Trichlor composition

[Phizy]

Hi.
No, I am not purposely seeking to use trichlor to maintain the chlorine level. I'm asking the following question for other reasons I can go into later.

I have read that for 1 ppm of chlorine generated by trichlor the CYA concentration increases by 0.6. However, I cannot make the math work so there is obviously something wrong in my assumptions.

So, I'm asking for corrections in my assumptions. Thank you.

Trichloroisocyanuric acid (aka trichlor) has a molecular weight of 232.41 g/mol.
Trichloroisocyanuric acid has 3 mol of Cl- per mol of acid

The dissociation of Cl- from the acid in contact with water is such that:
2 mol trichlor -> 2 mol CYA + 3 mol Cl2

The molecular weight of CYA is 129.07 g/mol and Cl2 is 70.9 g/mol.

Therefore, by weight in 1 L (aka ppm):

464.82 ppm trichlor -> 258.14 ppm CYA + 212.7 ppm Cl2

Dividing by 212.7 ppm Cl2 I get 1.2 ppm CYA for each 1 ppm Cl2

So.....what am I doing wrong??

Thanks.

 

[JoyfulNoise]

Your assumption about the chemical reaction is wrong. At normal pool pH, the dominant equilibrium reaction is this -

HClCY- + H2O H2CY- + HOCl
"Chlorine bound to CYA" + Water "CYA ion" + Hypochlorous Acid

You can't assume that all of the chlorine bound to the trichlor will immediately dissociate into chlorine and CYA; that's not what happens. There are 10 different chlorinated and unchlorinated cyanurate molecules possible and they have equilibrium concentrations depending on pH. An exact solution requires solving 13 simultaneous equations while iteratively adjusting the equilibrium constants for the ionic strength of the solution; not impossible to do on a computer but very tedious to do by hand.

See this post for details - Pool Water Chemistry

 

[Phizy]

Ah, yes, thank you. That's perfect.

 

[JamesW]

2 mol trichlor -> 2 mol CYA + 3 mol Cl2

Every molecule of trichlor creates 1 molecule of cyanuric acid and 3 atoms of chlorine in the +1 oxidation state, which is equivalent to 3 molecules of chlorine gas.

Trichloroisocyanuric acid has 3 mol of Cl- per mol of acid

You had it correct here. For every mole of trichlor, there is 1 mole of cyanuric acid and 3 moles of chlorine. Why did you do the calculation with 2 moles of trichlor?

C₃Cl₃N₃O₃ + 3H₂O --> C₃H₃N₃O₃ + 3 HOCl

 

[Phizy]

From what I understand the calculation for FAC is based on Cl2, hence why I multiplied by two. 3 moles of Cl- does not equal 3 moles of chlorine gas.

I need to review the link to see how it fits my purposes. But, the full answer for what I'm looking for is this:

If a season dissolved 50.3 kg of trichlor in 103,375 L (486.576 ppm trichlor) what was the total consumption of chlorine?

Every molecule of trichlor creates 1 molecule of cyanuric acid and 3 atoms of chlorine in the +1 oxidation state, which is equivalent to 3 molecules of chlorine gas.


You had it correct here. For every mole of trichlor, there is 1 mole of cyanuric acid and 3 moles of chlorine. Why did you do the calculation with 2 moles of trichlor?

C₃Cl₃N₃O₃ + 3H₂O --> C₃H₃N₃O₃ + 3 HOCl

 

[JamesW]

Cl₂ + H₂O --> H⁺ + HOCl + Cl^-

A molecule of chlorine gas has two chlorine atoms each with 7 valence electrons. They form a covalent bond.

As the chlorine gas dissolves, the atoms separate. One chlorine atom ends up with 8 valence electrons (chloride at -1) and one chlorine atom ends up with 6 valence electrons at +1.

If a season dissolved 50.3 kg of trichlor in 103,375 L (486.576 ppm trichlor) what was the total consumption of chlorine?

445 ppm chlorine measured as chlorine gas. The equivalent of 46 Kg chlorine gas.

46/50.3 = 0.91. That's why trichlor is labeled as 91% available chlorine. One kilogram of trichlor is equivalent to 0.91 kg chlorine gas.

1 mole of chlorine gas does equal one mole chlorine as found in trichlor. That's because only 1 of the chlorine atoms in chlorine gas becomes Cl⁺ while the other becomes a chloride ion Cl^-.

