Sciencemadness Discussion Board - Molarity Question

Molarity Question

ktw_100 - 10-5-2006 at 04:56

I have a solution of .1M NaOH. That's 4gr of NaOH in 1 liter H2O.

Here's the question: Ignoring the Na portion of the solution, what is the OH concentration of this .1M solution of NaOH? Is it also .1M? Also, how many moles of OH- would be in 1 ml of .1M NaOH?

Thanks
Keith

enhzflep - 10-5-2006 at 06:58

Are they hairy nosed Moles, or Mols?

Seriously though, 0.1M NaOH solution has a pH of 13, and since pH + pOH = 14,
and 10^-pOH = [OH-], one can clearly deduce that the answer to your question(homework?) is xxx.
Have a look at this site Defining pH & pOH It will explain it all to you as it just did for me

(again) - it's been over 10 years since I forgot

[Edited on 10-5-2006 by enhzflep]

ktw_100 - 10-5-2006 at 07:15

Thanks - will have a look!

ktw_100 - 10-5-2006 at 07:47

ok - I had a look at that site - interesting stuff! I shall study it profusely tonight!

Here's where I am at, trying to use a real-life situation:

I make biodiesel ...

Normally, you just titrate 1ml of waste oil with a .1M solution of NaOH, until you get to lasting pink with phenolphalein, which tells you how much NaOH you are going to need per liter of waste oil when processing it.

Of course, this works, and I know how much NaOH I need per liter. However, I have always wanted to be able to calculate the actual pH of the waste vegetable oil, as this is a good indicator of how 'good' the stuff is to begin with.

So I already have a ready supply of .1M NaOH made up.

.1M solution of NaOH is 4 gr in 1 liter. If I ignore the Na, and just consider the OH, then the solution concentration is 4/17 (atomic weight of OH only). Therefore, my solution is .23529M if just considering the OH.

So: I have 1ml of acidic substance (waste vegetable oil). It takes 5ml of this .23529M to bring it to a pH of 8.5 (using phenolphalein as indicator). I can use the pH= -Log[H+] or even pH=-Log[OH], as they are at equilibrium, and I get .0017645 moles. Therefore pH = -Log[.0017645] = 2.7534.

So can I just subtract this 2.7534 from 8.5 to get the actual pH of the acidic oil?

You can obviously tell I am not a chemist!

Quote:

Originally posted by enhzflep
Are they hairy nosed Moles, or Mols?

(again) - it's been over 10 years since I forgot

[Edited on 10-5-2006 by enhzflep]

Aderyn - 10-5-2006 at 08:25

First of all, you're calculation is wrong from the start. When calculating the number of moles of Sodium Hydroxide you have, you can't just ignore the sodium.

M(NaOH)=23+16+1=40 g/mol

Since you're volume of fluid is 1 liter, the concentration is equal to the number of moles.

[NaOH]=n(NaOH)=4/40=0.1 M

The equation for solvolysis of Sodiumhydroxide is simply:

NaOH <==> Na[+] + OH[-]

And simple stoichiometri tells us:

[Na+]=[OH-]=0.1 M

If you add 5 ml of this to 1 liter of waste oil you add

n(OH-)=[OH-]*5*10^-3=0.5*10^-3=0.5 mmol
[OH-]=n(OH-) (since the volume is ~1 liter)

to obtain an pH of 8.5. The added hydroxide neutralized an equal amount of oxonium ions:

pH_before=pH-log[OH-]=8.50-log(0.5*10^-3)=5.19

It's possible that I've overlooked something, but it's a lot more realistic than you're answer at least.

[Edited on 10-5-2006 by Aderyn]

Aderyn - 10-5-2006 at 08:46

This is, of course, based on the assumption that the oil is either dilute enough, or has similiar properties to water. On second thought, I'm not so sure about that. The entire idea about pH only works in water (at least without modification), since it is based on the waters ionic product.

ktw_100 - 10-5-2006 at 09:21

Thanks again .... glad someone is beginning to straighten me out!

I do know that there are a few things that make up waste vegetable oil, and are impurities added to the vegetable oil as it is used/heated/cooled/used again. One of those things is Free Fatty Acid (FFA). Seems the more the oil is used, the higher the Free Fatty Acid content. As I mentioned, and as is obvious, I know little about chemistry! But perhaps as this FFA is an acid, that is why you can measure pH of the oil. Maybe it's measuring that vs the oil? I dunno - just guessing. There is usually a certain amount of water in the used veg oil too, although for producing biodiesel, it has to be eliminated by heating, or settling.

Another byproduct of biodiesel is glycerin, which I extract from the non-biodiesel layer after making a batch. Dunno if this factors in either.

As I say, I don't know much, but I hate just doing stuff by instruction, and not understanding what is going on. Oh - I should have listened in chemistry class 40 years ago!

Anyway, I'll look over your much appreciated info, and see if I can learn a little from it!

Thanks again

Keith

unionised - 10-5-2006 at 10:44

enhzflep
What do you mean by "Are they hairy nosed Moles, or Mols?"?

Fleaker - 11-5-2006 at 15:11

Unionised: perhaps enhzflep did not know that a mol is the abbreviation for "mole" (not just a name for that small tunneling mammal!).

Keith: For the (I guess this would be a transesterification) of those vegetable oils, for all intents and purposes (at least on what you plan to do with it) the [OH-] concentration is that of your solution, 0.1M. When you go to make a solution, you must take into account the molar mass of the whole compound, not just the ion you're after. BTW, those free fatty acids are very very weak, on par with your indicator, which is itself a weak organic acid.

so if you did M=(grams/molar mass of OH-)/(liters solution0 then you were in error.

enhzflep - 11-5-2006 at 21:53

Fleaker: you're absolutely spot on. It's been that long that I did in fact, forget entirely that Mol is short for Mole. That said I'm not so sure that I ever knew this, so thanks.
Dammit - I hate it when I'm a smart-ass and I'm wrong...