# Homework Help: [Chemistry Q] Ionic Strength of a solution

1. Oct 29, 2009

### Remu

I got confused in dealing with ionic strength.

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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.

2. Oct 29, 2009

### Remu

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.

3. Oct 30, 2009

### Staff: Mentor

Yes.

No such thing as CaCl, perhaps you mean CaCl2.

Good, more texy:

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

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.

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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4. Oct 30, 2009

### Remu

I see my mistake!
Thanks for clarifying!

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