 

[Phizy]

Yes. I know. But, as I said, from what I've read when one refers to the ppm chlorine it is in the form of Cl2 without considering the dissociation and equilibrium. I assume this is correct?

Cl₂ + H₂O --> H⁺ + HOCl + Cl^-

A molecule of chlorine gas has two chlorine atoms each with 7 valence electrons. They form a covalent bond.

As the chlorine gas dissolves, the atoms separate. One chlorine atom ends up with 8 valence electrons (chloride at -1) and one chlorine atom ends up with 6 valence electrons at +1.

 

[JoyfulNoise]

From what I understand the calculation for FAC is based on Cl2, hence why I multiplied by two. 3 moles of Cl- does not equal 3 moles of chlorine gas.

I need to review the link to see how it fits my purposes. But, the full answer for what I'm looking for is this:

If a season dissolved 50.3 kg of trichlor in 103,375 L (486.576 ppm trichlor) what was the total consumption of chlorine?

So you know from your previous equation that 1 mole of trichlor produces 3 moles of HOCl. So calculate the ppm's of HOCl from your previous numbers and then multiple that number by the ratio of the molecular weights of (Cl2/HOCl) or (70.96/52.46) or 1.35 since the FC is measured in units of [Cl2]

 

[JamesW]

when one refers to the ppm chlorine it is in the form of Cl2 without considering the dissociation and equilibrium. I assume this is correct?

No, not correct.

The test essentially figures out the molarity of the active chlorine. Then it translates that to units of chlorine gas.

In other words, the test measures molarity and then asks how much chlorine gas would have had to have been added per weight of solution to get that molarity.

The units of Chlorine gas equivalents are just somewhat arbitrarily chosen.

The net effect is that adding one mole of trichlor is equivalent to adding three moles of chlorine gas.

 

[Phizy]

So, I'll ask this question straight out. If someone could check and correct, that would be great.

This coming year I am going to switch to liquid chlorine and I want to estimate how much liquid chlorine (12.5%) I will need to add. I am basing this on the total consumption of trichlor last year. I have a strict log of exactly how much extra liquid chlorine I added which I can account for later. But I need to estimate based on the consumption.

Total amount of trichlor consumed: 50.3 kg
Total volume = 103,375 L
This is 486.576 ppm trichlor.
This is equivalent to 222.66 ppm chlorine, as Cl2

The concentration of chlorine, as Cl2 in 12.5%w solution is 125433 ppm (mg/L) Cl2

To dose 222.66 ppm (mg/L) chlorine as Cl2 in a 103,375 L pool I need to add 183 L of 12.5%w solution of chlorine.

Is this correct? If not, what is the value?
Thanks,
Paul

 

[needsajet]

Yes and n/a. PoolMath says 222.6 ppm in 103375 L requires 183 L of 12.5% chlorinating liquid. - approved by the committee to elect PoolMath


Note: PoolMath calculates that 445 ppm FC was delivered with 50.3 kg of trichlor. For 222.66 ppm FC, the recommended amount in PoolMath is 25.15 kg (or about 55#).

 

[Phizy]

Thanks. Poolmath is great, but I can't see the calculations it uses. But that's OK. I've always been perplexed how concentrations of chlorine has never been written properly such as x ppm chlorine (as [state species]).

 

[JoyfulNoise]

Thanks. Poolmath is great, but I can't see the calculations it uses. But that's OK. I've always been perplexed how concentrations of chlorine has never been written properly such as x ppm chlorine (as [state species]).

Phizy (love the screen name!) -

A couple of things to remember. 12,5% LC is sold based on Trade % (volume percent) not weight percent. 10% LC is also "Trade %" while 8.25% and lower is sold as weight %. 12.5% LC has an "Available chlorine as Cl2" of 10.8% (that takes into account the specific density of the solution as well as the conversion of hypochlorite to Cl2). As JamesW stated earlier, tricolor is 91% available chlorine so you can simply take your PPM's of trichlor and multiply it by 0.91 to get your PPM's of Cl2. Then you can use that number along with the available chlorine value for the 12.5% LC to get the total volume needed.

As you'll see, it requires a heck of a lot of bottles of bleach to equal an equivalent amount of LC. So not only is trichlor more compact and easier to handle, it also tends to be (allowing for local price variations) the cheapest form of chlorine available on a PPM of Cl2 basis. It's also a lot easier on your back muscles to carry a bucket of pucks than it is gallons of LC.

But, we all know why pucks are NOT good to use long term....

