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Topic: Phosphate Buffer -- Ionic Strength  (Read 6216 times)

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Offline slvcs

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Phosphate Buffer -- Ionic Strength
« on: October 30, 2007, 01:32:54 AM »
Where do i start with this question I am kind of lost.


Given that the stepwise dissociation constants for phosphoric acid are:
Ka1 = 7.5×10-3; Ka2 = 6.2×10-8; Ka3 = 4.8×10-13

To prepare 1.20 L of a buffer solution having an ionic strength of 0.150 and a pH of 7.25 would require:

find (mass) of NaH2PO4(anhydrous) and find (mass) of Na2HPO4(anhydrous).


THANKS   

Offline AWK

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Re: Phosphate Buffer -- Ionic Strength
« Reply #1 on: October 30, 2007, 02:10:50 AM »
Use Henderson-Hasselbach equation to obtain ratio [H2PO4- ]/[HPO42- ]
An ionic strenth is equal to [H2PO4- ] + 3[HPO42- ].

Hence you have a set of equation of type
x/y=a
and
x+3y=0.15

This attempt is of course only (byt quite good) approximation

Use K2
AWK

Offline slvcs

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Re: Phosphate Buffer -- Ionic Strength
« Reply #2 on: October 30, 2007, 02:28:19 AM »
why An ionic strenth is equal to [H2PO4- ] + 3[HPO42- ].

1:3 ratio ?  is it given ?

Offline AWK

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Re: Phosphate Buffer -- Ionic Strength
« Reply #3 on: October 30, 2007, 03:09:17 AM »
I=1/2 sigma(ciZi2)= 2 x c1 x 12 + 2 x c2 x 12 + c2 x 22 = c1 + 3 x c2
AWK

Offline Borek

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Re: Phosphate Buffer -- Ionic Strength
« Reply #4 on: October 30, 2007, 04:39:08 AM »
why An ionic strenth is equal to [H2PO4- ] + 3[HPO42- ].

1:3 ratio ?  is it given ?

Check what is ionic strength definition, formula given by AWK simply follows.

NaH2PO4 is dissociated to Na+ and H2PO4- - in identical concentrations. Assume NaH2PO4 concentration is x - what will be its share in the ionic strength?

Do the same for Na2HPO4, assuming concentration is y - you should get 3y.

Thus total ionic strength in this solution is x + 3y.
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