Sciencemadness Discussion Board

Determination of HCl-concentration without titration (idea)

Diachrynic - 16-3-2018 at 12:49

This is just an idea that came to my mind.

How could I determine the concentration of hydrochloric acid without titration setup? I had the following idea. Please let me know what you think.

Say, our solution has x mol HCl in it.

Now, add y mol of sodium carbonate, such that a complete reaction is ensured. Better add a slight excess. Keep track of the mass you add.

Reaction:

2 HCl + Na2CO3 → 2 NaCl + H2O + CO2

Now boil the solution dry and measure weight difference to the added carbonate. This is Δm.

Given that for every mole of HCl half a mole of carbonate reacts, the remaining carbonate is given by (y - x/2) mol.

For every mole of HCl, one mole of salt is produced, which corresponds to x mol NaCl.

The total weight is

mend = mNaCl + mNa2CO3

mend = x * MNaCl + (y - x/2) * MNa2CO3

mend = x * MNaCl + y * MNa2CO3 - x/2 * MNa2CO3

mend - y * MNa2CO3 = x * MNaCl - x/2 * MNa2CO3

mend - mNa2CO3 = x * (MNaCl - 1/2 * MNa2CO3)


(Δm) / (MNaCl - 1/2 * MNa2CO3) = x

Δm * 0.183604 = x (in mol)

So, you just need the difference in the mass to determine the amount of hydrochloric acid.

Now, I want to know:

- Are my calculations correct?
- Is this a viable method for measuring concentration?
- Where could errors occur?
- How precise could this be?

aga - 16-3-2018 at 14:22

The assumptions are a bit vague, such as the lack of Volume or Mass of HCl solution used - without that the [M] is unobtanium.

Also, sodium carbonate is hygroscopic, so the exact weight of the dry powder will depend heavily on how hydrated it is.

The decahydrate will obviously be very much heavier (i.e. more water) than anhydrous, or anywhere in-between.

Standard titration methods work better, however your method would give a useful indication of the HCl concentration.

A source of error would be spattering as the solution was boiled dry - it would not go dry without a fight.

Edit:

Clearly this is random speculation, not something you Actually Did.
Trying it out for yourself would have been better.

[Edited on 16-3-2018 by aga]

mayko - 16-3-2018 at 14:55

My copy of Merck lists a (presumably saturated) solution of sodium carbonate as having a pH of ~11.6; this would introduce a small but non-negligible amount of bicarbonate ion alone, even before the addition of HCl:
http://ion.chem.usu.edu/~sbialkow/Classes/3650/Carbonate/Car...
(edit): this could be overcome if you make sure to roast it good before massing(/e)


If you want to do gravimetry, ammonia would have two advantages over sodium carbonate: you can add excess and then evaporate off the leftover, and the residual salt doesn't have any hydrated states. On the other hand, ammonium chloride tends to sublime, so evaporating the water off would need to be done carefully

[Edited on 16-3-2018 by mayko]

DavidJR - 16-3-2018 at 19:35

What you're describing effectively is a crude form of titration...

brubei - 16-3-2018 at 21:16

i agree with david, however dealing with evaporation is a source of mistake due loss of salt with vapor during evaporation and bad accuracy with the residual humidity in your final salt.

titrating how we learn it in chemistry class

1/ use NaOH instead of carbonate
2/take x mole of HCl (x is unknown) (about 10mL).
If it is too concentrate (>10%) dilute in 100mL of water.

3/measure n mole of NaOH and dissolve to 25 mL of water.
(n must be greater than x, you have to assume hypothetical value of x at this stage, if you think than your HCl is very concentrated assume than its a 30% solution, if not, just try 1%)


4/ add few drop of phenol red in your HCl (red phenol turn yellow to red à pH 7, so your solution is then yellow)
5/slowly add NaOH until your mixture turn red, do not add NaOH after color transition. If your hypothetical value of x was too high the solution will turn red very fast. If it was too low you'll probably add a lot of NaOH

6/when the solution just turn yellow to red you know you are at pH = 7, then you know you added exact quantity of each product n(NaOH) = n(HCl)
7/do the math

phenol red is available for water analysis for pool and etc.


clearly_not_atara - 16-3-2018 at 23:36

If you're dealing with just aqueous HCl, resistivity and density are both effective indicators. Boiling the solution down seems as tedious as titration imo.

Tsjerk - 16-3-2018 at 23:49

As DavidJR said, this is a back titration.

Diachrynic - 17-3-2018 at 00:27

Quote: Originally posted by aga  
The assumptions are a bit vague, such as the lack of Volume or Mass of HCl solution used - without that the [M] is unobtanium.

