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Offline Jolly- Joliver

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How do I calculate solution concentration
« Reply #25 on: 28/03/2011 12:02:11 »
Quote
I was quoting a scientist I read that stated "one dalton was one twelth the molecular weight of a carbon atom"

Which was entirely accurate, but that's the definition of a Dalton, and only applies to carbon atoms.

Well having no one to assit and the calculator no helping me as the Dalton kept messing up the figure I was looking for I thought Bi-carbon-ate. Umm devide the lot by twelve, It is a CARBON-ate.



Quote
That's really great, but how do you work out a dalton? You basically saving that 1 da is 1 au, or hyrdogen weights one dalton.

If so what is the dalton for the sulfate and bicarbonate?

OK, firstly, you wouldn't describe it as "working out a Dalton", you'd describe it as "working out the molecular weight of the substance" (well, you would if I'm right about the question I think you're asking.

But yes, plus-or-minus the exact definition, 1 Da = 1 a.u.

So for sodium bicarbonate you'd work out the molecular weight as follows:

Formula - NaHCO3
(So that's 1 atom of sodium, one each of hydrogen and carbon, and three of oxygen, per formula unit)

I'm going to work in moles, because I'm a chemist and it's what I do...

One mole of sodium has a mass of 22.99 g
One mole of hydrogen has a mass of 1.00 g
One mole of carbon has a mass of 12.01 g
One mole of oxygen has a mass of 16.00 g

So 22.99 + 1.00 + 12.01 + 3 * 16 = 84 g / mole

Equivalently, one molecular formula's worth of sodium bicarbonate has a mass of 84 Daltons.

I'll leave the other compound as an exercise for the reader.


Great the Dalton was 84 ok using that calcultor I came up with this figure eariler:-

For the whole 96 gallon mixture:-
84.01 molecular weight sodium bicarbonate, for a concentration at 9.92ppm, weight 370187.04064 grams.  

142.04.Molecular weight sodium sulfate, for a concentraion at 6.08ppm, weight 383612.54144 grams.

Just to me that seem like a lot of grams...


Now I know the dalton are 84 for sodium bicarbonate...

For one uk gallon I am told by the calculator I must add:- 3788.5418432 grams of sodium Bicarbonate to give me a 9.92ppm concentration, is that correct Rosy?
 

Offline Jolly- Joliver

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How do I calculate solution concentration
« Reply #26 on: 28/03/2011 12:13:40 »
Rosy another question, why is it 84?

Quote
I'm going to work in moles, because I'm a chemist and it's what I do...

One mole of sodium has a mass of 22.99 g
One mole of hydrogen has a mass of 1.00 g
One mole of carbon has a mass of 12.01 g

As stated before shouldnt the carbon be set at a dalton of 1 not 12?

Quote
One mole of oxygen has a mass of 16.00 g

So 22.99 + 1.00 + 12.01 + 3 * 16 = 84 g / mole

Equivalently, one molecular formula's worth of sodium bicarbonate has a mass of 84 Daltons.

I'll leave the other compound as an exercise for the reader.
Shouldnt it be:-

22.99 + 1.00 + 1.01 +  3 * 16 = 73 g / mole. I would devide the .01 by twelve but too much hassle.

Shouldnt the daltons be 73 for sodium bicarbonate? If not why?

« Last Edit: 28/03/2011 12:17:49 by Wiybit »
 

Offline rosy

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How do I calculate solution concentration
« Reply #27 on: 28/03/2011 12:27:00 »
You've been given the weight wanted, and there's no need to invoke the molecular weight or the molar concentration to do the calculation. You're making the whole thing unnecessarily complicated. But I'll ignore that for now.

No, the Dalton is defined as 1/12 the mass of a carbon-12 atom, so one carbon-12 has a mass of 12 Daltons (to keep it in line with the original definition of an atomic mass unit which as far as I recall was based on the hydrogen atom).

I have other stuff I should be doing now, so am going to have to leave you to it...
 

