Hi Mohammed,

1) 1.5 M refers to VOSO4 as it is written. If you will not change the amount of solid in exact volume, for molar concentration it does not play role, if this compound is alone in system or not. Examples:

a) If you take one mol of NaCl (1 mol of NaCl = 58,4398 g) and you dissolve in water and final volume will be 1 liter, you will have 1M solution of NaCl (and 1M solution of Na+ and 1M solution of Cl-)

b) If you take one mol of NaCl (1 mol of NaCl = 58,4398 g) and 1 mol of KCl (1 mol = 74,5483 g of KCl) and dissolve it in water and the final volume will be 1 liter, you will have again 1M solution of NaCl. But ofcourse this will be 1M solution of KCl too. According to previous, this will be 1M solution of Na+ or 1M solution of K+ but (!!!) 2M solution of Cl-.

2) 3N H2SO4 is the same as 1.5M H2SO4 (if we take acount H+) and it is the equal amount of H2SO4 molecules in unit volume. You can not prepare 3M H2SO4 from this solution. You can only dilute it and the concentration will be lower than 1.5M. If you try to dissolve VOSO4 hydrate in this (but remember the problems with solubility in acid described in article I sent you) you will dilute it with the water from hydrate and the concentration will be lower than 1.5M. You should get/buy more concentrated acid. For normality and molarity you can study links bellow. In another link there is an example of standard 0.5 M H2SO4 (1.0 N) to show the used way how concentration of H2SO4 can be expressed. Last link is online calculator that you can use for your further calculations.

3) This calculation is elucidated in capture 3.3.3. Vanadium (doesn't matter if V2+, V3+, VO2+, VO2+) concentration should be on both sides the same because vanadium is not migrating through membrane. If you have 3M H2SO4 in the presence of VOSO4 the concentration of H+ is two time higher than concentration of H2SO4, this means 6M. The last think, which should be explained, is 4.5M H+. When V2(SO4)3 is formed, part of sulfuric acid is spent and only part of free H2SO4 remains (and remaining protons can move). Imagine theoretical (simplified for understanding, not real) equation:

2 VOSO4 + H2SO4 --> V2(SO4)3 + H2 + O2

For this action, 2 mols of VOSO4 and 1 mol of H2SO4 (which is 2 mol of H+) are used. Generally, based on equation above, equal amount of moles of VOSO4 and H+ are needed to form V2(SO4)3. If you have initially 1.5 moles of VOSO4, you will spent 1.5 moles of H+ from initial 6 moles and therefore 4.5 moles remains.

The last think - the calculation of distributor:

Forget first his calculation and look to our calculation from yesterday which was:

2) VOSO4 with 4.3 H2O. Molar mass of this product will be: 163.00 + 4.3*((2*1.008)+15.999) = 240.46.

So in 1 mol of this product we have 240.46 - 163.00 = 77.46 g of water. Proportionally 250 g (your amount of hydrate) will contain (250*77.46)/240.46 = 80.53 g of water and 250-80.53=169.47 g of VOSO4.

If you have 1.5M VOSO4, you have 1.5*163.00 = 244.5 g VOSO4 in 1 liter. Proportionally your 169.47 g of VOSO4 will be in 169.47/244.5 = 0.693 liter of water (or in 693 g of water).

I suppose here the distributor made one mistake, he forget the water in your hydrate. This is already present and the amount of added water should be lower by this amount. Final amount of water which should be added is only 693-80.53 = 612.5 g of water (613 mL of water). But anyway, this theory - it is without acid, so don't do this and follow my yesterday advice.

Hope it helps you

Lukas

http://chemistry.about.com/od/chemistryterminology/a/What-Is-The-Difference-Between-Molarity-And-Normality.htm

http://www.sigmaaldrich.com/catalog/product/fluka/38294?lang=en®ion=CZ

http://www.sigmaaldrich.com/chemistry/stockroom-reagents/learning-center/technical-library/molarity-calculator.html

Please clarify your acid concentrations, it's a mess. Do you have concentrated one (96 % or 98 % = approx. 18 M) or diluted one 3 M (which is not concentrated)? And the concentration 1.5 M should be concentration of VOSO4.xH2O in final solution and the final concentration of H2SO4 in solution should be 1.5 M or 3 M? Better will be to attach the original instructions.

Thank you Dr. Richtera for your response and I am really looking for your help in that.

Actually I am not a chemical guy so I am not really sure whether the H2SO4 is concentrated or diluted one, but let me provide you the links of the products that I made orders from. Also, I have attached two photos for the products to look at them please.

For  H2SO4  (1 L)

http://www.alfa.com/en/catalog/43082

For  VoSO4. xH2O  ( 250 g)

https://us.vwr.com/store/catalog/product.jsp?catalog_number=AA11090-09#header

In fact  I would like to prepare same vanadium solution concentration that have been already made in another published paper, where he used an initial vanadium electrolyte of 1500 mol/ m3  PLUS  3000 mol/m3 of H2SO4. Eventually he got the following [ CALCULATED]  initial concentrations.

Initial vanadium concentration ( positive electrode) 1500 mol/m3.

