Calculating ionic strength for phosphate buffers

[hollislau]

Hi all,

Since my acid/base chemistry with tripotic systems is not quite up to snuff, I was hoping somebody could help me out with calculating the ionic strength of the phosphate buffers that I use for SEC. I have two related questions:

1) How do I calculate the ionic strength of a 100mM sodium phosphate, 200mM sodium chloride, pH 7.0 buffer? (Activity coefficients can be disregarded if the effect is small)

2) If I prepare the above buffer by adding NaOH to a solution of 100mM monobasic sodium phosphate, 200mM NaCl to reach pH 7.0, will I get the same ionic strength (and molar ratio of constituent ions) compared to a buffer prepared by mixing 100mM monobasic phosphate, 200mM NaCl with 100mM dibasic phosphate, 200mM NaCl to achieve pH 7.0? (My guess is yes)

Any responses would be greatly appreciated! Thanks!

[HW Mueller]

To 1), try

http://www.liv.ac.uk/buffers/buffercalc.html


To 2), how are you adding NaOH?

[HW Mueller]

Didn´t see this first time: If you add base it should be to the NaH2PO4, not the Na2HPO4. Also, at those concentrations there should be diffs between concentration and activitiy. Very tedious calcs if one has to do it by "hand".

[hollislau]

Hi, thanks for posting the link to the calculator - that should come in handy in the future. However, I would still like to understand how the ionic strength is calculated for my own edification. I know that the calculations are very tedious, but if anybody has the time or inclination I'd really love to learn the theory.

To your question, let's assume that I weigh out the appropriate amount of NaH2PO4 and NaCl for 100mM and 200mM respectively, dissolve in water (80% of my final volume), pH to 7.0 with 10N NaOH, then bring the solution up to my final volume with water. Or I suppose NaOH pellets could be added to adjust pH once the NaH2PO4 and NaCl have been dissolved in my final volume. Again, this is really just for my understanding - I'm just trying to figure out if both preparation methods would yield the same molar ratio of ions, and consequently same ionic strength, if the buffers were prepared perfectly.

[Uwe Neue]

Hmmh... is this a school project? Has been a while for me...

Question 1, ionic strength:
Ionic strength is defines as
I = 1/2 * sum (n(i)^2 * c(i))
where n(i) is the charge for every species i, and c(i) is its concentration.
In the first case, we have 100 mM sodium phosphate buffer, which I interpret as meaning that it is a 100 mM phosphate buffer, not 100 mM in sodium. At pH 7, you have an equal molar amount of (H2PO4)- and (HPO4)2-, 50 mM each. This translates to 50+100 mM Na+ from the buffer you also have 200 mM of Cl- and 200 mM of additional Na+ form the NaCl, which amounts to a total of 350 mM Na+

If I add this up, I get
I = 1/2 * (1*0.05 + 4*0.05 + 1*0.35 + 1*0.2)
for the singly charged phosphate, the doubly charged phosphate, the sodium and the chloride. I am too lazy to add this up.

Question 2:
You will get the same ionic strength, provided that your first preparation will have a FINAL concentration of 100 mM phosphate buffer, i.e. if you add solid NaOH to your 100 mM phosphate solution. Or, if you prepare a solution of phosphate that will be 100 mM after pH adjustment with NaOH and dilution to 1L.

[HW Mueller]

Maybe you should spend a few weeks with a text on chemical equilibrium.

I am not only too lazy, but it hurts to do these exactly, in my case it is even worse if I try to make all the assumptions often made to simplify the calcs.
For instance, I am at a loss to see the assumption that went into Uwe´s statement that at pH = 7 the mono and dibasic species are at equal concentrations. I put 0.1M Phosphate (memory tricked me, thought the example here was 0.1M) at pH =7, at 20°, ionic strength not specified, I got the following recipe:
Ionic strength = 0.223M
use 0.0382 mol acid (NaH2PO4)
use 0.06M base (Na2HPO4)

Now I can´t see how this will equilibrate to equal acid and base. I recommend to hollislau to start to play a bit by putting these figures into the ionic strength equation and see how it compares with Beynon´s figure.

