[SpellForce]

Hello,

I would like to ask quite basic question for buffer makers.
If I make a phosphate buffer, from certain amount of KH₂PO₄ and K₂HPO₄, mix them together, adjust pH and then add salts (NaCl) to adjust ionic strength... Or adjust ionic strength first and then pH... Anyway...

How do I calculate current ionic strength if I have both KH₂PO₄ and K₂HPO₄ present? I mean, who knows whats going on there, does it change in time?

I tried to do it, and for concentrations of
3.2·10^-3 M KH₂PO₄
4.8·10^-4 M K₂HPO₄
I get, I = 0.0175, because I used ions (K⁺, H₂PO₄^-, HPO₄^2-, H⁺) from KH₂PO₄ and (K⁺, HPO₄^2-, PO₄^3-, H⁺) from K₂HPO₄.
I am sure I didnt do it correctly, so please let me know what ions should I count in...

And then, the second part of my question would be to find out the concentration of salts, in order to adjust it to a certain ionic strength,
but I believe it would be something like:
I_wanted = 0.5 (Σc_i · z_i + 2x)
Where x is my concentration of NaCl I need to add?

Thanks.

[Babcock_Hall]

The formula for ionic strength I should include the square of the charge, which is z_i in your formula.  I = (0.5)Σc_i(z_i)².  For ions that have a charge-magnitude of 2 or more, the presence of the square becomes important.

[Borek]

Apart from a trivial (but important) mistake in the formula, you need to do calculations iteratively - ionic strength changes activities of ions, so it moves the equilibrium, when the equilibrium moves, you should recalculate ionic strength and so on. Usually two or threes steps are enough.

Buffer Maker does it automatically.

I used ions (K⁺, H₂PO₄^-, HPO₄^2-, H⁺) from KH₂PO₄ and (K⁺, HPO₄^2-, PO₄^3-, H⁺) from K₂HPO₄.

You are doing something strange. H₂PO₄^- and HPO₄^2- are just there, so using them is OK. PO₄^3- concentration is negligible. Looks liek you are adding dissociations products to each salt - you don't have to, in the first iteration step you can assume concentrations of all ions are those introduced into the solution. Next step requires recalculating equilibrium.

[AWK]

You can start with an approximation I = 3c(K₂HPO₄) + c(KH₂PO₄ + c(NaCl)).
This is exactly equivalent formula you used for your calculation (should be 0,0176; c(H⁺) is of order 10^-7 and can be safely neglected)

[SpellForce]

The formula for ionic strength I should include the square of the charge, which is z_i in your formula.  I = (0.5)Σc_i(z_i)².  For ions that have a charge-magnitude of 2 or more, the presence of the square becomes important.

Yea I forgot to put it in the formula, but I used it in calculations...

[SpellForce]

You are doing something strange. H₂PO₄^- and HPO₄^2- are just there, so using them is OK. PO₄^3- concentration is negligible. Looks liek you are adding dissociations products to each salt - you don't have to, in the first iteration step you can assume concentrations of all ions are those introduced into the solution. Next step requires recalculating equilibrium.

So you are saying that I should include only first steps of dissociation? What about the starting chemicals?
Could you just write me what ions should I use in calculations for KH₂PO₄ mixed with an order of magnitude less K₂HPO₄?

You can start with an approximation I = 3c(K₂HPO₄) + c(KH₂PO₄ + c(NaCl)).
This is exactly equivalent formula you used for your calculation (should be 0,0176; c(H⁺) is of order 10^-7 and can be safely neglected)

Ok, but why 3 concentrations of K₂HPO₄ and one of KH₂PO₄?
Thanks.

[Borek]

You are doing something strange. H₂PO₄^- and HPO₄^2- are just there, so using them is OK. PO₄^3- concentration is negligible. Looks liek you are adding dissociations products to each salt - you don't have to, in the first iteration step you can assume concentrations of all ions are those introduced into the solution. Next step requires recalculating equilibrium.

So you are saying that I should include only first steps of dissociation? What about the starting chemicals?

No, in the first step you use just a starting concentrations (that's what AWK did). Later, once you know ionic strength of the solution it changes value of the observed Ka, so you need to recalculate the equilibrium.

You can start with an approximation I = 3c(K₂HPO₄) + c(KH₂PO₄ + c(NaCl)).
This is exactly equivalent formula you used for your calculation (should be 0,0176; c(H⁺) is of order 10^-7 and can be safely neglected)

Ok, but why 3 concentrations of K₂HPO₄ and one of KH₂PO₄?

Assuming you have just a solution of K₂HPO₄ of concentration c, and there are no further shifts of equilibrium, can you express ionic strength of the solution using c? (Hint: just calculate concentrations of all ions from the obvious salt dissociation and plug them into the ionic strength formula).

For example, for a solution of NaCl with concentration C you have Na⁺ and Cl^- in the solution, both with concentration C, when plugged into the definition it yields

[tex]I = \frac 1 2 ( C\times 1^2 + C\times 1^2 )[/tex]

[SpellForce]

Well, if I write the dissociation with hydrolysis, then I get 3c, but without it (K₂HPO₄  2K⁺ + HPO₄^2-), then it is 5c?
If it is 2K⁺, then I use 2c·(1)¹ for K, so it would be 6c?

[Borek]

You are doing something wrong, please show your work step by step.

[SpellForce]

Buahaha, Im just not dividing it by two. Ok, so I get 3c for not hydrolyzed. Is this another assumption, that there is no hydrolysis? I dont understand the first one either (that we assume there is only K₂HPO₄).

[Borek]

What do you mean by hydrolysis?

And I have no idea what is the problem with the second salt, it is identical in principle.

[SpellForce]

I guess HPO₄^2- + H₂O  H₂PO₄^- + OH^- ?

So I just add first-step dissociation products of both chemicals to the equation assuming a lot of stuff and I got my ionic strength?

[Borek]

I guess HPO₄^2- + H₂O  H₂PO₄^- + OH^- ?

OK

So I just add first-step dissociation products of both chemicals to the equation assuming a lot of stuff and I got my ionic strength?

Yes, but in general that will be only the first approximation.

[SpellForce]

So basically, calculation of ionic strength (and many more chemical equations) are just rough guessing? Unless you spend additional time and nerves in looking at dissociation constants etc. Cause you dont really know what u have in solution, the distribution of many ions. Is it hydrolyzed or not, at what stage of ionization is it etc. Only if I make a buffer, do a mass spec and then calculate freakin ionic strength

[Borek]

Unless you spend additional time and nerves in looking at dissociation constants

There are programs for that, you don't have to solve it manually.

Only if I make a buffer, do a mass spec and then calculate freakin ionic strength

I don't think MS will help.