Fauzia,

Try this calculation and you be the judge of whether you believe it or not. I think this is what you want

First let's assign the repeat unit of the polymerized styrene sulfonate monomer as about 208 g/mole of repeat unit.  

What will I have if I made a solution of 0.4 grams of  PSS (which would be 2.2 g of your 18 wt% aqueous pSS stock polymer) in about 1000 mls of water?

0.4 grams/(208 g/mole repeat unit) = .00192 moles of repeat units.  That 0.00192 moles of repeat units would be in 1 liter of water so 0.00192 moles of repeats of PSS per liter of water is the same as 1.9 millimolar PSS or 1.9*10^03 M pss or 1.9e-3 PSS.

I think this is where your reference statement about NaCl solutions of pSS was going.

Best regards,

Mike

Dear all,

firstable, a polyelectrolyte is a polymer bearing ionizable groups (polyanion: - , polyaction: +, and polyawitterion: both - and +), and neutral polyelectrolyte bears polar neutral groups.

You have a concentration of 18 weight %, for 100ml volume, you have let us say 18g, if you devide this 18g by 75000 you will find the mol number of polymer chains in this 100ml. Know, knowing that each chain at averge contains nx(mass of reprating unit), this means 75000=nx(mass f repeating unit), this will allow you to go from the concentration expressed with respect to the molecular weight of the polymer chain to their equivalence with respect the the mass of the repeating unit forming the chain. Regards

Hallo Ms. Farheen,

I guess, your aimed concentration is 0.002 M with respect to repeating units of PSS in a NaCl solution.

The molecular weight of the repeating unit of PSS (polystyrene sulfonic acid) is 183 g/mol (take care: that of polystyrene sulfonics acid sodium salt is 206 g/mol). 

Then your 18% (weight per volume ?) stock solution of PSS (18 g solid in 100 ml = 180 g solid in 1 L) has a concentration of around 1 mol/L.

So to prepare solution of (around) 0.002 M of PSS (related to repeat units), just dilute your 18% stock solution by a factor of 500: Take 1 ml of stock solution in a beaker and fill up to 500 ml with NaCl solution.

Best regards,

Martin