initial moles are 0.1V₁+0.4V₂?
But how could i conserve the moles here, i need to something like initial moles=final moles, how should i go on making an equation for final moles?

 

[Infinitum]

So you mean the initial moles are 0.1V₁+0.4V₂?
But how could i conserve the moles here, i need to something like initial moles=final moles, how should i go on making an equation for final moles?

Yes, that's the initial moles. For final moles, you are already given the final concentration of chlorine, and volume doesn't tend to run away anywhere or manifest itself out of nothing (conserve volume!)

 

[Saitama]

Yes, that's the initial moles. For final moles, you are already given the final concentration of chlorine, and volume doesn't tend to run away anywhere or manifest itself out of nothing (conserve volume!)

0.1V₁+0.4V₂=0.34(V₁+V₂)

I have two variables, i need one more equation.

 

[Infinitum]

0.1V₁+0.4V₂=0.34(V₁+V₂)

I have two variables, i need one more equation.

Well, this equation gives you a relation between V1 and V2. Now you also have a limited amount of NaCl and CaCl2. Use these constraints!

 

[Saitama]

Well, this equation gives you a relation between V1 and V2. Now you also have a limited amount of NaCl and CaCl2. Use these constraints!

I get a relation $V_1=\frac{23}{12}V_2$ which is approximately $V_1=2V_2$.

I still can't figure out, how these constraints are going to help me?

EDIT: One relation i get is that V2 should be less than or equal to 1/2 since we do not have more than 1L of NaCl, but still i don't understand how this will help me.

 

[Borek]

Technically you are diluting CaCl₂ solution, and you want to prepare as much of the final solution as possible. That means you need to use all 2L of the CaCl₂ solution, doesn't it?

 

[Infinitum]

I get a relation $V_1=\frac{23}{12}V_2$ which is approximately $V_1=2V_2$.

Recheck your calculations, this seems to be incorrect.

One of the volumes will get exhausted because you just don't have enough, meaning that CaCl2 will be used up fully, while from that relation(after you get the calculations correct! ) you can find the amount of NaCl to go along with it.

 

[Saitama]

Technically you are diluting CaCl₂ solution, and you want to prepare as much of the final solution as possible. That means you need to use all 2L of the CaCl₂ solution, doesn't it?

Before doing the algebraic stuff and making equations, i thought of this but then i got stuck on finding the volume we need to use from NaCl.

 

[Saitama]

Recheck your calculations, this seems to be incorrect.

One of the volumes will get exhausted because you just don't have enough, meaning that CaCl2 will be used up fully, while from that relation(after you get the calculations correct! ) you can find the amount of NaCl to go along with it.

Oh yes, sorry about that, the relation is V₁=V₂/4.

Therefore the volume from NaCl is 0.5L?

 

[Infinitum]

Oh yes, sorry about that, the relation is V₁=V₂/4.

Therefore the volume from NaCl is 0.5L?

Yep!

 

[Saitama]

Yep!

But what if the question asked about only 0.1 M Chlorine ions at the end?

 

[Infinitum]

But what if the question asked about only 0.1 M Chlorine ions at the end?

Isn't it obvious then??

CaCl2 would give you 0.1M Cl concentration for half liter of water. NaCl gives you 0.1 M for one liter of water. And you need maximum volume...

 

[Saitama]

Isn't it obvious then??

CaCl2 would give you 0.1M Cl concentration for every half liter of water. NaCl gives you 0.1 M for one liter of water. And you need maximum volume...

Oops, sorry to ask you such a foolish question.