0.15M Borate Buffer pH 8.3 - (Nov/04/2009 )

I decided to create this buffer based on a few different protocols I'd read. There seem to be so many different ways of making this buffer - so I decided to try work out how to make it using the good old classic equations. Does anyone have a better recipe? Please point out any flaws in the method I used (below).

0.15M Borate Buffer pH 8.3

pH = pKa + log /
= +
In this case: Borax (MW = 381.37) is the salt or weak base, Boric Acid (MW = 61.83) is the weak acid.

pKa of Boric Acid = 9.14
Desired pH = 8.3
Desired = 0.15M

8.3 = 9.14 + log /
antilog(8.3 - 9.14) = /
0.145 = /

To obtain ratio of salt to acid...
1/0.145 = 6.9
Therefore...
1/6.9 = 0.145
Apply these values to: = +
0.15M = 1x + 6.9x
1 + 6.9 = 7.9
x = 0.15/7.9 = 0.0189
0.015M = 0.019M + 0.131M

Determine the weight of Borax and Boric Acid required using:
n = C V/1000
n x MW = mass
(n, number of moles; mass, mass in grams; MW, molecular weight; C, molar concentration; V, volume in millilitres)

For example if you want to create 100ml of 0.15M Borate Buffer pH 8.3:
Borax (MW = 381.37)
0.019 x 100 /1000 = 0.0019
0.0019 x 381.37 = 0.72g

Boric Acid (MW = 61.83)
0.131 x 100 /1000 = 0.0131
0.0131 x 61.83 = 0.81g

i.e. dissolve 0.72g Borax and 0.81g Boric Acid in 100ml of water.

i would just take boric acid and adjust to the desired pH with base (naoh or koh, whichever i want).

mdfenko on Nov 4 2009, 09:41 PM said:

Thanks for your reply - your method is certainly much easier - however would this method affect the properties of the buffer e.g. buffer capacity etc.?

BMandelsonn on Nov 6 2009, 07:14 AM said:

not at all. you are just converting some of the "acidic component" into the "basic component". the final concentration of the buffering species will determine capacity.