Subject:  Exp 19 help

I usually receive a number of questions dealing with exp 19.  I hope this
answers some of them.

To get moles of a substance from molarity you need to multiply the molarity
by the volume in liters,

        moles = M * V (in L)

When doing a dilution problem,

        moles after dilution = moles before dilution

            M2 * V2          =      M1 * V1

Using this eqn you don't have to convert the volumes to liters, they
can remain as milliliters.

This will apply in most cases for exp 19.  The only place it doesn't
is for that last row in Part A for the [HCl] (see below).


Part A:
-------

The easiest way to handle the [SbCl3] is to pretend you have 5 different
test tubes.  To the first test tube you add 4 mL and use M2*V2 = M1*V1.
Then you are adding another 4.0 mL of H2O for a total of 8 mL.  That's like
having a 2nd test tube in which you add 8 mL of H2O.  Thus, your V1 = 5.0 mL,
M1 = 0.10 M and V2 = 13.0 mL.  You can do this for the entire column for SbCl3.

You can do the same for the HCl until that last line.  For most of the conc.
you simply need to calculate the [HCl] using M2*V2 = M1*V1 (except for the
last line).  Remember, the HCl completely ionizes so the conc. of H+ and Cl-
will be the same as the conc. of HCl.  The last line is a little different.
For this line you have to take into account that you are adding some
additional HCl.  So you have to calc. the # moles of HCl you are adding
to the # moles of HCl you had originally and then divide by the total
volume.  That's what it means in the lab manual by "the moles of H+ and Cl-
must be combined before dividing by the total volume".  You can do this by
using a slightly modified dilution eqn.,

   moles after adding additional HCl = original moles HCl + moles HCl added

                M2 * V2              =       Mo * Vo      +      M1 * V1


Part C:
-------

Do a similar thing as I explained above for the SbCl3 in part A.  For the
Cl- (NaCl) you are starting with an M1 of 6.19 M and V1 of 1.0 mL and mixing
in 5.0 mL of the copper solution (although you actually add the NaCl to the
copper solution).  That would be for line 2 of table C2 (the big table).  The
next line is the same except V1 is 2.0 mL of 6.19 M NaCl being mixed with 5.0
mL of the copper soln (V2 changes appropriately), and so forth.  Your [Cl-]
should be increasing.

For table C3 the smaller table below the big table in Part C, you take the
conc. you got for the last line in table C2 (the big table) and use that as
your starting conc., M1, and then do your dilution calculations.


Hopefully this helps.


Dr. Zellmer