[Manie28]

Hi,

I want to prepare a solution with an ionic strength of 0,01 M NaCl and pH 3. I think it can´t be possible because pH 3 is equivalent 10^-3 M = 0,001 M HCl.  So if I had a solution with 0,01 M NaCl the error will be too large if I´d try to adjust pH 3. Is this true or did I forget something? Is there a possibility to adjust pH3 with an ionic strength of 0,01 M without getting to big errors?

Thanks in advance for the *delete me*

Manie28

[enahs]

NaCl has nothing to do with the acidity of HCl or your solution, at least not in the concentrations you are talking about, and not with the assumption that the pH is the -log of the H+ concentration .

[Manie28]

I understand what you mean, but I just don´t have any idea how to prepare this solution correctly without producing big errors (> 2 % deviation of the volume and ionic strength). How would a chemist proceed?

[Borek]

Depending on the situation 2% error can be acceptable. Do you need soultion pH = 3, or pH = 3.000? Ionic strength of 0.01 or 0.01000?

[Manie28]

Hi Borek,

It´s good enough when the deviation is pH 2.8-3.2 and the ionic strength deviates max. 2 % of 0,01 M. How would you solve this?

[enahs]

There are a couple of ways.

First make your solution pH 3 with HCl, this will give you 1x10^-3 mols Cl^-.
You can then add 9x10^-3 mols of NaCl.
This will give you 0.01 mols of Cl^- and 0.009 mols of Na⁺.
Your Na⁺ will be off by 10%.

Or,
You can make the pH say 3.05, giving 8.91x10^-4 mols of Cl^-.
You can then add 9.10x10^-3 mol of NaCl. You are no off in Na⁺ by 8.91%.

Rinse and repeat until you find a error in the pH and Na⁺ you find acceptable in your application.


Alternatively, you can do it the first way, and make up for the lack of Na⁺ in solution with another chemical whose counter ion will not interfere with whatever you are attempting to do.


The easiest way, would be of course to make the NaCl solution, and acidify it with something that does not have Cl^- as a counter ion, or that other acids counter ion will interfere with whatever you are trying to do.



Those are probably the most direct and basic ways.


But no, you can not do that exactly with just HCl and NaCl.

[Borek]

What about solution containing 0.001 HCl and 0.009 NaCl? Remember, that HCl increases ionic strength as well.

[Manie28]

HI Borek,

I also thought about your suggestion. If I want 50 ml of this solution, I calculate this:

50 ml * 0,001 M HCl = x ml *0,1 M HCl (stock solution)
x = 0,5 ml of 0,1 M HCl are needed for 50 ml solution to get 0,001 M HCl

50 ml * 0,009 M NaCl = x ml * 0,1 M NaCl (stock solution)
x = 4,5 ml of 0,1 M NaCl are needed for 50 ml solution to get 0,009 M NaCl

So I put 0,5 ml of 0,1 M HCl in a test-tube and add 4,5 ml of 0,1 M NaCl and fill this up to 50 ml.
Sounds pretty easy, but is it correct or did I forget something?

[Borek]

Sounds OK, the only thing that bothers me is that it is pretty difficult to fill up to 50 mL in the test tube, better use some kind of volumetric glass

[AWK]

To be more precise - calculate activity coefficient f(H3O+) for ionic strength 0.01, then solve equation for concentration of H3O+ [H3O+]
a(H3O+)= 0.001 = f(H3O+) x [H3O+]

assuming [HCl] = [H3O+]
add enough NaCl to obtain
[HCl] + [NaCl] = 0.01

[Manie28]

OK, thanks a lot for your *delete me*

[Borek]

In general AWK is right, but if you are satisfied with pH 2.8-3.2 activity coefficients won't change pH that much.

[AWK]

Borek is right. My example allows calculation of pH within 0.01-0.02 unit of pH or concentrations with error about 1% in the first iteration .
Note ~5 % error in concentration is equivalent to ~0.03 unit of pH

[Borek]

to be exact...

[AWK]

Good result, but assuming the ionic force comes from eg KNO3, but not NaCl+HCl (this aproximation is, of course, quite sufficient to solve student problems)