Offline Jolly- Joliver

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How do I calculate solution concentration
« Reply #28 on: 28/03/2011 12:47:13 »
Im going to have a go at sodium sulfate

Sodium dodecyl sulfate C12H25SO4Na

Cabons 12. carbon weight 12.01g x 12 = 144.12 g / 12       = 12.01 daltons.

Hydrogen 25. mass of 1.00 g 25 x 1.00 = 25, AU equal to da = 25 daltons.

sulfer 1.  32.065g         au equal to da                  = 32.065 daltons

Oxygen 4. mass 16.00g. 16 x 4 = 64.00g mass total          = 64.00 Daltons

Sodium 1. mass 22.99g        au equal to da                = 22.99 daltons


12.01 + 25 + 32.065 + 64 + 22.99 = 126.056 daltons.

Is that correct Rosy?

looking at the post you made before as I was writting this, the answer is no.

Yet the molecular weight I found before was 142 for sodium sulfate, and if I add 144. instead of 12 the daltons will be 258.

So what is wrong here?

144.12 + 25 + 32.065 + 64 + 22.99 = 258.166 daltons
« Last Edit: 28/03/2011 13:00:27 by Wiybit »
 

Offline Jolly- Joliver

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How do I calculate solution concentration
« Reply #29 on: 28/03/2011 13:06:05 »
You've been given the weight wanted, and there's no need to invoke the molecular weight or the molar concentration to do the calculation. You're making the whole thing unnecessarily complicated.

It's not me it's the calculator, it demands Daltons, I try to tell it, but it wont listen,

"Give me daltons" it's says "or you're never know what weight you'll need to get 16.0ppm in 96 gallons(uk) of RO water, nerr nerr"(I'm sure if it hand fingers it would wiggle them at me) 
« Last Edit: 28/03/2011 13:26:08 by Wiybit »
 

Offline rosy

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« Reply #30 on: 28/03/2011 13:40:33 »
Ah, well no.. Sodium dodecyl sulfate is not the same as sodium sulfate. The correct formula for sodium sulfate is Na2SO4.

Do you want to try again?
 

Offline Jolly- Joliver

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How do I calculate solution concentration
« Reply #31 on: 28/03/2011 13:43:32 »
Ok Rosy so thanks to what you told me and based on the calculator

I get:

3788.5418 grams of sodium Bicarbonate to give me a 9.92ppm concentration in one gallon(uk)

and 7135.6256 grams Sodium Sulfate to give a 6.08ppm concentration in one Gallon(uk)

Combining the two:

10.92416 kg of sodium Bicarbonate 62% + Sodium Sulfate 38% to give a 16.0ppm concentration in one Gallon(uk).

It's just under 11kg solids, in one Gallon, to get a 16.0ppm does that sound right?




 

Offline Jolly- Joliver

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How do I calculate solution concentration
« Reply #32 on: 28/03/2011 13:44:34 »
Ah, well no.. Sodium dodecyl sulfate is not the same as sodium sulfate. The correct formula for sodium sulfate is Na2SO4.

Do you want to try again?

Yeah I will now
 

Offline Jolly- Joliver

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« Reply #33 on: 28/03/2011 13:54:07 »
Sodium sulfate
Na2SO4.

Na Sodium 2. 22.99g mass x 2 equals = 45.98 unit equal da             = 45.98 daltons

S sulfer 1.  32.065g         au equal to da                           = 32.065 daltons

O Oxygen 4. mass 16.00g. 16 x 4 = 64.00g mass total. equal to da      = 64.00 Daltons

45.98 + 32.065 + 64.00 =   142.045 OMG!


 

Offline Jolly- Joliver

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« Reply #34 on: 28/03/2011 14:05:39 »
So the first one I posted was with Sodium dodecyl Sulfate:-

3788.5418 grams of sodium Bicarbonate to give me a 9.92ppm concentration in one gallon(uk)

and 7135.6256 grams Sodium dodecyl Sulfate to give a 6.08ppm concentration in one Gallon(uk)

Combining the two:

10.92416 kg of sodium Bicarbonate 62% + Sodium dodecyl Sulfate 38% to give a 16.0ppm concentration in one Gallon(uk).

