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calculation of salts in potassium phosphate buffer


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#1 Makosad

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Posted 29 May 2013 - 05:02 AM

hello guys

I made potassium phosphate buffer, 1 M stock solution as indicated:

First I made 1 M stock solutions of KH2PO4 and 1M K2HPO4.3H2O seperately. (book says use K2HPO4 but we have the trihydrate salt in lab, so I used the one available, i.e. K2HPO4.3H2O) (Here I found a link to recipe, http://www.unl.edu/c...hate buffer.pdf)

To make buffer of pH of 7.4, I need to add 802 ml of K2HPO4. 3H2O and 198 ml of KH2PO4., well, i needed buffer of 7.5 pH so I added a bit more of K2HPO4 till the pH was set at 7.5. In principle, the molarity of both solutions is same so final molarity will stay the same, i.e. at 1 M.

Now, this is my 1M stock and usually I dilute it as I require, usually 50mM or 20 mM.

I make a buffer solution and make my enzyme preparation in this buffer.
Questions is:
If I use 20 mM buffer, e.g 5 ml of 20 mM phosphate buffer solution, how much actual salts are present in there?

it is important for me b/c I am lyophilizing enzyme in buffer solution, and I want to calculate how much w/w percent of salts is present.

thanks in advance.
kind regards

P.S. what you think after lyophilization, the K2HPO4.3H2O is reduced to K2HPO4 (without waters of hydration, or they will still be there)?

#2 mdfenko

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Posted 29 May 2013 - 06:59 AM

in answer to your first question (although not asked), using the trihydrate is fine as long as you used the formula weight for the trihydrate, not the anhydrous.

the concentration of "salt" present is the concentration of the phosphate buffer plus any other salts added. to determine the weight of "salt" you just have to back-calculate (to be precise, you have to know the proportion of mono and dibasic, their formula weights, concentration and volume).

i believe you can remove the waters of hydration by heating. i don't think lyophilization will remove it.
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#3 El Crazy Xabi

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Posted 29 May 2013 - 10:01 PM

K2HPO4 and K2HPO4.3H2O are the same if you use them in solution, just check the molecular weight, 1M of the first is the same to 1 M of the second once in aqueous solution

Preparation of phosphate buffer is based on the mixing of equimolar solutions of both salts, knowing the proportions you can know the pH
http://www.uslims.ut.../po4buffers.php

In 20mM you can consider density equal to the water, just calculate how much salts you have in solution (forget about the hydration)

#4 mdfenko

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Posted 30 May 2013 - 06:57 AM

K2HPO4 and K2HPO4.3H2O are the same if you use them in solution, just check the molecular weight, 1M of the first is the same to 1 M of the second once in aqueous solution

Preparation of phosphate buffer is based on the mixing of equimolar solutions of both salts, knowing the proportions you can know the pH
http://www.uslims.ut.../po4buffers.php

In 20mM you can consider density equal to the water, just calculate how much salts you have in solution (forget about the hydration)

i am well aware of these points (i've been preparing phosphate buffers since the early 70's) but makosad was asking about calculating actual mass, not concentration.

nice tables, though

Edited by mdfenko, 30 May 2013 - 06:59 AM.

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#5 Makosad

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Posted 10 June 2013 - 01:20 AM

Thanks for the replies.

It turns out, next time I make the buffer, I have to keep track of total volume, as the final volume is not 1L but few ml extra because of the pH adjustment I was trying to achieve. If I just mix the two solutions as it is mentioned in the table, which I actually did in the beginning, the pH is not exactly the same everytime so it needs some adjustment by adding one or the other solution.
Now I will try to lyophlize the empty buffer (without my enzyme) in same volume and see how much does it weigh. Probably, this can tell me how much salts is present in a given amount of buffer.

#6 mdfenko

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Posted 10 June 2013 - 04:18 AM

unless you are preparing the buffer directly into the tube that you want to weigh, the total volume of the phosphate buffer stock you prepare is unimportant (since you decided to weigh rather than calculate the mass).
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