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.