[Ballistic]

Hi there, I decided to make some Benedict's Solution with things I had to hand.
The Sodium Carbonate is in a tub for swimming pool chemicals and does not state
the hydration, e.g. Anhydrous, Decahydrate etc.
I wonder what the effect of using too much in Benedicts would be if I add the 270g required for decahydrate as it isn't a constituent in the process of reducing the copper, but rather to provide the alkaline conditions?

I did trying cooking it up in an enamelled pan, but after half hour, it had only lost 3 grams in weight (out of a starting weight of 154g) and the instructions to dehydrate Sodium Carbonate from the web states 110 degrees C for one hour in a lab oven. And to be honest I had to give up because of the smell, so either an enamel pan is not correct to use or some other reason something was being given off (I did burn wooden spatula).


So... does anyone know what the effect of adding too much would be OR do you have a better technique to do with density or some other way of working out the hydration?
thanks in anticipation.

[Arkcon]

Probably they mean the most common hydrated form, you can find  the most common form,often on Wikipedia, or other online source.  i say probably, for a number of reasons. Briefly, the anhydrous form is usually insoluble in water.  Yes, you heard right, the insoluble form doesn't dissolve well,it hydrates first, then it dissolves. So if for some reason, they want to specify the recipe with the anhydrous form, then they've made it more difficult to make, for practically no reason. Still,these things are sometimes written this way, because people feel like it, I guess.

Also, remember, this is a learning forum. you can amuse yourself by computing the concentration with both forms of salts.  What molarity would you get each way?  The difference can't be insignificant, but you've hit on the point yourself, maybe it doesn't affect the reaction?  Do the math and show us.

[Ballistic]

Thank you for your reply. The reason I haven't done any maths (apart from being rubbish at Math) is that I have the following two versions of instructions:

1. Can make 1 litre with 100g Anhydrous Sodium Carbonate.
or
2. Can make 1 litre with 270g Decahydrate Sodium Carbonate.

So all I really need to do is determine what I actually have here. As you can see there is quite a bit of difference in how much Sodium Carbonate you will actually be adding if the hydration is unknown.

It would be easier if, say the solution were to be saturated or some other easy factor.

Another clue I have is that apparently the Anhydrous form readily takes up moisture to become Mono-hydrate whereas some of the forms with more water decompose at relatively low temperatures.

What I don't know is enough about the PH values and how these would be affected or whether it's even important as discussed.

[Arkcon]

Then you'll have to find a better recipe, or consult an expert, or just follow someone's recipe blindly.  Or search the peer-reviewed journal wherein Benedict himself^* first published his^* reaction.

Or you could work with what I said and what you've said just now -- its likely the decahydrate, because its easier to find, keep, and have around.

*I'm usually more gender sensitive, but c'mon, this was probably determined a century ago, and the world was even more paternalistic back then.

[Ballistic]

OK, I just threw it in a measuring cylinder and it came out at 112g for 100ml, and compared this with Wikipedia which shows the density should gradually drop as follows:
2.54 g/cm3 (25 °C, anhydrous)
 2.25 g/cm3 (monohydrate)[1]
 1.51 g/cm3 (heptahydrate)
 1.46 g/cm3 (decahydrate)

As my sample wasn't settled or compressed, I think we can assume with a small margin of error that it is as you suggest the decahydrate at 1.12 g/cm3.

[sjb]

...I'm usually more gender sensitive, but c'mon, this was probably determined a century ago, and the world was even more paternalistic back then.

Indeed, see https://en.wikipedia.org/wiki/Stanley_Rossiter_Benedict

[Ballistic]

OK, well fortunately I didn't mix the other chemicals, just the Sodium Carbonate, and there's about an inch of crystal Sodium Carbonate in the bottom by morning. It was agitated every time I passed it yesterday and I'm pretty sure it's over saturated.
It is pretty cold in there, probably about 15 degrees C so less will dissolve, but the figures for solubility are as follows:
Decahydrate:
 7 g/100 mL (0 °C)
 16.4 g/100 mL (15 °C)
 34.07 g/100 mL (27.8 °C)
 Heptahydrate:
 48.69 g/100 mL (34.8 °C)
 Monohydrate:
 50.31 g/100 mL (29.9 °C)
 48.1 g/100 mL (41.9 °C)
 45.62 g/100 mL (60 °C)
 43.6 g/100 mL (100 °C)[3]
I'm not sure why wiki doesn't give the anhydrous figure?