Also, sodium carbonate is hygroscopic, so the exact weight of the dry powder will depend heavily on how hydrated it is.

The decahydrate will obviously be very much heavier (i.e. more water) than anhydrous, or anywhere in-between.

Standard titration methods work better, however your method would give a useful indication of the HCl concentration.

A source of error would be spattering as the solution was boiled dry - it would not go dry without a fight.


Well, that's true. It also might be hard to get it anhydrous, as sodium carbonate forms hydrates. Too bad I don't have a normal titration setup.
Quote:

Clearly this is random speculation, not something you Actually Did.
Trying it out for yourself would have been better.


It is, it was just an idea I had.
Quote: Originally posted by mayko  
My copy of Merck lists a (presumably saturated) solution of sodium carbonate as having a pH of ~11.6; this would introduce a small but non-negligible amount of bicarbonate ion alone, even before the addition of HCl:
http://ion.chem.usu.edu/~sbialkow/Classes/3650/Carbonate/Car...
(edit): this could be overcome if you make sure to roast it good before massing(/e)


Hmmm, would using bicarbonate from the get go be better? It is also not as hygroscopic. Then again when heated above 50 °C it turns into hygroscopic carbonate.
Quote:

If you want to do gravimetry, ammonia would have two advantages over sodium carbonate: you can add excess and then evaporate off the leftover, and the residual salt doesn't have any hydrated states. On the other hand, ammonium chloride tends to sublime, so evaporating the water off would need to be done carefully


That sounds interesting! Maybe boil it down quite a bit, then dry over sodium carbonate as a desiccant? Oh, the irony...
Quote: Originally posted by DavidJR  
What you're describing effectively is a crude form of titration...


Really? Cool. But I meant titration in the way of an indicator and dripping a liquid into another.
Quote: Originally posted by brubei  
i agree with david, however dealing with evaporation is a source of mistake due loss of salt with vapor during evaporation and bad accuracy with the residual humidity in your final salt.

titrating how we learn it in chemistry class

[...]


Thats true. Maybe ammonia as suggested would be better in that aspect.

I know how a titration works, I just don't have the equipment to measure it precisely. (I.e. burettes, full-pipettes, volumetric flasks, ...)
Quote: Originally posted by clearly_not_atara  
If you're dealing with just aqueous HCl, resistivity and density are both effective indicators. Boiling the solution down seems as tedious as titration imo.


I thought about that, too. But the density ranges from 1.05 to 1.15 in the range 10 to 30%. So, quite small. Which means either a lot of volume or very precise measurement of volume. The first can't go over 250 ml because of my scale. The latter comes back to the reasons why I try to avoid a titration.
Quote: Originally posted by Tsjerk  
As DavidJR said, this is a back titration.


Thank you for that piece of info. I didn't know this could also be called a titration. My picture of titration is just that of a liquid dripping into another and measuring how much dripped in.

Tsjerk - 17-3-2018 at 01:07

If you pull your definition a bit wider it becomes what you are getting to do here: putting a compound in a liquid and measuring how much you dripped in (or dripped in too much, as it is a back titration).

I think it is cool though you come up with your own protocol. The fact it is sort of not new doesn't matter, as it was new too you when you thought of it.

DavidJR - 17-3-2018 at 01:18

You don't need to buy a burette for the occasional titration - a graduated pipette is also quite usable, is cheaper, and is useful for other things too. Then all you need is an appropriate indicator.

aga - 17-3-2018 at 01:58

Titration is weighting/measuring/reacting up to some certain point, then calculating, so your idea is a titration.

A variation would be to add sodium carbonate in small quantities until it stops fizzing - that would be the 'end-point'.

The moles of HCl = 2x moles of Na2CO3 in this reaction, so if :
gH = grammes of HCl
gN = grammes of carbonate
mwH = molecular weight of HCl
mwN = m.w. of sodium carbonate
mH = moles of HCl, mN = moles of carbonate

$$mH = 2 mN$$
$$mH = \frac {gH}{mwH} , mN = \frac{gN}{mwN}$$
$$\frac{gH}{mwH} = \frac{2gN}{mwN}$$
$$gH = gN \frac {2mwH}{mwN}$$

I get mwH = 36.458, mwN = 105.9874, so

$$gH = 0.688 gN$$

Practically speaking, heat some sodium carbonate to get it as dry as possible.
Measure out X ml of your acid and weight it (Y grammes)
Weight the carbonate (sN grammes)
Add the carbonate to the acid until no more fizzing happens.
Weight the remaining carbonate (eN grammes).

The w% of the acid is given by :
$$\frac{68.8(sN - eN)}{Y}$$

the [M] of the acid will be :
$$\frac{688(sN-eN)}{36.458X}$$
$$= 18.9 \frac{sN-eN}{X}$$

I hope the maths are right - best check before using !

