I ran out of 0.2 normal Sodium Hydroxide for my acid test kit, and ordered
some from my preferred on-line supplier. My order arrived today, and I
received 0.1 normal NaOH. Rather than bother with returning a 4 ounce
bottle, can I simply adjust the procedure?

Specifically, the procedure instructs me to

"Draw 10cc of Sodium hydroxide into the syringe. Add 1 cc at a time to test
solution. Swirl sample as drops are added..."

With the 01 normal solution, would I not add *2* cc at a time?

Or, am I being too simple about this?

Thanks in advance,

Bart

"bwesley7" > wrote in message news:<Tyfgd.185060$a85.31369@fed1read04>...
> I ran out of 0.2 normal Sodium Hydroxide for my acid test kit, and ordered
> some from my preferred on-line supplier. My order arrived today, and I
> received 0.1 normal NaOH. Rather than bother with returning a 4 ounce
> bottle, can I simply adjust the procedure?
>
> Specifically, the procedure instructs me to
>
> "Draw 10cc of Sodium hydroxide into the syringe. Add 1 cc at a time to test
> solution. Swirl sample as drops are added..."
>
> With the 01 normal solution, would I not add *2* cc at a time?
>
> Or, am I being too simple about this?
>
> Thanks in advance,
>
> Bart

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

"bwesley7" > wrote in message news:<Tyfgd.185060$a85.31369@fed1read04>...
> I ran out of 0.2 normal Sodium Hydroxide for my acid test kit, and ordered
> some from my preferred on-line supplier. My order arrived today, and I
> received 0.1 normal NaOH. Rather than bother with returning a 4 ounce
> bottle, can I simply adjust the procedure?
>
> Specifically, the procedure instructs me to
>
> "Draw 10cc of Sodium hydroxide into the syringe. Add 1 cc at a time to test
> solution. Swirl sample as drops are added..."
>
> With the 01 normal solution, would I not add *2* cc at a time?
>
> Or, am I being too simple about this?
>
> Thanks in advance,
>
> Bart

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

(JEP) wrote:
> > I ran out of 0.2 normal Sodium Hydroxide for my acid test kit, and ordered
> > some from my preferred on-line supplier. My order arrived today, and I
> > received 0.1 normal NaOH. Rather than bother with returning a 4 ounce
> > bottle, can I simply adjust the procedure?
> >
> > Specifically, the procedure instructs me to
> >
> > "Draw 10cc of Sodium hydroxide into the syringe. Add 1 cc at a time to test
> > solution. Swirl sample as drops are added..."
> >
> > With the 01 normal solution, would I not add *2* cc at a time?
> >
> > Or, am I being too simple about this?
> >
> > Thanks in advance,
> >
> > Bart
>
> 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.

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

(JEP) wrote:
> > I ran out of 0.2 normal Sodium Hydroxide for my acid test kit, and ordered
> > some from my preferred on-line supplier. My order arrived today, and I
> > received 0.1 normal NaOH. Rather than bother with returning a 4 ounce
> > bottle, can I simply adjust the procedure?
> >
> > Specifically, the procedure instructs me to
> >
> > "Draw 10cc of Sodium hydroxide into the syringe. Add 1 cc at a time to test
> > solution. Swirl sample as drops are added..."
> >
> > With the 01 normal solution, would I not add *2* cc at a time?
> >
> > Or, am I being too simple about this?
> >
> > Thanks in advance,
> >
> > Bart
>
> 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.

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

(JEP) wrote in message m>...
> "bwesley7" > wrote in message news:<Tyfgd.185060$a85.31369@fed1read04>...
> > I ran out of 0.2 normal Sodium Hydroxide for my acid test kit, and ordered
> > some from my preferred on-line supplier. My order arrived today, and I
> > received 0.1 normal NaOH. Rather than bother with returning a 4 ounce
> > bottle, can I simply adjust the procedure?
> >
> > Specifically, the procedure instructs me to
> >
> > "Draw 10cc of Sodium hydroxide into the syringe. Add 1 cc at a time to test
> > solution. Swirl sample as drops are added..."
> >
> > With the 01 normal solution, would I not add *2* cc at a time?
> >
> > Or, am I being too simple about this?
> >
> > Thanks in advance,
> >
> > Bart



>
> 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


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 =120 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.