------

And now actual sodium sulfate:-

3788.5418 grams of sodium Bicarbonate to give me a 9.92ppm concentration in one gallon(uk)

and 3926.0783 grams Sodium Sulfate to give a 6.08ppm concentration in one Gallon(uk)

Combining the two:

7.7146 kg of sodium Bicarbonate 62% + Sodium Sulfate 38% to give a 16.0ppm concentration in one Gallon(uk).

-------------

Rosy are they ok? It's just under 8kg in one gallon, the calculator says that's the weights needed.

This is the calculator I'm using
http://www.currentprotocols.com/tools/solution-concentration-calculator

One gallon(uk) I calculate to be 4546ml
ppm at 9.92 or 6.08- M. as stated on the calculator
and then the Daltons

Just under 8kg for one gallon tho, there are 96 gallons total just seems like a lot of solid for 16ppm.

If I choose the smaller concentration setting mM for the bicarbonate it states I need 3.79grams to get 9.98. but I felt the ppm was the highier setting M as the guy who questioned was using 36.6kg to get 0.8ppm.

Rosy what's the difference between the M and mM settings which represents ppm? I can see mM deals with smaller parts.

Definition ppm
Quote

"This is a way of expressing very dilute concentrations of substances. Just as per cent means out of a hundred, so parts per million or ppm means out of a million. Usually describes the concentration of something in water or soil. One ppm is equivalent to 1 milligram of something per liter of water (mg/l) or 1 milligram of something per kilogram soil (mg/kg)."

Would I be right in thinking that M is parts per cent?
and mM is ppm?

Nightmare gotta re-do all the calculations
« Last Edit: 28/03/2011 14:53:38 by Wiybit »
 

Offline Jolly- Joliver

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« Reply #35 on: 28/03/2011 15:02:08 »
OK...

If mM is the repersentation for ppm then I get


3.79 grams of sodium Bicarbonate to give me a 9.92ppm concentration in one gallon(uk)

and 3.93 grams Sodium Sulfate to give a 6.08ppm concentration in one Gallon(uk)

Combining the two:

7.72g of sodium Bicarbonate 62% + Sodium Sulfate 38% to give a 16.0ppm concentration in one Gallon(uk).

7.72g x 96 = 741.12g for the whole 96 gallons gives 16.0 ppm.

The maths is easier actaully if mM is the ppm.

But how can that be? The guy that posted claimed he was getting 0.8ppm for 36.6kg in 96 gallons?

Rosy is that right?

----------------------------------

If M is ppm

then:-

3788.5418 grams of sodium Bicarbonate to give me a 9.92ppm concentration in one gallon(uk)

and 3926.0783 grams Sodium Sulfate to give a 6.08ppm concentration in one Gallon(uk)

Combining the two:

7.7146 kg of sodium Bicarbonate 62% + Sodium Sulfate 38% to give a 16.0ppm concentration in one Gallon(uk).

if not then the above should read ppc parts per cent instead of ppm!

But I kinda think it's pretty clear.
« Last Edit: 28/03/2011 15:36:04 by Wiybit »
 

Offline Jolly- Joliver

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« Reply #36 on: 28/03/2011 15:20:14 »
I work at a water bottling plant and I'm just trying to figure out how to test our mineral solutions.  I'd appreciate it if anyone can help me with this one.

One blend of water uses a sodium sulfate/sodium bicarbonate solution which is injected in RO water at a rate of .75ml per 1 gallon.  The mineral solution is made of 96 gallons RO water, 30.72 lbs sodium sulfate and 50 lbs sodium bicarbonate.  TDS on RO averages approx 0.8 ppm and the desired TDS on finished product is 12 - 18 ppm.  What should the TDS on the mineral solution be to give me the desired result on finished product. 



DEAR Hogied,

Thankyou for helping me understand Daltons. Although really Rosy did, you did play some part in that.

As a final answer the TDS total disolved solids or total dry solids for the whole mixture should be about 741.12grams.

