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Author Topic: calculation problem in the preparing of phosphate buffer  (Read 20082 times)

Offline wlyne

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i have NaH2PO4,molecular weight is 156.01 and the NaOH with 1.0M...
if i want to prepare 25mM phosphate buffer, pH7.4
how much of the solutions i have to use to add up the total volume is 100mL?
after prepare the phosphate buffer,i need to add 1.0mM EDTA (MW = 292.25 ), 0.1mM PMSF (MW = 174.19)and 0.1mM DTT (MW = 154.25) into the phosphate buffer to add up until 50mL...
how much of EDTA, PMSF and DTT i need to add in to the phosphate buffer?


 

Offline dentstudent

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calculation problem in the preparing of phosphate buffer
« Reply #1 on: 31/07/2007 14:41:59 »
*reads post, quietly shuts door hoping nobody saw him come in........slips note under the door saying*

Hi Wlyne, and good luck!
 

Offline Cut Chemist

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calculation problem in the preparing of phosphate buffer
« Reply #2 on: 31/07/2007 15:45:45 »
In our lab we make phosphate buffer using both mono and dibasic sodium phosphate.

monobasic = NaH2PO4*H2O, M.W. = 137.99

dibasic = Na2HPO4, M.W. = 141.96

0.026 M of monobasic and 0.041 M of dibasic will produce a buffer with a pH of 7.2 - 7.4

But I believe you can make a buffer with just the monobasic. 
(I'm using using your molecular weights. Make sure they're correct.)

25 mM = 0.025M * 156 (g/mol) = 3.90 g/L * 0.100 L = 0.390 g of NaH2PO4

After you add the water and mix thouroughly, you would have to adjust the pH using a pH meter, adding NaOH until the pH reaches 7.4.  Note: it doesn't matter how much NaOH you add, it will not change the molarity of the phosphate, but adding more and more volume will dilute it.  Ideally, the total volume after adjusting the pH should be 100 mL.

« Last Edit: 01/08/2007 04:07:17 by Cut Chemist »
 

Offline Cut Chemist

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calculation problem in the preparing of phosphate buffer
« Reply #3 on: 31/07/2007 16:00:24 »
You use the same equations to calculate the weights of the other chemicals

EDTA:

1.0 mM = 0.001 M = 0.001 mol/L * 292.25 g/mol = 0.29225 g/L * 0.050 L = 0.0146 g EDTA

Note: 0.050 L = 50 mL (...I think that's how much you needed.)

Also 0.0145 g = 14.5 mg = a very small amount to weigh out!!  If you don't have a very sensitive/accurate balance I would suggest making a stock solution and then dilluting.

C1 * V1 = C2 * V2

« Last Edit: 31/07/2007 16:06:25 by Cut Chemist »
 

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calculation problem in the preparing of phosphate buffer
« Reply #3 on: 31/07/2007 16:00:24 »

 

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