[hopes]

how does citric acid (C6H8O7*H2O)and sodium phosphate (Na2HPO4*7H2O) react with each other to act as a buffer system? according to;
pH= pKa+log [base]/[acid], i guess "acid" must be the citric acid, but what is the "base"? sodium phosphate can not be the "base" since it is not the conjugate base of citric acid. hope you can explain this.

thanks.

p.s i have to make citrate phosphate buffer 0.1M, pH 5.6. can you explain how you make it? thanks again.

[Demotivator]

This is a mixed buffer of two systems commonly used in biology. Citric is the acid. Na2HPO4 is the base (it's not the conjugate base of citric but it does act as a weak base nevertheless ie it accepts protons to form NaH2PO4).
For ph 5.4
Add 557.5 mL of 0.2 M Na2HPO4 to 442.5 mL of 0.1 M citric acid.
Test the ph. If off, adjust citric as needed.
From
http://www.rockystar.com/home/chemistry/chem_prep.htm

[hopes]

For ph 5.4
Add 557.5 mL of 0.2 M Na2HPO4 to 442.5 mL of 0.1 M citric acid.
Test the ph. If off, adjust citric as needed.
From
http://www.rockystar.com/home/chemistry/chem_prep.htm

Demotivator,
many thanks to you for your advice, but i hope you and the others will keep guiding me because i have spent alot of time with these questions but can't find any good answers:

my buffer has to be at 0.1M, pH 5.6.

does this mean that:

1.  the sum of the final concentration of citric acid and sodium phosphate together must be 0.1M?

2. when i add 557.5 mL of 0.2 M Na2HPO4 to 442.5 mL of 0.1 M citric acid. what is the final concentration of citric acid and sodium phosphate?

my calculation:
final concentration of citric acid:( 0.1M X0.4425L)/ 1L=0.04425M

final concentration of sodium phosphate: (0.2M X 0.5575L)/1L= 0.1115M

according to my answers then after adding the solutions to each other, the buffer concentration now is: 0.04425M + 0.1115M= 0.1556M. is this correct?

to get the concentration of the  buffer to 0.1M, then i have to dilute it with dH20:

the buffer volum presently : 1L
the buffer concentration presently: 0.1556M

0.1556Mx1L/X=0.1M
X=1,556L

1.556L-1L=0.556L

this means i have to add 556ml of dH20. is this the way thing works out or what?

i appreciate for any replies and ideas!
thanks again so much!

[Demotivator]

You're correct, the sum should be 0.1M so dilute it (Or start with scaled down quantities for a final volume of 1L).
Then, If the ph is a little low, adjust with a few drops of conc NaOH.

[hopes]

hmmmm....are you sure that everything is right from the top to the bottom? also the calculations? Gosh, i can't believe it!  :jump: :jump: :jump:

thanks a bunch!

[hopes]

This is a mixed buffer of two systems commonly used in biology. Citric is the acid. Na2HPO4 is the base (it's not the conjugate base of citric but it does act as a weak base nevertheless ie it accepts protons to form NaH2PO4).

Demotivator ,

even if i did this right, but i don't find it logically, because when we mix citric acid and Na2HPO4, then the sodium phosphate will act as a base and takes a proton from the citric acid. this means the reaction between citric acid and  sodium phosphate will act this way;


let us call citric acid as HCA,
start mol of HCA:  0.1M X0.4425L= 0.044mol.
start mol HPO4 (2-): 0.2M X 0.5575L= 0.11mol

         HCA   +    HPO4(2-)   ->   CA-   +    H2PO4-

start:  0.044mol + 0.11mol -> 0mol + 0mol

equilibrium: 0mol+0.066mol -> 0,044mol + 0.044mol

the total volum when we mix 557.5 mL of 0.2 M Na2HPO4 to 442.5 mL of 0.1 M citric acid is 1L

at equilibrium the concentration of citric acid (HCA): 0mol / 1L =0M.

at equilibrium the concentration of HPO4 (2-): 0.066mol/1L =0.066M

this mean when i mix the solutions together, then the buffer concentration is 0.066M+0M= 0.066M, which is different than  what i calculated before (0.1556M)


do you have any explanations of this, please?

thanks alot!

[Demotivator]

Citric is triacidic, pKa = 3.15, 4.77, 5.19.
The phosphate reacts with it and reduces its acidity probably to the monacidic form. The end result is a still a weak acid and base.
Whatever the species are at equilibrium, they add up to the same quantity as the starting materials.

[hopes]

Whatever the species are at equilibrium, they add up to the same quantity as the starting materials.

[buffer ] at equilibrium is ; [HCA]  +   [HPO4(2-)]  +  [CA-]  +    [H2PO4-]

0M+0.066M + 0,044M + 0.044M= 0.154M,
which is the buffer's  concentration. is this what you mean?
 :g:

[Demotivator]

I think it's more like (the numbers are only an approximation because we're employing the assumption of a complete reaction):
[buffer] at equilibrium is ;[H3CA] + [H2CA-] + [HCA(2-)]  +  [HPO4(2-)]    +    [H2PO4-] +  [CA-]
0 + 0? + .044 + .022 + .088  + 0? = .154

where 0? means maybe or close to. (If greater than 0 then the other species would be of lesser amount to balance to the same total.) To get the actual numbers at equilibrium, you'd have to solve the actual equlibrium equations for many steps and that is complicated for this system. But there is no reason to actually solve the equations unless you want to predict the PH theoretically.

[hopes]

my last question  :angel2:


so instead of calculating the concentrations of all species at equilibrium to find the buffer's concentration, we should instead find the final concentration of the starting materials (citric acid and sodium phosphate) as i did?

hope you don't lose your patience, but i think i have got the point now   :rolleyes:

[Demotivator]

Yes, provided you have the mixing recipe.
OK, you're looking for more convincing.
Here is an example of a 0.07M citrate phosphate buffer prep from a web site(scroll down).
http://www.precisionenv.com/PDFS/dustmitestudy.pdf
A calculation proves that the sum of the ingredients is the concentration of the stated buffer:
[0.1(.147) + 0.2(.103)]moles/0.5 liter= 0.07M
for the buffer as given:
70mM citrate-phosphate, ph 4.2
Solution A= 0.1M anhydrous citric acid, 19.21g/L
Solution B= 0.2M Dibasic Na Phospate. 7H2O, 53.65 g/L
For 500ml buffer, mix 147ml A + 103ml B and make up to 500ml with deionized H2O  

BTW, Also this site has a better table of the citrate phosphate buffer recipies.
http://www.istonline.org.uk/Handbook/21-22.pdf

[hopes]

thank you very much, Demotivator! you are a real   :angel1:  

thanks alot!

[Borek]

Calculating the recipe by hand is out of the question, but see the video:

http://www.youtube.com/watch?v=DLJjXVuvqsI

http://www.chembuddy.com/?left=Buffer-Maker&right=buffer-calculator