Although I think you want the part per cent and not part per million. but then your using a 75ml solution to add the material, go figure.
« Last Edit: 28/03/2011 15:22:44 by Wiybit »
 

Offline rosy

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« Reply #37 on: 28/03/2011 15:54:05 »
Just quickly whilst I've got a moment:

There are two ways of measuring concentration.. one is by weight-per-solvent, the other by molar concentration.

Weight-per-volume is self explanatory, at least in principle. I think this is what is meant here, because it's the sort of thing you see written on the side of mineral water bottles. One ppm or part-per-million is one milligram of whatever the solid (or more generally solute) is per litre of water (since on litre of water weighs 1 kg, at room temperature), and one milligram is one millionth of a gram. So if the water contained 1 mg of sodium carbonate it would be one ppm sodium carbonate. This is why I said originally that we didn't need to worry about the molecular weights to do the calculations.


The molar concentration is to do (in effect) with the number of molecules (of whatever it is) in the solution. As I explained earlier, one mole of a substance is 6.02 x 1023 whatever the substance. And a molar concentration, a concentration of 1 M is 1 mole of the solute dissolved per litre. So if the molar mass of sodium bicarb is 84 g/mol, the mass of one mole of sodium bicarb is 84 g and to make a 1 M solution we'd put 84 g of sodium bicarbonate into a container, make up the volume to 1 L and bingo, a 1 M solution. A 1 mM solution has a concentration of 0.001 M (1/1000 of 1 M), so we'd use 84 mg or 0.084 g of sodium bicarb.

To make a 2 M solution of sodium hydroxide (Mw = 23 + 1 + 16 = 40) we'd use 2 x 40 = 80 g/L.

Is this starting to make sense?
 

Offline Jolly- Joliver

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« Reply #38 on: 28/03/2011 16:28:14 »
Just quickly whilst I've got a moment:

There are two ways of measuring concentration.. one is by weight-per-solvent, the other by molar concentration.

Weight-per-volume is self explanatory, at least in principle. I think this is what is meant here, because it's the sort of thing you see written on the side of mineral water bottles. One ppm or part-per-million is one milligram of whatever the solid (or more generally solute) is per litre of water (since on litre of water weighs 1 kg, at room temperature), and one milligram is one millionth of a gram. So if the water contained 1 mg of sodium carbonate it would be one ppm sodium carbonate.

So 16ppm is 16 mg.



This is why I said originally that we didn't need to worry about the molecular weights to do the calculations.


The molar concentration is to do (in effect) with the number of molecules (of whatever it is) in the solution. As I explained earlier, one mole of a substance is 6.02 x 1023 whatever the substance. And a molar concentration, a concentration of 1 M is 1 mole of the solute dissolved per litre. So if the molar mass of sodium bicarb is 84 g/mol, the mass of one mole of sodium bicarb is 84 g and to make a 1 M solution we'd put 84 g of sodium bicarbonate into a container, make up the volume to 1 L and bingo, a 1 M solution. A 1 mM solution has a concentration of 0.001 M (1/1000 of 1 M), so we'd use 84 mg or 0.084 g of sodium bicarb.

Simply put to make a 16M or 16ppm solution of sodium bicarbonate we add- 84 miligrams which is 1M x 16 = 1344miligrams or 1.344grams, for each litre. if not it's 1.344kg to a litre, at 84g times 16.

One thing a mil is a thousand, not a million. A kg is a thousand grams not a million grams, but it is a million miligrams.

So why does the calculator ask for Daltons? I assumed it was because the molecular weight effected the solvency. but as you said
Quote
the molar mass of sodium bicarb is 84 g/mol
you need the daltons to know the single ppm unit or M unit.


To make a 2 M solution of sodium hydroxide (Mw = 23 + 1 + 16 = 40) we'd use 2 x 40 = 80 g/L.

Is this starting to make sense?

80g per litre. One unit is 40, 2 units 80. 2ppm. I think it is.


Sorry I never studied this at school, and the guy maths just through me off completly.
« Last Edit: 28/03/2011 16:46:02 by Wiybit »
 

Offline rosy

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« Reply #39 on: 28/03/2011 16:53:50 »
Could you give me a link to the calculator? Might help me explain what it's asking for!

