You can tell I'm not a chemist. The Wine Thief article was using a test
kit and there wasn't any info on the chemicals used, just how much NAOH
had been used so I felt that this was probably why it was different.
Both you and Jeff Cox specify the .1N NAOH but he used a 15ml sample.
Because it was said that dilution of the material if red wine was OK to
do to make it easier to read, I didn't pick up on the sample size
difference.

Sorry I quoted your multiple wrong. I realized that I'd put the . in
the wrong place as I hit send. Thanks for the help.

Dick



Lum Eisenman wrote:
Dick,
There are two variables. The size of the wine sample and the
"normality" of the sodium hydroxide.
If a 5 ml wine sample and if the normality of the sodium hydroxide is
0.1, then the TA equals the amount of sodium hydroxide in milliliters
used in the titration multiplied by 0.15.
Lum

"Dick Heckman" wrote in message
...
I'm confused. I've read 3 test procedures and gotten 3 ways to do the
test. Lum says to multiply the ml of naoh by 1.5. Jeff Cox says
divide by 2 and somebody posted a procedure here from the Wine Thief
which I copied some time ago which says the the amount used is the TA
in ml/1000. Help!

Dick