So to my 2 litres of water I added 540g of what I thought was decahydrate.

Oh flip, is it just that it's too cold??? :O

Which leaves me with one question, bearing in mind I haven't added the other chemicals yet, why would my product crystalise out due to the temperature drop over-night but another small amount of benedict's I have left here, small amount from a proprietary supplier does never crystalise out despite temperature drops?

[billnotgatez]

I am following this thread and out of curiosity wondering what is the source of the  Na₂CO₃.
Is it Arm and Hammer washing soda?

Edit:
Sorry, I now notice you said swimming pool stuff.
Could you give the brand name?

[Ballistic]

It's a tub from swimmingpoolchemicals.co.uk It was sealed so I don't think it would have absorbed a lot of CO2 over time, possibly a little. It has an EC number but unfortunately no CAS no.
As my old threads on this site show, I'm always making things from old things lying around or from car boot sales, it's really important to what I'm all about.

[billnotgatez]

I was about to post this before you answered, but I will post it anyway == just to get it off my chest.

So I confuse myself more by reading WIKI

One litre of Benedict's reagent can be prepared from 100 g of anhydrous sodium carbonate, 173 g of sodium citrate and 17.3 g of copper(II) sulfate pentahydrate.[3]

and

Sodium carbonate (also known as washing soda, soda ash and soda crystals), Na2CO3, is the water-soluble sodium salt of carbonic acid.

It most commonly occurs as a crystalline heptahydrate, which readily effloresces to form a white powder, the monohydrate. Pure sodium carbonate is a white, odorless powder that is hygroscopic (absorbs moisture from the air). It has a strongly alkaline taste, and forms a moderately basic solution in water.

[Ballistic]

OK yes, to counter your source for the Anhydrous with a formula for the Decahydrate:

http://www.camlab.co.uk/benedicts-solution-qualitative-p15193.aspx
"One litre of solution contains Cupric sulphate 5H2O 17.3g Sodium carbonate 10H2O 270g tri-Sodium citrate 2H2O 173g"

With regard to the comment about the most common type found, are you suggesting I should re evaluate what exactly it is I'm using. Maybe the density idea wasn't very good after all as I don't know what a professional evaluation of density would entail other than temperature.

[billnotgatez]

Just to confuse us more
Here is a MSDS I found
http://www.swimmingpoolchemicals.co.uk/msds/MSDS_pH_increaser.pdf

it says anhydrous

It came from
http://www.swimmingpoolchemicals.co.uk/healthsafe.php


I am adding these WIKI links for easy access
https://en.wikipedia.org/wiki/Benedict's_reagent
https://en.wikipedia.org/wiki/Sodium_carbonate
https://en.wikipedia.org/wiki/Copper%28II%29_sulfate
https://en.wikipedia.org/wiki/Sodium_citrate
https://en.wikipedia.org/wiki/Trisodium_citrate

[Ballistic]

oh... Am I allowed to swear now.

[billnotgatez]

Sorry for the confusion
Even the WIKI pages have some inconsistencies as to what the actual sources of  Sodium carbonate.

Do you think it is worth trying Titration.

https://en.wikipedia.org/wiki/Titration

[Ballistic]

I could try that. From what I can see it's loads and loads of work! I'm wondering that because diluting Sodium Carbonate is an exothermic reaction, that could explain why it dissolved so easily initially. If I bring the temperature up a bit and see what happens? It's a shame you can't simply take the ph with paper to work it out.

Can I simply bring the temperature up to 25 degrees and pour off the clear part and treat it as a saturated solution or no? or do the math as if I've over added reagent and add more water to compensate?