Exp 19 questions

[robert zellmer]

I'm getting a lot of the same questions about calculating the
molarity of the H+ and Cl- in part A and quite a few for the
first table in part C.

1) Part A and B:

     "How do I calculate the conc. of H+ and Cl-?", "Are they the same?"
     "Should I be calculating the conc. of the HCl or the conc. of the 
ions?"
     "What about the Cl in the SbCl3?"

     a)  HCl is a STRONG acid (not the caps because it's strong). You
         learned in 1210 a strong acid is one which ionizes completely.
         That means for something like HCl ALL the HCl molecules ionize
         (come apart) to form H+ and Cl-.  Look at the stoichiometry for
         this,

                 HCl(aq) --->  H+(aq)  +  Cl-(aq)

         So for 1 HCl molecule you get 1 H+ and 1 Cl-.  None of it is 
actually
         in the form of HCl but as the ions.  Thus, whatever the conc. 
of HCl
         that is the conc of both the H+ and Cl-.  In the table you need 
one
         number (the conc. of H+ and Cl- which is the same as the conc. 
of HCl,
         you do NOT multiply the conc. of HCl by two).

     b)  The SbCl3(aq) is not ionic so it does not ionize in water to form
         Sb^3+ and 3 Cl- ions.  It REACTS with the water to form SbOCl and
         H+ and Cl-.  Thus there are no Cl- ions from the SbCl3 present in
         the original solution (which by the way is 0.50 M in SbCl3 and
         6.0 M in HCl).  If you started with just SbCl3 and put it in 
water you
         would eventually reach an equilibrium in which you would produce
         SbOCl(s), H+(aq) and Cl-(aq) due to a reaction, not an 
dissociation
         or ionization of SbCl3.

     c) Someone asked about the questions after part A.  I can't answer them
         directly.  However, I did address the reaction in lecture. When you
         started with the solutions containing SbCl3 and HCl the 
reaction was
         not only not at equilibtium, it wasn't even occurring.  When you
         added water you were diluting everything until you got to the 
proper
         conc. of SbCl3 and HCl so the reaction would occur.  Once you added
         enough water and diluted everything to the correct conc. for 
the reaciton
         to occur it then started occurring and proceeding to the right 
until you
         reached equilibrium.  In Part B I stated you should add the 
water until
         it was just slightly cloudy, meaning the reaction occurred and you
         reached an equilibrium and then I told you to add some more water
         and see what happened.  If you did what did you see, more ppt or
         did the ppt dissolve?

2)  Part C:

         You need to look at the reaction on page 60 and pay attention to
         which Co complexes are pink or blue.

     a) You are starting with Co(H2O)_6 ^2+ and adding 12.0 M HCl to 
introduce
         Cl- ions in the first big table.  You need to calculate the 
conc. of Cl-
         (i.e. HCl) as you add it.  Think of it as adding the 5.0 mL of 
Co^2+ solution
         to the 12.0 M HCl and diluting the HCl.  For instance, using 
made up
         numbers, if I had 7.0 mL of  12.0 M HCl and I added 6.0 mL of 
any other
         solution (or even just water) to it I would dilute the HCl and 
M1 would be
         12.0 M, V1 would be 7.0 mL and V2 would be (7.0 mL + 6.0 mL = 
13.0 mL).
         Then I would solve for M2.

For the questions in parts C and D dealing with the colors in the 
different solvents
and why less Cl- is required to get the blue color in the last part of 
Part C and
how the colors change in part D based on the temp think about the 
reaction on
page 60.  Also, for Part C think about the colors you saw for the solids 
in the first
part.

Hopefully this clears up some of these questions.

Dr. Zellmer
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