I am trying to make a 0.0667N NAOH solution for titration to find TA.
I have .2N NAOH solution for going the phenolpthalein route but don't
like using that technique because of inaccuracy. I want to use pH
meter to find end point. I know the grams/liter of NAOH needed to
make the solution but would have to buy the NAOH from chem supply.
One supplier wanted $23.50 for 200 grams of lab quality NAOH and on
top of that wanted $37.50 to ship it! Forget that! All I want to
know is how many milliliters of distilled/deionized water I need to
add to the .2N NAOH solution to make it a 0.0667N NAOH solution.
Anyone?

In article >, jestar > wrote:
>I am trying to make a 0.0667N NAOH solution for titration to find TA.
>I have .2N NAOH solution for going the phenolpthalein route but don't
>like using that technique because of inaccuracy. I want to use pH
>meter to find end point. I know the grams/liter of NAOH needed to
>make the solution but would have to buy the NAOH from chem supply.
>One supplier wanted $23.50 for 200 grams of lab quality NAOH and on
>top of that wanted $37.50 to ship it! Forget that! All I want to
>know is how many milliliters of distilled/deionized water I need to
>add to the .2N NAOH solution to make it a 0.0667N NAOH solution.
>Anyone?

The volume of water you need to add depends on the volume of solution you have
to start with, or, alternatively, the volume of solution you want to end up
with.

Easiest to do it as a simple ratio calculation. The solution you have now is
(0.2 / 0.0667) = 3 times as strong as you need, so you need to dilute it to
one-third of its present strength -- which means that you add *two* parts
water to one part solution.

On Sep 1, 7:07*pm, (Doug Miller) wrote:
> In article >, jestar > wrote:
>
> >I am trying to make a 0.0667N NAOH solution for titration to find TA.
> >I have .2N NAOH solution for going the phenolpthalein route but don't
> >like using that technique because of inaccuracy. *I want to use pH
> >meter to find end point. *I *know the grams/liter of NAOH needed to
> >make the solution but would have to buy the NAOH from chem supply.
> >One supplier wanted $23.50 for 200 grams of lab quality NAOH and on
> >top of that wanted $37.50 to ship it! *Forget that! *All I want to
> >know is how many milliliters of distilled/deionized water I need to
> >add to the .2N NAOH solution to make it a 0.0667N NAOH solution.
> >Anyone?
>
> The volume of water you need to add depends on the volume of solution you have
> to start with, or, alternatively, the volume of solution you want to end up
> with.
>
> Easiest to do it as a simple ratio calculation. The solution you have now is
> (0.2 / 0.0667) = 3 times as strong as you need, so you need to dilute it to
> one-third of its present strength -- which means that you add *two* parts
> water to one part solution.

That's way too easy. Can't you misplace a decimal point or two? I
have to confess that I didn't even do the simple math. I knew that 41
g/l was a 1N solution of NaOH and 2.73 g/l was a 0.0667 solution but
the numbers did not obviously appear to me to be a simple ratio. I
slapped myself on the head. I won't be so lazy next time. Especially
with decimal points and SO2.