16 ppm is only 16 mg if you're making 1 L of solution, because 1L water = 1 kg water = 100 g = 1,000,000 mg, so you've got 16 "parts solute" per million "parts solvent". If you're talking, as here, about mass-per-mass, you simply don't need to think about the molar concentration, and thus the molecular mass (in units of Daltons or any other unit) just doesn't come into it. There's no need to consider it.

The molar concentration is, numerically, completely different to the concentration expressed in ppm weight-for-weight. You're trying to conflate two unrelated things!

To make a 16 ppm solution, we take 16 mg of stuff, here sodium bicarb, and dissolve it in 1 L water (1000000 mg water).

To make a 16 mM solution (millimolar, 0.001 M) we want 0.016 moles of sodium bicarb per litre. 1 mole of sodium bicarb has a mass of 84 g, so 84 x 0.016 = 1.34 g, we dissolve 1.34 g of sodium bicarbonate in our 1 L of water.

 

Offline rosy

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« Reply #40 on: 28/03/2011 17:03:55 »
Hm. I think I'm playing whack-a-rat with a whole lot of misconceptions here. Each time I correct one, I create a new one.

I'm going to throw in the towel and suggest you get hold of a GCSE chemistry book and work through the section on solutions, because you're obviously interested but my explaining skills aren't up to doing this as Q&A in a text-only medium when I'm supposed to be doing something else... there are GCSE text books in your local library, and on the internet too. I reckon that's the level you want to start off with.
 

Offline Jolly- Joliver

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« Reply #41 on: 28/03/2011 17:37:08 »
Could you give me a link to the calculator? Might help me explain what it's asking for!

This is the calculator
http://www.currentprotocols.com/tools/solution-concentration-calculator
although I already linked to it.



16 ppm is only 16 mg if you're making 1 L of solution, because 1L water = 1 kg water = 100 g = 1,000,000 mg, so you've got 16 "parts solute" per million "parts solvent". If you're talking, as here, about mass-per-mass, you simply don't need to think about the molar concentration, and thus the molecular mass (in units of Daltons or any other unit) just doesn't come into it. There's no need to consider it.

Yeah I think the problem is conception of numbers. Metric follows for each kilo metre is 1000 metres, metres a humdred centmetres, and centmetres 10 minimetres.

so 100 centimetres is 100,000 milimters. 10cm 100 milimetres.

Kilogram 1000 grams, gram 1000 miligrams, but a 100g would be miligrams x 100= 100,000mg not 
Quote
100 g = 1,000,000 mg
that's a kilo.

isnt it? are there not a 1000mg in a gram? and 100 centigrams in a gram? and 100,000 centigrams in a kilogram? a 1,000,000mg is one kilogram.

http://www.metric-conversions.org/weight/milligrams-to-grams.htm

http://www.unitconversion.org/weight/centigrams-to-grams-conversion.html



The molar concentration is, numerically, completely different to the concentration expressed in ppm weight-for-weight. You're trying to conflate two unrelated things!

Yet as you said the mass/mol figure gives you 1m or 1ppm, the daltons give you the 1m unit.




To make a 16 ppm solution, we take 16 mg of stuff, here sodium bicarb, and dissolve it in 1 L water (1000000 mg water).

To make a 16 mM solution (millimolar, 0.001 M) we want 0.016 moles of sodium bicarb per litre. 1 mole of sodium bicarb has a mass of 84 g, so 84 x 0.016 = 1.34 g, we dissolve 1.34 g of sodium bicarbonate in our 1 L of water.



I think I got it milmolers are 1000s of a miligram so there should be 1,000,000MM in one gram.

that to me
Quote
(millimolar, 0.001 M)
Is one miligram. expressed as such 1,Gram 001miligram 001Minimoler one gram one miligram one minimoler surely should be expressed in number form as 1.001001

if not a miligram and a milimoler are the same thing .001

Or your shifting the point back three places from gram 1.000(to here) 1.001 one miligram one milimoler. 

