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Specialty Chemistry Forums => Biochemistry and Chemical Biology Forum => Topic started by: SpellForce on June 23, 2014, 11:32:17 AM

Title: Ionic strength question
Post by: SpellForce on June 23, 2014, 11:32:17 AM
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-, HPO42-, H+) from KH2PO4 and (K+, HPO42-, PO43-, 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:
Iwanted = 0.5 (Σci · zi + 2x)
Where x is my concentration of NaCl I need to add?

Thanks.
Title: Re: Ionic strength question
Post by: Babcock_Hall on June 23, 2014, 02:06:05 PM
The formula for ionic strength I should include the square of the charge, which is zi in your formula.  I = (0.5)Σci(zi)2.  For ions that have a charge-magnitude of 2 or more, the presence of the square becomes important.
Title: Re: Ionic strength question
Post by: Borek on June 23, 2014, 02:19:20 PM
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 (http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator) does it automatically.

I used ions (K+, H2PO4-, HPO42-, H+) from KH2PO4 and (K+, HPO42-, PO43-, H+) from K2HPO4.

You are doing something strange. H2PO4- and HPO42- are just there, so using them is OK. PO43- 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.
Title: Re: Ionic strength question
Post by: AWK on June 24, 2014, 02:36:34 AM
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)
Title: Re: Ionic strength question
Post by: SpellForce on June 24, 2014, 04:35:13 AM
The formula for ionic strength I should include the square of the charge, which is zi in your formula.  I = (0.5)Σci(zi)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...
Title: Re: Ionic strength question
Post by: SpellForce on June 24, 2014, 04:48:05 AM
You are doing something strange. H2PO4- and HPO42- are just there, so using them is OK. PO43- 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.



Title: Re: Ionic strength question
Post by: Borek on June 24, 2014, 05:17:41 AM
You are doing something strange. H2PO4- and HPO42- are just there, so using them is OK. PO43- 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.

Quote
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]
Title: Re: Ionic strength question
Post by: SpellForce on June 24, 2014, 05:50:21 AM
Well, if I write the dissociation with hydrolysis, then I get 3c, but without it (K2HPO4  :rarrow: 2K+ + HPO42-), then it is 5c?
If it is 2K+, then I use 2c·(1)1 for K, so it would be 6c?
Title: Re: Ionic strength question
Post by: Borek on June 24, 2014, 06:08:05 AM
You are doing something wrong, please show your work step by step.
Title: Re: Ionic strength question
Post by: SpellForce on June 24, 2014, 07:12:34 AM
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).
Title: Re: Ionic strength question
Post by: Borek on June 24, 2014, 07:39:41 AM
What do you mean by hydrolysis?

And I have no idea what is the problem with the second salt, it is identical in principle.
Title: Re: Ionic strength question
Post by: SpellForce on June 24, 2014, 09:12:39 AM
I guess HPO42- + H2O  ::equil:: 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?
Title: Re: Ionic strength question
Post by: Borek on June 24, 2014, 10:32:42 AM
I guess HPO42- + H2O  ::equil:: H2PO4- + OH- ?

OK

Quote
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.
Title: Re: Ionic strength question
Post by: SpellForce on June 25, 2014, 11:19:28 AM
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 :D
Title: Re: Ionic strength question
Post by: Borek on June 25, 2014, 12:32:36 PM
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.

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

I don't think MS will help.
Title: Re: Ionic strength question
Post by: SpellForce on June 27, 2014, 07:35:10 AM
And what is the best source to look for pKa values? I found somewhere that pKa2 of phosphoric acid is 7.2, and somewhere its 6.8...
Title: Re: Ionic strength question
Post by: SpellForce on June 27, 2014, 10:50:01 AM
Never mind, I found a way to calculate it :D
Title: Re: Ionic strength question
Post by: Borek on June 27, 2014, 12:33:27 PM
And what is the best source to look for pKa values? I found somewhere that pKa2 of phosphoric acid is 7.2, and somewhere its 6.8...

Most likely these are values listed form different ionic strengths. For IS=0 it should be 7.199 if memory serves me well.
Title: Re: Ionic strength question
Post by: Babcock_Hall on September 16, 2016, 05:49:54 PM
Buffer Solutions: The Basics, by Beynon and Easterly has the best intermediate-level discussion of the effect of ionic strength on pKa that I have found.  They present a formula for estimating the shift in pKa with ionic strength.