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I'm getting a lot of the same questions about calculating the <br>
molarity of the H+ and Cl- in part A and quite a few for the <br>
first table in part C. <br>
<br>
1) Part A and B:<br>
<br>
"How do I calculate the conc. of H+ and Cl-?", "Are they the
same?" <br>
"Should I be calculating the conc. of the HCl or the conc. of
the ions?" <br>
"What about the Cl in the SbCl3?" <br>
<br>
a) HCl is a STRONG acid (not the caps because it's strong). You
<br>
learned in 1210 a strong acid is one which ionizes
completely. <br>
That means for something like HCl ALL the HCl molecules
ionize <br>
(come apart) to form H+ and Cl-. Look at the stoichiometry
for <br>
this, <br>
<br>
HCl(aq) ---> H+(aq) + Cl-(aq) <br>
<br>
So for 1 HCl molecule you get 1 H+ and 1 Cl-. None of it is
actually <br>
in the form of HCl but as the ions. Thus, whatever the
conc. of HCl <br>
that is the conc of both the H+ and Cl-. In the table you
need one <br>
number (the conc. of H+ and Cl- which is the same as the
conc. of HCl, <br>
you do NOT multiply the conc. of HCl by two). <br>
<br>
b) The SbCl3(aq) is not ionic so it does not ionize in water to
form <br>
Sb<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>3</sup>+
and 3 Cl- ions. It REACTS with the water to form SbOCl and <br>
H+ and Cl-. Thus there are no Cl- ions from the SbCl3
present in <br>
the original solution (which by the way is 0.50 M in SbCl3
and <br>
6.0 M in HCl). If you started with just SbCl3 and put it in
water you <br>
would eventually reach an equilibrium in which you would
produce <br>
SbOCl(s), H+(aq) and Cl-(aq) due to a reaction, not an
dissociation <br>
or ionization of SbCl3.<br>
<br>
c) Someone asked about the questions after part A. I can't
answer them<br>
directly. However, I did address the reaction in lecture.
When you<br>
started with the solutions containing SbCl3 and HCl the
reaction was<br>
not only not at equilibtium, it wasn't even occurring. When
you<br>
added water you were diluting everything until you got to
the proper<br>
conc. of SbCl3 and HCl so the reaction would occur. Once
you added<br>
enough water and diluted everything to the correct conc. for
the reaciton<br>
to occur it then started occurring and proceeding to the
right until you<br>
reached equilibrium. In Part B I stated you should add the
water until<br>
it was just slightly cloudy, meaning the reaction occurred
and you<br>
reached an equilibrium and then I told you to add some more
water<br>
and see what happened. If you did what did you see, more
ppt or<br>
did the ppt dissolve?<br>
<br>
2) Part C: <br>
<br>
You need to look at the reaction on page 60 and pay
attention to <br>
which Co complexes are pink or blue. <br>
<br>
a) You are starting with Co(H2O)<sub>6</sub><sup>2+</sup>and
adding 12.0 M HCl to introduce <br>
Cl- ions in the first big table. You need to calculate the
conc. of Cl- <br>
(i.e. HCl) as you add it. Think of it as adding the 5.0 mL
of Co<sup class="moz-txt-sup"><span
style="display:inline-block;width:0;height:0;overflow:hidden">^</span>2</sup>+
solution <br>
to the 12.0 M HCl and diluting the HCl. For instance, using
made up <br>
numbers, if I had 7.0 mL of 12.0 M HCl and I added 6.0 mL
of any other <br>
solution (or even just water) to it I would dilute the HCl
and M1 would be <br>
12.0 M, V1 would be 7.0 mL and V2 would be (7.0 mL + 6.0 mL
= 13.0 mL). <br>
Then I would solve for M2. <br>
<br>
For the questions in parts C and D dealing with the colors in the
different solvents <br>
and why less Cl- is required to get the blue color in the last part
of Part C and <br>
how the colors change in part D based on the temp think about the
reaction on <br>
page 60. Also, for Part C think about the colors you saw for the
solids in the first <br>
part. <br>
<br>
Hopefully this clears up some of these questions. <br>
<br>
Dr. Zellmer
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