If you are doing that its abit confusing because and one miligram is expressed 0.001 and one minimoler is also expressed in the same way 0.001
« Last Edit: 28/03/2011 18:08:38 by Wiybit »
 

Offline Jolly- Joliver

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« Reply #42 on: 28/03/2011 18:15:32 »
Sorry thats a microgram 1000 of a miligram so,

Just to clarify one kilogram, one gram, one centigram, one miligram, and one microgram I Believe should be written in number form as so:

1,001.011001

Kilo. gram. centigram, miligram, microgram.

But looking into it, a milimole is 1000 of a mole and a mole is, http://en.wikipedia.org/wiki/Mole_(unit)

Quote
"1 mol of calcium-40 has an approximative mass of 40g,
so 1 minimole of calcium-40 is 0.04 or 4 centigrams.

http://wiki.answers.com/Q/1_mole_is_how_many_millimoles

Quote

1 mole = 1000 millimoles, 1 millimole = 0.001 moles
and here it is expressed in the same way as a miligram.

It appears to me that a mole and a milimole are a totally different system of calculation, doing a differnt job to miligrams.
« Last Edit: 28/03/2011 18:44:33 by Wiybit »
 

Offline Jolly- Joliver

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« Reply #43 on: 28/03/2011 18:56:27 »
I work at a water bottling plant and I'm just trying to figure out how to test our mineral solutions.  I'd appreciate it if anyone can help me with this one.

One blend of water uses a sodium sulfate/sodium bicarbonate solution which is injected in RO water at a rate of .75ml per 1 gallon.  The mineral solution is made of 96 gallons RO water, 30.72 lbs sodium sulfate and 50 lbs sodium bicarbonate.  TDS on RO averages approx 0.8 ppm and the desired TDS on finished product is 12 - 18 ppm.  What should the TDS on the mineral solution be to give me the desired result on finished product. 


Last go, 4.546 litres, 9.98 ppm Daltons of 84 for sodium bicarbonate, 84mg is one unit, I seek 10 more or less 10 x 84 miligrams 840mg of sodium bicarbonate per litre.

840mg x 4 = 3.360grams. then .5 of a litre need half a unit = 420mg + 3.360grams = 3.780grams of bicarbonate of sodium for 4.546 litres should give 9.98-10ppm more or less, it's roughtly that I think.

Rosy is that correct for the bicarbonate of sodium?
 

Offline Bored chemist

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« Reply #44 on: 28/03/2011 19:12:56 »
"It's not me it's the calculator, it demands Daltons, I try to tell it, but it wont listen,"
get a better calculator.


OK, I guess I should have done this earlier before Wiybit wasted so much of his time.

The question is "I work at a water bottling plant and I'm just trying to figure out how to test our mineral solutions.  I'd appreciate it if anyone can help me with this one.

One blend of water uses a sodium sulfate/sodium bicarbonate solution which is injected in RO water at a rate of .75ml per 1 gallon.  The mineral solution is made of 96 gallons RO water, 30.72 lbs sodium sulfate and 50 lbs sodium bicarbonate.  TDS on RO averages approx 0.8 ppm and the desired TDS on finished product is 12 - 18 ppm.  What should the TDS on the mineral solution be to give me the desired result on finished product.  "

OK, the water starts with about 1 ppm of TDS and they want about 15 ppm so they need to add about 14 ppm.
They plan to do that by adding 0.75 ml/ of solution per gallon.
I guess its a US gallon. that's 3.79 litres.
0.75 ml in 3.79 litres is a dilution of about 5050 to one so the solution added needs to be about 5050 times 14 ppm
That's about 7% w/v

So Just to check.
Start with a solution containing 7% of whatever. That's 7 g in 100 ml or 7000 mg in 100 ml or 70 mg in 1 ml
Take 0.75 ml of it so 52.5 mg
Dissolve that in 3.79 litres
13.85 mg/ litre i.e 13.85 ppm
There's already 0.8 mg/l present so that makes a total of 14.65 mg / litre
That's nicely in the middle of the range (12 to 18) they asked for.

See, no Daltons and no moles. Barely any maths and certainly not several postings and several pages of stuff.

If you want that in UK gallons then it's 7*4.54/3.79
about 8.4%


Of course, it depends on how you measure TDS too.
 

