Tuesday, May 31, 2011

How do you make a buffer solution with a PH of 7?

need to prepare 250 mL of the solution.



50 mL of the buffer solution will need yo be able to absorb 0.1 moles of additional H+ or OH- ions without changing th pH by more than 0.01 units.



Must design a procedure for preparing and testing the buffer.



available chemicals

12M HCl

15M NH3

15M Hc2H3O2

NaOH(s)

NaC2H3O2(s)

NH4Cl

Distilled WaterHow do you make a buffer solution with a PH of 7?
Buffers have their highest capacity (and thus should be used) for the range pH= pKa +/- 1.

Acetic acid has pKa=4.75

NH3 has pKb=4.75, so the conjugate acid (NH4+) has pKa=14-4.75 =9.25



Thus you cannot use a weak and a strong electrolyte, you have to use two weak ones. If you combine acetic acid (HC2H3O2) and NH3 in an equimolar fashion, you will get the salt CH3COONH4. This hydrolyzes and gives approximately

[H+]= squareroot (Kw*Ka(HC2H3O2)/Kb (NH3))

But for pKa (HC2H3O2)= pKb (NH3) =%26gt; Ka (HC2H3O2)= Kb(NH3) and [H+]=squareroot (Kw) =squareroot (10^-14) =10^-7 =%26gt;pH =7



You want to add 0.1/0.05= 2 mole/L H+ or OH- and have a change of less than 0.01 in pH.

Thus the buffering capacity should be

尾=dC/dpH= 2/0.01= 200



I think that according to

http://www.chembuddy.com/?left=pH-calcul



尾=2.303* (Kw/[H+] + [H+] + CaKa[H+] / ((Ka+[H+])^2) + CbKb[OH-] / ((Kb+[OH-])^2))



and since Ka=Kb, [H+]=10^-7=[OH-] and we need Ca=Cb so that we have only the salt remaining in the solution, we have



尾=2.303* (Kw/[H+] + [H+] + 2CaKa[H+] / ((Ka+[H+])^2)) =%26gt;

200 =2.303* (10^-7+10^-7+ 2C*10^-4.75*10^-7/((10^-4.75+10^-7)^2)) =%26gt;

100/2.303 -10^-7= C*10^-11.75 / ((10^-4.75+10^-7)^2) =%26gt;

C=((10^-4.75+10^-7)^2)*10^11.75*(100/2?=%26gt;

C=7808

This is an unrealistic value which means that maybe I didn't use the formula correctly or simply you are asking for extraordinary buffering capacity.



If we had a simple acetate buffer and we wanted the same buffering capacity at pH=pKa=4.75, we would need



200= 2.303*(10^-14/10^-4.75 + 10^-4.75 + C*10^-4.75*10^-4.75/ (10^-4.75+10^-4.75)^2) =%26gt;

200/2.303 = 10^-9.25+10^-4.75+ C/4 =%26gt;

C= 347 M. So you see that the buffering capacity you want is not feasible.How do you make a buffer solution with a PH of 7?
the site could answer your question with the examples on the following chemicals you give



http://www.csudh.edu/oliver/chemdata/buf
This intrigues me:

First of all pHs are almost impossible to read or adjust at .01 ???

Wow, to adjust a buffer to trim out at a range7.01 to 6.99 would be quite a stretch! I would guess that .01 Normal solutions of both acids and bases would have to be used to stabilize a solution at this level of accuracy.

Even a little ambient air CO2 migrating into the solution would drop the pH a smidgen. If you made the buffer with a lot more salt(from the acid-base reaction) and brought it accurately to 7.00 pH; I believe you would still not be able to hold it that stable.

We call his spurious accuracy.



Oh, I almost forgot, the two that you would use to make the attempt would be HCl and NaOH.
Add an Acidic compound
A salt formed due to neutralisation of weak acid with weak base behaves as simple buffer at pH = 7.

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