[Edited on 17-3-2018 by aga]

Diachrynic - 17-3-2018 at 02:27

Thank you aga for your effort. Your math is correct as far as I can tell. Yes, adding until it stops fizzing is indeed a possibility I guess. I remember when I made NaOAc the liquid would bubble and fizz even after three times as much (had a calculation error) carbonate was added than needed (on heating). Probably dissolved carbon dioxide. Maybe it was because acetic acid is a weak acid?

unionised - 17-3-2018 at 03:40

You can get a rough idea of the acid concentration by adding it to a solution of bicarbonate and measuring the loss of weight of CO2
You need to correct for water lost as vapour along with the CO2.

You can also get a fairly good indication of the HCl concentration by measuring the density.

mayko - 17-3-2018 at 06:49

Titration is a volumetric technique; what's described here is a gravimetric one :cool:

Tsjerk - 17-3-2018 at 07:21

You have volumetric and gravimetric titration, only the term titration doesn't specify between the two.

Just the term titration only specifies the fact you are measuring something that is a result of changing something.

I once titrated the membrane potential of a bacterium by measuring fluorescence after adding a potential dissipating compound

aga - 17-3-2018 at 08:12

If the volume of CO2 was measured, then that'd be volumetric ;)

[Edited on 17-3-2018 by aga]

DavidJR - 17-3-2018 at 16:30

That could definitely work, and wouldn't be affected by the unknown hydration of the carbonate.

I titrated sodium hypochlorite (bleach) using Nile Red's method recently, which basically consisted of reacting a known volume of the bleach solution with an excess of hydrogen peroxide, and measuring the volume of oxygen gas produced using an inverted graduated cylinder filled with water.

AJKOER - 20-3-2018 at 18:54

Add an excess of alkaline diluted chlorine bleach to the HCl:

xHCl + NaOCl = xHOCl + xNaCl + (1-x)NaOCl

Distill to collect the volatile HOCl/Cl2O. Then add an excess of H2O2 with NaCl and record mainly O2 volume over hot salted water:

HOCl + H2O2 --> HCl + O2 (g) + H2O

Every mole of oxygen created (22.4 liters) equates to one mole of HCl.

Any HClO3 created (from HOCl) will be reduced (to gaseous ClO2 diluted in O2, see https://www.researchgate.net/publication/230093604_Mechanism...).

[Edited on 21-3-2018 by AJKOER]

[Edited on 21-3-2018 by AJKOER]

Texium - 20-3-2018 at 19:31

Quote: Originally posted by AJKOER  
Add an excess of chlorine bleach to the HCl:

xHCl + NaOCl = xHOCl + xNaCl + (1-x)NaOCl

Distill to collect the HOCl. Then add an excess of H2O2 with NaCl and record mainly O2 volume over hot water:

HOCl + H2O2 --> HCl + O2 (g) + H2O

Every mole of oxygen created (22.4 liters) equates to one mole of HCl.

Any HClO3 created (from HOCl) will be reduced (to gaseous ClO2 diluted in O2, see https://www.researchgate.net/publication/230093604_Mechanism...).
Just... no. Do you really have to find a way to try and use hypochlorite or Fenton's for everything?

This suggested procedure has so much more potential for error than regular titration or Diachrynic's procedure, not to mention being much more difficult to carry out.

ninhydric1 - 20-3-2018 at 20:07

And chances are the HClO will break down into Cl2 anyway. A pretty bad idea, considering you're wasting reagents and you're going to be exposed to chlorine gas.

Tsjerk - 21-3-2018 at 02:30

You can't distill HOCl anyway

AJKOER - 21-3-2018 at 04:02

Thanks for the comments, and while the process I suggested has a major, yet some unaddressed issues, I have since edited my original comment to add words like 'alkaline' , ‘dilute’ and ‘volatile HOCl/Cl2O’.

First, per Watts Dictionary of Chemistry (please see comments and links in a prior SM thread at https://www.sciencemadness.org/whisper/viewthread.php?tid=17... ), the more volatile than water HOCl/Cl2O is driven largely off first, so the concentration of the hypochlorous acid can interestingly nearly be doubled upon boiling off half the solution and the unreacted NaOCl/NaCl presence removed.

Note, the concept of 'dilute' here is impotant as concentrated HOCl is increasingly unstable and should be cooled, free from strong light and used promptly (see comments/links at https://www.sciencemadness.org/whisper/viewthread.php?tid=17... ). However, note that the decomposition reaction:

HOCl + hv --> HCl + O

is not much of a concern here as the freshly created HCl reforms HOCl in the presence of excess NaOCl:

HCl + NaOCl --> HOCl + NaCl

Similarly, upon boiling a possible disproportionation reaction forming chloric acid:

3 HOCl --> 3 H+ + 2 Cl- + ClO3-

would also regenerates, in excess NaOCl, three HOCl.