---------------------------------------------------------------------------------------------------------------------

2. 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

(JEP) wrote in message m>...
> "bwesley7" > wrote in message news:<Tyfgd.185060$a85.31369@fed1read04>...
> > I ran out of 0.2 normal Sodium Hydroxide for my acid test kit, and ordered
> > some from my preferred on-line supplier. My order arrived today, and I
> > received 0.1 normal NaOH. Rather than bother with returning a 4 ounce
> > bottle, can I simply adjust the procedure?
> >
> > Specifically, the procedure instructs me to
> >
> > "Draw 10cc of Sodium hydroxide into the syringe. Add 1 cc at a time to test
> > solution. Swirl sample as drops are added..."
> >
> > With the 01 normal solution, would I not add *2* cc at a time?
> >
> > Or, am I being too simple about this?
> >
> > Thanks in advance,
> >
> > Bart



>
> 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


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 =120 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.

---------------------------------------------------------------------------------------------------------------------

2. 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

(JEP) wrote in message m>...
> "bwesley7" > wrote in message news:<Tyfgd.185060$a85.31369@fed1read04>...
> > I ran out of 0.2 normal Sodium Hydroxide for my acid test kit, and ordered
> > some from my preferred on-line supplier. My order arrived today, and I
> > received 0.1 normal NaOH. Rather than bother with returning a 4 ounce
> > bottle, can I simply adjust the procedure?
> >
> > Specifically, the procedure instructs me to
> >
> > "Draw 10cc of Sodium hydroxide into the syringe. Add 1 cc at a time to test
> > solution. Swirl sample as drops are added..."
> >
> > With the 01 normal solution, would I not add *2* cc at a time?
> >
> > Or, am I being too simple about this?
> >
> > Thanks in advance,
> >
> > Bart



>
> 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


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 =120 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.

---------------------------------------------------------------------------------------------------------------------

2. 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

(JEP) wrote in message m>...
> "bwesley7" > wrote in message news:<Tyfgd.185060$a85.31369@fed1read04>...
> > I ran out of 0.2 normal Sodium Hydroxide for my acid test kit, and ordered
> > some from my preferred on-line supplier. My order arrived today, and I
> > received 0.1 normal NaOH. Rather than bother with returning a 4 ounce
> > bottle, can I simply adjust the procedure?
> >
> > Specifically, the procedure instructs me to
> >
> > "Draw 10cc of Sodium hydroxide into the syringe. Add 1 cc at a time to test
> > solution. Swirl sample as drops are added..."
> >
> > With the 01 normal solution, would I not add *2* cc at a time?
> >
> > Or, am I being too simple about this?
> >
> > Thanks in advance,
> >
> > Bart
>
> 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

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

(JEP) wrote in message m>...
> "bwesley7" > wrote in message news:<Tyfgd.185060$a85.31369@fed1read04>...
> > I ran out of 0.2 normal Sodium Hydroxide for my acid test kit, and ordered
> > some from my preferred on-line supplier. My order arrived today, and I
> > received 0.1 normal NaOH. Rather than bother with returning a 4 ounce
> > bottle, can I simply adjust the procedure?
> >
> > Specifically, the procedure instructs me to
> >
> > "Draw 10cc of Sodium hydroxide into the syringe. Add 1 cc at a time to test
> > solution. Swirl sample as drops are added..."
> >
> > With the 01 normal solution, would I not add *2* cc at a time?
> >
> > Or, am I being too simple about this?
> >
> > Thanks in advance,
> >
> > Bart
>
> 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

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

> > 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
>
> 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

(Ben Rotter) wrote in message om>...
> > > 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
> >
> > 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