June 12, 2021, 10:57:57 AM
Forum Rules: Read This Before Posting


Topic: i am out of ideas!  (Read 11982 times)

0 Members and 1 Guest are viewing this topic.

light

  • Guest
i am out of ideas!
« on: September 07, 2004, 05:19:21 PM »
Hello everybody!

This is one of the biggest challenges I have ever faced  :-X. I hope that you have dealt with this matter before and hope that you can give me some advice and help.
Thanks very much in advance!

the recipe:

Sodium phosphate:   Each ml contains 276mg of monobasic Naphos (NaH2PO4), and 142mg of dibasic naphos (Na2HPO4). 93 mg of phosphorus/ml = 3 mmol.  92 mg of sodium per ml/23 = 4 meq/ml.  Osmolarity: 7000 mOsm/L

my problems:

a) as the recipe states 1 ml solution contains 276mg of monobasic Naphos (NaH2PO4), and 142mg of dibasic naphos (Na2HPO4). then how come they only get 93 mg of phosphorus/ml = 3 mmol??
It seems like we are dealing with a buffer ( NaH2PO4 and Na2HPO4).

b) again, how come they only get 92 mg of sodium per ml/23 = 4 meq/ml???

c) how many particles do i have in this solution?

My point:
NaH2PO4 -> Na+    +    H2PO4-
I am not sure if the dissociation stops here or if H2PO4- will dissociate more and give more particles????


Na2HPO4-> 2Na+     +    HPO4(2-)
like the above, will H2PO4- dissociate more and give more particles????

then what is the total number of particles we have in solution? How can you calculate the osmolarity when you have the combination of  NaH2PO4 and Na2HPO4?


hope very much for advices!

Demotivator

  • Guest
Re:i am out of ideas!
« Reply #1 on: September 07, 2004, 07:23:34 PM »
276mg/120 mg/mmole = 2.3 mmoles : rounded off = 2 mmoles = 2 mmole Na and 2 mmole H2PO4. 2 + 2 = 4 mOsmole
142 mg/ 142 mg/mmole = 1 mmole = 2 mmole Na and 1 mmole HPO4. 2 + 1 = 3 mOsmole
Grand total particles: 4 + 3  = 7 mOsm/ml = 7000 mOsmoles/L

Th 93 mg P is a different  technique of counting that leads to the same mmoles. There are 3 mmoles phosphorous, 2 from the H2PO4 give 2x31 = 62 mg
and 1 from the HPO4 gives 31 mg. Total is 93 mg.
(31 is the atomic weight of phosphorous)
« Last Edit: September 07, 2004, 07:36:22 PM by Demotivator »

light

  • Guest
Re:i am out of ideas!
« Reply #2 on: September 08, 2004, 08:34:01 AM »
thanks so much!

i would never be able to solve this matter by my own  :P so thanks to you!

but i wonder in adjusting pH we can either use HCl or NaOH dependent on the pH values, but what about osmolarity? what can i generallly use to adjust the osmolarity?

like when i want to increase the osmolarity, what should i use? should i use any kind of salts/compounds as long as they aren't the specified ones? since adding more of them would change the concentration which is specified/needed??


and when i want to dilute/decrease the osmolarity what should i use?

thanks again!
« Last Edit: September 08, 2004, 08:57:23 AM by light »

Demotivator

  • Guest
Re:i am out of ideas!
« Reply #3 on: September 08, 2004, 10:01:02 AM »
To increase osmolarity one would add salts that don't interfere with the PH or affect the system by reacting with it. NaCl is a typical adjuster.
I don't think you can decrease osmolarity other than by diluting it.

light

  • Guest
Re:i am out of ideas!
« Reply #4 on: September 08, 2004, 04:38:10 PM »
hi Demotivator!

thanks for your great help, but i have one doubt left ::)
i hope you don't mind answering this question, since this is quite important for me!

 if i want to make a solution of 1litre. after i have added all the ingredients and added water to for instance 500ml. the osmolarity shows to be for instance 200mOsm (using an osmometer) in this 500ml. do you know how i can calculate to find what the total osmolarity will be when i have added the other 500ml to make the total volume? the osmolarity will decrease when we add more water, but how can i calculate it? this is important since i have to know it for adjusting.

hope very much for replies!
thank you so much!
« Last Edit: September 08, 2004, 04:53:00 PM by light »

Demotivator

  • Guest
Re:i am out of ideas!
« Reply #5 on: September 08, 2004, 06:26:38 PM »
To calculate Use the dilution formula
M1V1 = M2V2
(assuming you meant 200 mOsm/Liter, not 200 mOsm):
(200 mOsm/Liter)(500 ml) = M2(1000 ml)
M2 = 100 mOsm/Liter  Makes sense that when the volume is doubled the concentration is halved.

Sponsored Links