Initial vanadium concentration ( negative electrode) 1500 mol/m3.

Initial proton concentration ( positive electrode) 6000 mol/m3.

Initial proton concentration ( negative electrode) 4500 mol/m3.

So, I would like to get same those initial concentrations as an input to the model.

I have attached this paper as well and all these values are mentioned in table 4  please.

I really appreciate any kind of help of how to start preparing the electrolyte solution in the right way.

Best regards.

You are welcome. For non-chemist it is not so trivial. Lets start with the VOSO4 first. The problem is the real amount of VOSO4 in your product (and amount of water in this hydrate which makes aditional weight). You write that the amount of water is 4.3, but if you check it according to the "Certificate of analysis" (first link below), you will get different results.

In this certificate it is written that the amount of vanadium in product is 22.98 % (impurities are not important). For better understanding imagine:

1) VOSO4 without water. Molar mass will be: 50.9415 + 32.06 + 5*(15.999) = 162.9965 which is 163.00. The amount of vanadium is: 50.9415*100/163.00 = 31.25 %

2) VOSO4 with 4.3 H2O. Molar mass of this product will be: 163.00 + 4.3*((2*1.008)+15.999) = 240.46. The amount of vanadium is: 50.9415*100/240.46 = 21.19 %

3) VOSO4 with x H2O where x corresponds to 22.98 % of vanadium. This means that 50.9415 corresponds to 22.98 % (and 100 % is 221.6775 which is molar mass of product according to the "Certificate of analysis"). From this is 163.00 VOSO4 and the rest (221.6775-163.00 = 58.6775) is water. One molecul of water has molar mass 18.015 and therefore 58.6775 correspond to 3.26 molecules of water (58.6775/18.015 = 3.26)

From this you can see that it is not clear which mass you should use (you should check once again your "Certificate of analysis" or ask distributor).

And now answer to the first part of your question. You have to prepare solution of VOSO4 with concentration of 1500 mol m-3 which is 1.5 mol dm-3. This means you must have 1.5 mol of VOSO4 in one liter of final electrolyte.

If the case (2) is correct and your product can be described as VOSO4·4.3H2O, than you should have in one liter of electrolyte 1.5 * 240.46 = 360.69 g of hydrate.

If the case (3) is correct and product correspond to calculated formula VOSO4·3.26H2O, than you should have in one liter of electrolyte 1.5 * 221.68 = 332.52 g of hydrate.

For lower final volumes of electrolyte, e.g. 100 cm3, you will use proportionally smaller amounts - 36.07 or 33.25 g of hydrate.

Now the sulfuric acid. I have bad news. Your solution should be 3M (in one liter you should have 3 mol of sulfuric acid). I am little afraid that you must get more concentrated acid. 3N H2SO4 is 1.5M - so it is to much diluted. You should buy concentrated acid like 98 % (which is approx. 18 M).

After you will get the more concentrated acid, you can take (very approximately) 167 cm3 of 98 % H2SO4 (it can be calcculated more precisely). Than you can dilute it carefully by slowly pouring into the 0.3 liter of water (not in the reverse order). Cooling and mixing is appropriate. After diluting hydrate in 0.45 liter of water you can mix the solutions and replenish to the final volume of 1 liter to have desired concentration.

Concentrations in reference [16] are different (2M) - see the second link.

Some other useful information can be found under the last link.

It is rather late - it would be good if someone could check my opinions.

https://us.vwr.com/store/asset?assetURI=https://us.vwr.com/stibo/hi_res/std.lang.all/68/11/14006811.pdf

http://www.sciencedirect.com/science/article/pii/S037877539702692X

Article High-performance microfluidic vanadium redox fuel cell

Hi Dr. Richtera,

I really liked the way of your explanation and I agree with the Uncertainties about the product concentrations.

I have been in contact with the distributor last time and for the same question he asked me to do this:-

'' Product #11090 does not have a chemical formula of VOSO4, it has a chemical formula of VOSO4•xH2O. The number of waters varies from lot to lot. Lot #R08B021 contained 4.3 waters of hydration, so has a formula weight of 240.3. If you only have 250 grams then dissolve the 250 grams in 0.69 liters (690 milliliters) to get a 1.5 molar solution. If you add any water to the solution then the sulfuric acid concentration will be less than 3.0N and the VOSO4•4.3H2O will be less than 1.5 molar. '' And I do not know how he gets this 0.69 liters.

I would like to mix 1.5 M of VOSO4.xH2O  and 3 M of H2SO4.

Three more things I still confused with please.

1 ]  As long as I am looking for a final product ( vanadium solution )1.5 molar solution, does the 1.5 M refer to the concentration of VOSO4.xH2O or  H2SO4 OR both are the same? In another word,  after mixing is the H2SO4 concentration still 3 M or 1.5 M?

2 ] Is there any differences for the H2SO4 concentration being 1.5 M, 3N H2SO4 is 1.5M, as you mentioned or being 3M on the final solution concentration? Suppose it is both of 1.5 and 3M , what is the amount in (ml) do I need for each case to make the final solution 1.5 M, please?

3] For the published one that I have mentioned in previous post and is posted below again.

