0.1 vs 0.2

I think you're being too simple about this. How much you add at a time will not effect the test results, it's the total amount of Sodium Hydroxide you add that matters. I think you could either reduce the amount of wine to 5 CC or take the volume of Hydroxide added and divide by 2.

Andy

I recommend you do the later of the above, since reducing the sample volume will minimise accuracy.

You can use the formula for any case:

TA = (75 * Cn * Vn) / Vs

where TA = titratable acidity (g/l as tartaric) Cn = NaOH concentration (normality) Vn = Volume NaOH added (ml) Vs = Volume of sample (ml)

Ben

Here are two concepts that will make your life and Acid Testing much simpler. They did mine:

1. When using .2 Normal NaOH (Sodium Hydroxide) reagent, the number of drops of reagent added will always equal the resultant TA% AS LONG AS THE SAMPLE SIZE IS 15 ML'S. Although one can draw this implication from reading many such instructions, if one fails to notice that what they all have in common is the 15 ml sample, a wine maker could mistakenly assume that the amount of NaOH dropped always equals the amount of titratable acid regardless of the sample size!

EXAMPLE:

Formula for titration is:

VOLUME OF NaOH ADDED X NaOH STRENGTH X .75 ____________________________________________

DIVIDED BY VOLUME OF THE SAMPLE

Sample computations:

6 x 20 x .75 = 90 divided by 15 = 6 TA

8 x 20 x .75 0 divided by 15 = 8 TA

Of course this only works with a 15 ML sample. A 10 ML sample would give a different result, i. e., the amount of NaOH dropped would not equal the TA result.

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1. If using .10Normal NaOH then different sample amounts have different math correction factors attached to them.

For example, if you are using .10 Normal NaOH reagent and taking a 3 ML sample, the math correction factor is .25. (See Presque Isle test instructions).

Whereas, if you are taking a 5 ML sample and using the same strength reagent, you would use a .15 factor to arrive at TA%. (See Crosbey & Baker test instructions).

A 10 ML sample of the same strength reagent (0.10N) would require that a math correction factor of .075 be applied.

There is no logical progression, there just appears to be one. You'd logocally think that 4 being betwwen 3 and 5, the correction factor would be .2 Not so. In fact if "x" is the math factor you seek, then when you solve the equation for "x", the formula is:

"X" = NaOH strenght X 7500 divided by 1000 X the sample size.

Applying this formula, the solutions for "X" (the math factors used to arrive at TA%) are as below:

At 1 ml, use .750, 2 ml use .375. 3 ml use .250, 4 ml's use .1875,

5 ml's use .150, 6 ml's use .125, 7 ml's use .1071428, 8 ml's use .09375, 9 ml's use .083333, and at 10 ml's use .075.

There is an advantage to using .10N reagent, you can be more discreet in your additions. Since .1N is 1/2 the strenght of .2N, you need to add twice as much and can therefore make more controlled additions and you are also then less likely to overshoot your reagent addition.

SIMPLER STILL IS TO BUY A pH METER AND TITRATE TO 8.2. WELL WORTH THE INVESTMENT; IT TAKES AWAY ALL OF THE ANGST OF LOOKING FOR COLOR CHANGE. Hope this helps. Vincent

Andy: Since he's using .10N (in place of .20N) would he not have to multiply by two rather than divide by 2?

No, the amount of NaOH required to titrate with 0.1 N is twice the amount required when using 0.2 N. So to get the amount would be required if using a 0.2 N NaOH solution but when actually using a 0.1 N solution means the result has to be divided by two.

Ben

Your absolutely correct, what was I thinking? Vincent