>Actually Randy I just can't seem to understand how so little could buffer
it, and would like MORE chemistry.

Thanks, Stan--you beat me to the punch.  I took the erroneous number that I
posted from a graph that Dr LeBlanc had hastily made for me at my request.
Immediately after I sent the post, I too questioned the tiny amount, so
looked more carefully at the graph, and realized that there was an
error--the amount of NaHCO3 went up as pH went up!  Obviously the inverse of
what would be expected.

So I fired off a request to Dr LeBlanc, who quickly returned a revised graph
(he'd made a simple omission in the formula that he used on the
spreadsheet).

The new figures make more sense.  For HFCS type 55 at ph 4, the amount of
NaHCO3 necessary to buffer to neutrality would be 2.77g.

So I wondered why this figure differed from yours, since your calcs looked
right to me.  Upon second inspection of your calcs, I realized that you
neglected to account for the amount of solids in the solution, which would
not effect pH.  So if we take the 66% solids out, you have 33% water.  Your
figure of 8.4g x .33 = 2.77g, so it appears that we have agreement between
the two means of calculation!!!

The amount would go up by a factor of 10 for each integer drop in pH, and
correspondingly down for less acidic syrups.  However, I will wait for Dr.
LeBlanc to publish the actual amounts.

Thanks for your keen eye, Stan!  I should have trusted my gut and questioned
the figure before I fired off the post.

Randy Oliver

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