[Chemistry Q] Ionic Strength of a solution

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    Ionic Strength
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The discussion centers on the calculation of ionic strength in two different solutions containing NaCl and CaCl2. Participants clarify that ionic strength does depend on the composition and concentrations of the ions present in the solution. Initial calculations suggested both solutions had the same ionic strength of 0.3, but further analysis revealed that the actual ionic strengths differ due to varying concentrations after mixing. The importance of accurately accounting for the dilution and correct formulas for the ions is emphasized. Ultimately, the conclusion is that composition and concentration significantly affect ionic strength.
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I got confused in dealing with ionic strength.

Homework Statement



In calculating ionic strength, do we consider the composition of the solution?

Take for example:

Solution 1 has
20 ml of 0.1 M NaCl and 35ml of 0.2 M CaCl

Solution 2 has
25 ml of 0.1 M NaCl and 30 ml of 0.2 M CaCl

Would the ionic strength be different for the two or be the same?
In other words, does ionic strength depends on the composition?

Homework Equations



Ionic Strength:
I = 0.5*sum(c*z^2)

The Attempt at a Solution



Ionic Strength:
I = 0.5*sum(c*z^2)
I = 0.5*(0.1*(+1)^2+0.2*(+1)^2+0.1*(-1)^2+0.…
I = 0.5*(0.6)
I = 0.3

So, are the two solutions going to have an ionic of 0.3 or would they have different ionic strengths?

Thanks in advance.
 
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I just realize that it doesn't depend on that issue. It's the matter of whether the ions are present or not. Since they have the same ions present in their solution, regardless of how much of the ions are present or regardless of their different amount in each solution, the ionic strength are the same. The ionic strength for solution 1 and solution 2 both have the value of 0.3.
 
Remu said:
In calculating ionic strength, do we consider the composition of the solution?

Yes.

20 ml of 0.1 M NaCl and 35ml of 0.2 M CaCl

No such thing as CaCl, perhaps you mean CaCl2.

Ionic Strength:
I = 0.5*sum(c*z^2)

Good, more texy:

I = \frac 1 2 \sum c_i z_i^2

I = 0.5*(0.1*(+1)^2+0.2*(+1)^2+0.1*(-1)^2+0.…

You forgot that concentrations of ions after mixing have changed - for example concentration of Na+ is now 0.036M, that's because of the volume change.

Remu said:
I just realize that it doesn't depend on that issue. It's the matter of whether the ions are present or not. Since they have the same ions present in their solution, regardless of how much of the ions are present or regardless of their different amount in each solution, the ionic strength are the same. The ionic strength for solution 1 and solution 2 both have the value of 0.3.

No, you got it completely wrong - composition matters, concentrations matter, charges matter, neither of these solutions have ionic strength of 0.3 and their ionic strengths are different. Try to calculate using correct formula for CaCl2 and taking dilutions into account.

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I see my mistake!
Thanks for clarifying!
 
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