Offline Jolly- Joliver

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« Reply #45 on: 28/03/2011 19:59:29 »
"It's not me it's the calculator, it demands Daltons, I try to tell it, but it wont listen,"
get a better calculator.


OK, I guess I should have done this earlier before Wiybit wasted so much of his time.

The question is "I work at a water bottling plant and I'm just trying to figure out how to test our mineral solutions.  I'd appreciate it if anyone can help me with this one.

One blend of water uses a sodium sulfate/sodium bicarbonate solution which is injected in RO water at a rate of .75ml per 1 gallon.  The mineral solution is made of 96 gallons RO water, 30.72 lbs sodium sulfate and 50 lbs sodium bicarbonate.  TDS on RO averages approx 0.8 ppm and the desired TDS on finished product is 12 - 18 ppm.  What should the TDS on the mineral solution be to give me the desired result on finished product.  "

OK, the water starts with about 1 ppm of TDS and they want about 15 ppm so they need to add about 14 ppm.
They plan to do that by adding 0.75 ml/ of solution per gallon.
I guess its a US gallon. that's 3.79 litres.
0.75 ml in 3.79 litres is a dilution of about 5050 to one so the solution added needs to be about 5050 times 14 ppm
That's about 7% w/v

So Just to check.
Start with a solution containing 7% of whatever. That's 7 g in 100 ml or 7000 mg in 100 ml or 70 mg in 1 ml
Take 0.75 ml of it so 52.5 mg
Dissolve that in 3.79 litres
13.85 mg/ litre i.e 13.85 ppm
There's already 0.8 mg/l present so that makes a total of 14.65 mg / litre
That's nicely in the middle of the range (12 to 18) they asked for.

See, no Daltons and no moles. Barely any maths and certainly not several postings and several pages of stuff.

If you want that in UK gallons then it's 7*4.54/3.79
about 8.4%


Of course, it depends on how you measure TDS too.


Cool thankyou bored chemist.

I think what I miss understood form the begging was the 0.8, as far as I understood it he was saying they wanted between 12-18 but were only getting a 0.8, so what did he need to do to increase the solution level to get between 12-18? You've added a secound time to the same RO water after the first 75ml injuction. 

I think it is Uk gallons, he's based in Britain.

Quote
0.75 ml in 3.79 litres is a dilution of about 5050 to one so the solution added needs to be about 5050 times 14 ppm That's about 7% w/v

Right so the 75ml solution it to be 5050 times 14ppm, he is using sodium bicardonate and sodium sulfate and the ratios are 62% and 38% per ppm.

The 7% w/v is water volume I take it? 7% of the 75ml solution should be 14ppm, to achieve the correct increase?

The thing I have trouble with is 75ml isnt much liquid, is really possible to add the 14ppm needed for the gallon, to that? You claim it is, but it just seems very tight to me, do you get what I'm saying? It 75ml of solution to increase the ppms of a gallon by 14ppm.

I dont know I'm not Chemist you are.

Thanks for the reply.

One last thing the equasion I placed in my last post, what would that achieve in a uk gallon mix if you did it, in terms of ppm, adding 3.780 grams of sodium bicarbonate, what would that raise the ppm to, if it was plain RO water as I assumed, no ppm present?
I just want to know how I did?   



 

Offline Jolly- Joliver

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How do I calculate solution concentration
« Reply #46 on: 28/03/2011 20:33:03 »
Sorry Bored Chemist, stupid as always,

I'll just asked the frigging calculator. So silly!

Looks alright for mM.

I get 9.89881mM
and 0.0831500 v/W

or

uM 9898.81
ppm 831.500, so that's well off.

or

per uM3 5.96120e+6
ppb 831500

Moler 0.00989862

Mass against, molecular concentrations.

http://www.calctool.org/CALC/chem/molecular/solution

Anyway gonna give up on chemisty I think.
« Last Edit: 28/03/2011 20:53:51 by Wiybit »
 

Offline Bored chemist

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How do I calculate solution concentration
« Reply #47 on: 28/03/2011 22:20:45 »
"One last thing the equasion I placed in my last post, what would that achieve in a uk gallon mix if you did it, in terms of ppm, adding 3.780 grams of sodium bicarbonate, what would that raise the ppm to, if it was plain RO water as I assumed, no ppm present?"

