April 20, 2024, 12:37:40 AM
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Topic: Chemistry Help on the subject of: % concentration, mg/mL, osmoles, osmolarity  (Read 19715 times)

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Offline boostar

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Your understanding of z is also correct, and also the 10 + 120 is correct calculated.
And how ‘many’ Percent instead of 90% are those?
Do you see already the general formula for calculating that for different z? z?

Hmmm now I'm confused... I'll show you why:

-  In case of   80%   Dissociation:
         50 molecules (z = 2)    *  0.8             --------> =    80 Ions      =  80 % Species.
         50 molecules (z = 1)    * (1.0 - 0.8 )   --------> =    10  Particles or Molecules.
                                                             Total       =    90 Species  ~ 90 % Species.

Ok so everything is solvated, so we can say there is 100% solvation? In the above, we said that 80% of the species dissociated, and than we said that 10 particles from the original compound make up 10% of the total species. Why is this? If everything is solvated, shouldn't there be 100% species solvated - not just 90%?

Offline ARGOS++

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Dear Boostar;

You are doing a very miner mistake in reasoning. - 
Not “of the total Species”!    ─     But “of the thpNoS” instead!

100% Dissociation is giving us the “theoretical possible Number of dissolved Species” (thpNoS), in this case all are Ions. This theoretical value we are easy able to calculate!

80% Dissociation is only a general value telling us that in real we only have 80 % in the ionic form.
So the remaining Species must be dissolved as Molecules, and so we have two different but dissolved Species.

Now we are able to express all in %age of theoretical possible Number of Ions (= % thpNoS):
a.)  80% dissociated Ions we can take directly  =  80.0 % thpNoS
b.)  20% not dissociated Ions we must
      convert back:  20% / z  = 20% / 2             =  10.0 % thpNoS
c.)                                                  Total     =  90.0 % thpNoS

In case all would be MgCl2 (z =3) the line b.) must look:
       ==>  20% / z  = 20% / 3                         =   6.66 % thpNoS

How big is now the Error I have done for simplification, approximately?


Good Luck!
                    ARGOS++


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