He used an initial vanadium electrolyte of 1500 mol/ m3  PLUS  3000 mol/m3 of H2SO4. Eventually he got the following [ CALCULATED]  initial concentrations.

Initial vanadium concentration ( positive electrode) 1500 mol/m3.

Initial vanadium concentration ( negative electrode) 1500 mol/m3.

Initial proton concentration ( positive electrode) 6000 mol/m3.

Initial proton concentration ( negative electrode) 4500 mol/m3.

And as you know the '' H2SO4 '' is dissociated to ( HSO4 and H ) and the '' HSO4 '' into ( SO4 and H ).

How do I calculate and get the initial concentrations above (as he did)? I might use different initial concentrations for VOSO4.xH2O and H2SO4, so I would like to know the way of getting PROTON ( H ), VANADIUM, HSO4, SO4 concentrations in positive and negative electrodes please as he did.

I know I asked quite a lot, but I am working alone on this new topic and I hope this will not bother you!

Best regards,

Mohammed 

Hi Mohammed,

1) 1.5 M refers to VOSO4 as it is written. If you will not change the amount of solid in exact volume, for molar concentration it does not play role, if this compound is alone in system or not. Examples:

a) If you take one mol of NaCl (1 mol of NaCl = 58,4398 g) and you dissolve in water and final volume will be 1 liter, you will have 1M solution of NaCl (and 1M solution of Na+ and 1M solution of Cl-)

b) If you take one mol of NaCl (1 mol of NaCl = 58,4398 g) and 1 mol of KCl (1 mol = 74,5483 g of KCl) and dissolve it in water and the final volume will be 1 liter, you will have again 1M solution of NaCl. But ofcourse this will be 1M solution of KCl too. According to previous, this will be 1M solution of Na+ or 1M solution of K+ but (!!!) 2M solution of Cl-.

2) 3N H2SO4 is the same as 1.5M H2SO4 (if we take acount H+) and it is the equal amount of H2SO4 molecules in unit volume. You can not prepare 3M H2SO4 from this solution. You can only dilute it and the concentration will be lower than 1.5M. If you try to dissolve VOSO4 hydrate in this (but remember the problems with solubility in acid described in article I sent you) you will dilute it with the water from hydrate and the concentration will be lower than 1.5M. You should get/buy more concentrated acid. For normality and molarity you can study links bellow. In another link there is an example of standard 0.5 M H2SO4 (1.0 N) to show the used way how concentration of H2SO4 can be expressed. Last link is online calculator that you can use for your further calculations.

3) This calculation is elucidated in capture 3.3.3. Vanadium (doesn't matter if V2+, V3+, VO2+, VO2+) concentration should be on both sides the same because vanadium is not migrating through membrane. If you have 3M H2SO4 in the presence of VOSO4 the concentration of H+ is two time higher than concentration of H2SO4, this means 6M. The last think, which should be explained, is 4.5M H+. When V2(SO4)3 is formed, part of sulfuric acid is spent and only part of free H2SO4 remains (and remaining protons can move). Imagine theoretical (simplified for understanding, not real) equation:

2 VOSO4 + H2SO4 --> V2(SO4)3 + H2 + O2

For this action, 2 mols of VOSO4 and 1 mol of H2SO4 (which is 2 mol of H+) are used. Generally, based on equation above, equal amount of moles of VOSO4 and H+ are needed to form V2(SO4)3. If you have initially 1.5 moles of VOSO4, you will spent 1.5 moles of H+ from initial 6 moles and therefore 4.5 moles remains.

The last think - the calculation of distributor:

Forget first his calculation and look to our calculation from yesterday which was:

2) VOSO4 with 4.3 H2O. Molar mass of this product will be: 163.00 + 4.3*((2*1.008)+15.999) = 240.46.

So in 1 mol of this product we have 240.46 - 163.00 = 77.46 g of water. Proportionally 250 g (your amount of hydrate) will contain (250*77.46)/240.46 = 80.53 g of water and 250-80.53=169.47 g of VOSO4.

If you have 1.5M VOSO4, you have 1.5*163.00 = 244.5 g VOSO4 in 1 liter. Proportionally your 169.47 g of VOSO4 will be in 169.47/244.5 = 0.693 liter of water (or in 693 g of water).

I suppose here the distributor made one mistake, he forget the water in your hydrate. This is already present and the amount of added water should be lower by this amount. Final amount of water which should be added is only 693-80.53 = 612.5 g of water (613 mL of water). But anyway, this theory - it is without acid, so don't do this and follow my yesterday advice.

Hope it helps you

Lukas

http://chemistry.about.com/od/chemistryterminology/a/What-Is-The-Difference-Between-Molarity-And-Normality.htm

http://www.sigmaaldrich.com/catalog/product/fluka/38294?lang=en®ion=CZ

http://www.sigmaaldrich.com/chemistry/stockroom-reagents/learning-center/technical-library/molarity-calculator.html

Hi Dr. Richtera,

I really appreciate your time spending with me and now I got better understanding but of course I need to go back over all what you have explained and attached.

Thank you again for all the details.

Best regards.

Dear Mohammed, you are welcome. I hope your experiments will be successful.

Lukas