Now it is correct that the species will be the same with the prep. methods mentioned here if everything is done nearly perfectly. For instance, it is usually close enough if one titrates ~perfectly to pH = 7 , here, with a strong base or acid and then topping it off: If you overshoot your titration you will have to add more acid or base, that changes the conc. of buffer and thus its ionic strength. If you use equal moles of mono and dibasic (moles phosphate) you can overshoot the titration a theoretically unlimitet times and still not change the ionic strength. Nevertheless, it has been pointed out many times here that the best method is to have the calculator figure out the M and dilute to the 1L mark.

[Uwe Neue]

Hans, I see that you are an analytical chemist
But the temperature is not specified, and maybe at the target temperature the pKa of phosphate is at 7.0... After all, he is getting a pH of 7.0 when he mixes equimolar quantities of mono and dibasic phosphate, right?

I would not rely on silly calculators at a university website, if I can do the calculation myself.

[Don_Hilton]

The "Easy" equilibrium calculation is the Henderson–Hasselbalch equation:

pH = pKa + log[base]/[acid]

And, the pKa of H2PO4- is 7.21.

at pH7, the log of [HPO4--] to [H2PO4-] is -0.21, giving a ratio of 0.62:1 base to acid.

If you want to get more acurate with the answer, you can skip over the assumptions made in the Henderson–Hasselbalch equation, such as with that shown in http://www.chembuddy.com/?left=pH-calcu ... -acid-base or http://www.chembuddy.com/?left=pH-calcu ... t-solution. Now you can use the pKa of H3PO4 (2.12) and HPO4 (12.32) - which will have minor contributios at pH 7.

Unfortunately, the web site with the calculator does not give their method of calculation.

The only question is how many places after the decimal do you want (or need).

[Don_Hilton]

And - the sodium from the NaCl enters in - or not... Which I forgot to account for as I was looking up the equilibrium equation. And it increases the math.... I'll need to come back to that later. There are days...

[Uwe Neue]

OK, reading it again, it did not say that you want to mix equimolar amounts of monobasic and dibasic buffers to get pH 7.0; it only said that you want to mix 100 mM solutions of both to get pH 7.0. So Don is right. You need to plug Don's concentrations into the ionic strength equation.

[HW Mueller]

Don, your link and your statement, "... the web site with the calculator does not give their method of calculation." sort of seconds what I tried to say: The calcs are not trivial.

For instance,
the H-H equation is merely a restatement (rearrangement) of the equation for the equilibrium constant. As such it ignores the contribution of the H2O equilibrium (Kw). So the assumption nr. 1 is that at 50mM the Kw doesn´t contribute, apparently not too bad. But at such concentrations the deviation of activity and concentration may not be negligible.
Also the assumption nr. 2 neglects H3PO4 and PO4---, also probably not bad at pH 7.
Thus: I am glad such silly calculators exist, I don´t have to figure out such approximations.

The catch with the calculator of my link: There is a book one needs to buy if one wants to find out how the calcs are done. I know from the author personally that he doesn´t ignore activities.

Fortunatly for chromatography this is not all that important, one has to find a pH at which ones stuff works, and now the important part, you and others have to be able to repeat this buffer prep.

(Incidentally: Can one imagine today that you rearrange a common equation, publish that and get your name attached to it for all times?)

[zokitano]

Don, your link and your statement, "... the web site with the calculator does not give their method of calculation." sort of seconds what I tried to say: The calcs are not trivial.

The catch with the calculator of my link: There is a book one needs to buy if one wants to find out how the calcs are done. I know from the author personally that he doesn´t ignore activities.

Dear Hans,

could you share with us the name of the book you're refering to in your previous post?

Thanks

[HW Mueller]

Check "References" at the calculation site or maybe this works:

http://www.liv.ac.uk/buffers/book.html

[zokitano]

Check "References" at the calculation site or maybe this works:

http://www.liv.ac.uk/buffers/book.html

Thanks for the information

Regards

[hollislau]

Thanks so much to everybody for the great responses! I learned a lot.