The seeming advantage of my approach is that unlike the acid/carbonate approach, O2 is far less soluble than CO2 in warmed salted water and there is no measuring issue with a moisture loaded carbonate, or in determining when the CO2 bubbles actual stop forming, not just locally, but in the entire solution upon stirring/shaking. The only precise measure required in the hypochlorite method is the volume of oxygen liberated (recommend rapid recording over distilled, not tap, water).
------------------------------

There is still a possible unaddressed issue(s) with my suggested approach which I will leave for the more astute readers to cite (clue: the major reagent and also read my referenced source on the HClO3/H2O2 reaction).
-----------------------------------------------

Note, one cannot heat dry NaHCO3 to remove moisture without the possible transformation (via CO2 loss) to Na2CO3 to some extent.

[Edited on 21-3-2018 by AJKOER]

unionised - 21-3-2018 at 12:08


OK, so you know about this reaction
HOCl + hv --> HCl + O

and, in the real world it doesn't actually need light- heat will do teh job just fine..
So you make your HOCl and , while you try to distil it, it decomposes to give oxygen- which is lost.
And you end up with a dilute solution of HCl in the receiver.
Then you add H2O2- and not a lot happens- except, maybe the H2O2 decomposes a bit and gives some O2- which you measure and attribute to the original HCl in the solution.

I'm really not sure I have ever seen a less useful proposal.

aga - 21-3-2018 at 12:13

Quote: Originally posted by unionised  
... I'm really not sure I have ever seen a less useful proposal.

Hey ! Have you never seen any of my posts ? ;)

AJKOER - 21-3-2018 at 16:03

Quote: Originally posted by unionised  

OK, so you know about this reaction
HOCl + hv --> HCl + O

and, in the real world it doesn't actually need light- heat will do teh job just fine..
So you make your HOCl and , while you try to distil it, it decomposes to give oxygen- which is lost.
And you end up with a dilute solution of HCl in the receiver.
Then you add H2O2- and not a lot happens- except, maybe the H2O2 decomposes a bit and gives some O2- which you measure and attribute to the original HCl in the solution.

I'm really not sure I have ever seen a less useful proposal.


Well, there is a small amount of substance in your scenario as to quote from Watts' Dictionary of Chemistry, Volume 2, page 16 (scroll to p. 16 at this link: https://books.google.com/books?id=Wt3nAAAAMAAJ&printsec=... ):

"A dilute solution of HClO may be distilled with partial decomposition, the distillate is richer in HClO; Gay-Lussao found that, on distilling a dilute solution to one-half, the distillate contained five-sixths of the total HClO (C. R. 14, 927)"

So, dilute HOCl by itself (not in the presence of NaOCl) will suffer partial decomposition on heating, but with NaOCl, much less to none, given the interaction of any created HCl (or HClO3 via disproportionation of the hypochlorous acid) with NaOCl reforming HOCl, as I detailed previously above.

Actually, a mix of NaOCl/HOCl is a quite useful path to chlorate in the added presence of a source hydroxyl radicals, as I have demonstrated in one of my chlorate threads (see https://www.sciencemadness.org/whisper/viewthread.php?tid=34... ).

[Edited on 22-3-2018 by AJKOER]

Diachrynic - 22-3-2018 at 08:52

HOCl seems too dangerous and too complicated.

I guess I stay with soda and baking soda...

(Not to mention I have a hard time finding bleach with no crap in it, it is always a sirupy liquid, like thin honey, dyed and all that shit. And I don't have a destillation setup. So, that's a no for me...)

aga - 22-3-2018 at 09:10

Pool supply shops sometimes have 30% hypochlorite with no crap in it.

Looking into sodium carbonate and bicarbonate, it appears that the bicarbonate takes on less water and is very easily available, so that'd be the better candidate for this titration.

Write it up into a 'lesson' with a practical example - call it "The Diachrynic Acid Titration" ;)

Seriously, for an absolute beginner, this would be a great (and easily done) exercise.

It introduces moles & molarity, and will give a good indication of acid strength.

Diachrynic - 22-3-2018 at 11:41

Hey, that sounds like a really neat idea, aga!

aga - 22-3-2018 at 12:06

Go for it !

It's your idea, and a Good one, so get writing.

MrHomeScientist's recent "Thermite Primer" is a great example of clear formatting.

Doesn't need to be in Kemistry Skoool - just give the post a descriptive title so people can find it.

U2U if you need any help.