OK
3.78g in 1 gal
3.78g in 4.54 litres
0.83 g/litre
830 milligrams/ litre
830 ppm
Not close.
Why do you keep talking about daltons, molarities and moles?
Try the calculator here
http://www.calctool.org/CALC/chem/molecular/solution
(It's the one you cited.)
Put in 3.78g in 1 Imp gallon and it tells you that it's about 800 ppm.
Put any number you like in the molecular mass box ( and any units). It even  gives the right answer in ppm when you put a negative molar mass (which is, of course, impossible) because it doesn't need to know it.

« Last Edit: 28/03/2011 22:25:31 by Bored chemist »
 

Offline Jolly- Joliver

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How do I calculate solution concentration
« Reply #48 on: 29/03/2011 07:22:29 »
"One last thing the equasion I placed in my last post, what would that achieve in a uk gallon mix if you did it, in terms of ppm, adding 3.780 grams of sodium bicarbonate, what would that raise the ppm to, if it was plain RO water as I assumed, no ppm present?"

OK
3.78g in 1 gal
3.78g in 4.54 litres
0.83 g/litre
830 milligrams/ litre
830 ppm
Not close.
Why do you keep talking about daltons, molarities and moles?

Well I came in knowing nothing really and then after looking into it a bit the concentration calculator stated it needed daltons to assess the ppm needed.




Try the calculator here
http://www.calctool.org/CALC/chem/molecular/solution
(It's the one you cited.)

You did say find a better one ;)

The first one I used would only work having a dalton figure.



Put in 3.78g in 1 Imp gallon and it tells you that it's about 800 ppm.
Put any number you like in the molecular mass box ( and any units). It even  gives the right answer in ppm when you put a negative molar mass (which is, of course, impossible) because it doesn't need to know it.




It's werid I get 1 ppm as 4,547micrograms or 0.004547grams or 4.547miligrams

PPm is parts per milionth of a litre correct?

4.5miligrams give a ppm of 1 actually it's 4,546.1 mircograms or 0.0045461grams

Giving 0.0727376 grams nes for a 16ppm it's 72,737.6 micrograms. 72.7376mg

73mg in 75ml of water goes I think, almost 1 for 1 mix, solids to liquid.
« Last Edit: 29/03/2011 10:13:13 by Wiybit »
 

Offline Jolly- Joliver

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How do I calculate solution concentration
« Reply #49 on: 29/03/2011 08:56:33 »
So knowing I need to add arround 73mg to the gallon of water to get a ppm of 16ppm.

He already has 0.8

73mg / 100 = 0.73 x 62 = 45.26mg of sodium bicarbonate

73mg - 45.26mg = 27.74mg of sodium sulfate

so making a 75ml RO soultion with 27.74mg sodium sulfate and 45.26mg sodium bicarbonate

then adding that to one gallon with a 0.8ppm

Should give a gallon RO with 16.8ppm

Ofcourse I have increase the mg to 73mg an increase of 0.2624mg

The calculator tells me 0.2624mg will increase the ppm by 0.0577199ppm

So the actual ppm of the Gallon Ro should be if I did that 16.8577199ppm

Almost 17ppm

Is that correct? I think that is.

The 75ml shouldn't effect the ppm too much should it?

To have 0.8pmm in the Ro gallon he already had 0.0036368grams

Adding 75ml to the gallon gives me a litres of 4.621

Adding 0.0036368grams to the 73mg gives me 0.0766368grams in 4.621 litres

Or 76.6368mg in 4,621ml

The ppm I get is = 16.5845 Umm

I made it 16.85 before but then I changed from Gallon to litres and added the 75ml.

So the effect of the 75ml in the gallon looks to be about 0.21 something ppm
« Last Edit: 29/03/2011 09:29:26 by Wiybit »
 

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How do I calculate solution concentration
« Reply #49 on: 29/03/2011 08:56:33 »

 

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