As for the choice of units, yes it doesn't seem logical but it is. In a lab, you often need to compare things on a unit basis that is common. So you have to choose materials that are readily available and easy to measure. Believe it or not, there are very simple and easy ways to produce chlorine gas on a lab scale in very precise amounts that is easily measured and chlorine gas is pure which means it does not add "anything" else to the measurement. Chlorine (Cl+) as an atomic species does not really exist in any form that is measurable in the lab. So it makes sense that chlorine containing products are all normalized to chlorine gas.

This is similar to TA and CH measurements which are all put into units of "ppm of CaCO3" even though TA can be comprised of many different types of alkaline species (carbonates, borates, phosphates, sulfates, cyanurates, etc, etc). Caclium carbonate is fairly easy to find (most of the rocks outside contain it) and it's relatively easy to dissolve and measure. So it tends to be the unit of choice for those types of standards. Borates are often measured in units of the actual element boron (B) rather than some other boron containing species. So when TFP recommends that pool owners can use up to 50ppm "borates" what is really meant is 50ppm in units of B. The actual concentration of "borates" is really about 250ppm when you measure out the weight of borax or boric acid and add it to water. Why is "boron" chosen? Well, most of the available literature on boron interactions with plants and algae and such are all normalized to units of boron and so everyone follows the same convention.

Welcome to fun world of chemistry....(most engineering types hate it!) and remember the mantra about science experiments-

If it wiggles, it's biology;
If it stinks, it's chemistry;
If it doesn't work, it's physics;
Ask an engineer, they can fix anything....

 

[Phizy]

Thanks. I'm actually a chemist (analytical) which is why I'm trying to go through these calculations and why I'm frustrated with the lack of proper units at times. It makes it very hard to verify anything

 

[Phizy]

One additional thing to add and this proves my point. You said 12.5% is volume percent which i don't think it is, however when I calculate the composition based on wt% and then run through the calculations to convert to ppm as Cl2, I get the appropriate amount based on the label and based on analyzing after addition.

 

[JoyfulNoise]

Thanks. I'm actually a chemist (analytical) which is why I'm trying to go through these calculations and why I'm frustrated with the lack of proper units at times. It makes it very hard to verify anything

Yeah, pool water chemistry is definitely not as precise as your works of analytical chemistry. I can understand the frustration. But it has to be accessible to the layman so the tests are fairly simple to perform and the nomenclature can't be bogged down with a lot of technical jargon or else it's too difficult for the non-specialist to practice.




Sent from my iPhone using Tapatalk

 

[JoyfulNoise]

One additional thing to add and this proves my point. You said 12.5% is volume percent which i don't think it is, however when I calculate the composition based on wt% and then run through the calculations to convert to ppm as Cl2, I get the appropriate amount based on the label and based on analyzing after addition.

See these posts -

Bleach concentration per unit volume and other calculations

pool calculator, which % to use

I'll just pull out the relevant quote form Richard as I almost always mix up "trade" "weight" and "volume" percent (and you're right, Trade % is % available chlorine by volume, not the volume % of sodium hypochlorite).

% Available Chlorine by Weight < Weight % Sodium Hypochlorite < % Available Chlorine by Volume (aka Trade %)

For typical 6% bleach and 12.5% chlorinating liquid, this is as follows because bleach is typically listed by the weight % of sodium hypochlorite while chlorinating liquid is normally spec'd by Trade % which is % Available Chlorine by Volume:

5.71% Available Chlorine by Weight < 6% Weight Sodium Hypochlorite < 6.17% Available Chlorine by Volume (Trade %)
10.78% Available Chlorine by Weight < 11.31% Weight Sodium Hypochlorite < 12.5% Available Chlorine by Volume (Trade %)

 

[Phizy]

Whatever it is, the calculations I did worked perfectly. Calculated 12.5 as weight %, used the density for NaOCl, converted to Cl2 and obtained the label claim.

 

[Caco]

Welcome to fun world of chemistry....(most engineering types hate it!) and remember the mantra about science experiments-

If it wiggles, it's biology;
If it stinks, it's chemistry;
If it doesn't work, it's physics;
Ask an engineer, they can fix anything....

Lmao Matt, as an engineer I have to agree on this, except... I like chemistry, I find it fascinating but my brain just doesn't get it

 

[kimkats]

Y'all be You go to Pool Math, plug in your test results, add what it tells you to and go swim! You buy more chlorine when your jugs are empty! DONE!

Kim