How Much Solution Can Be Prepared with Specific Chloride Concentration?

  • Thread starter Thread starter Saitama
  • Start date Start date
  • Tags Tags
    Concentration
Click For Summary

Discussion Overview

The discussion revolves around determining the maximum volume of a solution that can be prepared with a specific chloride ion concentration of 0.34M using given solutions of NaCl and CaCl2. The context includes mathematical reasoning and constraints based on the available volumes and concentrations of the solutions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Participants explore the possibility of mixing the two solutions to achieve the desired chloride concentration.
  • Some participants calculate the initial moles of chloride ions from both NaCl and CaCl2, noting that NaCl provides 0.1 moles and CaCl2 provides 0.8 moles, totaling 0.9 moles of Cl- ions.
  • There is a discussion about the implications of evaporation, with participants agreeing that it is not a viable option.
  • One participant suggests using the equation 0.1V1 + 0.4V2 = 0.34(V1 + V2) to relate the volumes of NaCl and CaCl2 needed.
  • Another participant points out the need for a second equation due to having two variables, V1 and V2.
  • Constraints based on the maximum available volumes of NaCl and CaCl2 are discussed, with one participant noting that the volume of CaCl2 should be fully utilized.
  • There is a correction regarding the relationship between V1 and V2, with a participant stating it should be V1 = V2/4.
  • Participants also consider a hypothetical scenario where the target concentration is only 0.1M, discussing the implications for the volumes of NaCl and CaCl2 needed.

Areas of Agreement / Disagreement

Participants generally agree on the need to conserve moles and volumes in their calculations, but there is no consensus on the final solution or the exact volumes to be used from each solution. Disagreements arise regarding the relationships between the variables and the implications of the constraints.

Contextual Notes

Participants express uncertainty about how to formulate equations for the final moles based on the initial moles and the constraints of the available solutions. There are also limitations regarding the assumptions made about the mixing and the final volume calculations.

Saitama
Messages
4,244
Reaction score
93

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. :smile:
 
Physics news on Phys.org
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?
 
You have 2L of .2 M CaCl2 but what concentration of Cl- do you have for this reagent?
 
Evaporation is most likely out of the question.
 
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.
 
Pranav-Arora said:
From CaCl2 we have 0.4M concentration of Cl- ions.

Yep. :approve:

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.
 
Infinitum said:
Yep. :approve:

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?
 
Pranav-Arora said:
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 :wink:
 
Infinitum said:
Yes, but that is the moles for both the solutions together. You only need to conserve moles of the volume you are using :wink:

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?
 
  • #10
Pranav-Arora said:
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 :biggrin: (conserve volume!)
 
  • #11
Infinitum said:
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 :biggrin: (conserve volume!)

0.1V1+0.4V2=0.34(V1+V2)

I have two variables, i need one more equation. :confused:
 
  • #12
Pranav-Arora said:
0.1V1+0.4V2=0.34(V1+V2)

I have two variables, i need one more equation. :confused:

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!
 
  • #13
Infinitum said:
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 [itex]V_1=\frac{23}{12}V_2[/itex] which is approximately [itex]V_1=2V_2[/itex].

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.
 
Last edited:
  • #14
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?
 
  • #15
Pranav-Arora said:
I get a relation [itex]V_1=\frac{23}{12}V_2[/itex] which is approximately [itex]V_1=2V_2[/itex].

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! o:)) you can find the amount of NaCl to go along with it.
 
  • #16
Borek said:
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.
 
  • #17
Infinitum said:
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! o:)) 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?
 
  • #18
Pranav-Arora said:
Oh yes, sorry about that, the relation is V1=V2/4.

Therefore the volume from NaCl is 0.5L?

Yep! :approve:
 
  • #19
Infinitum said:
Yep! :approve:

But what if the question asked about only 0.1 M Chlorine ions at the end?
 
  • #20
Pranav-Arora said:
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...
 
  • #21
Infinitum said:
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.
 

Similar threads

Calculate the concentration (in M) of Cl- ions in solution C
  • · Replies 2 ·
Replies
2
Views
5K