Subject:  Exp 5 (LCP) help

I'm getting a number of questions dealing with exp 5 (LCP).  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 5.  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 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 mL,
M1 = 0.50 M and V2 = 13 mL.  You can do this for the entire column for SbCl3.

You can do the same for the HCl until that last line.  First of all 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 B:

For the conc. in this part after adding the water from the titration you
use the simple dilution eqn. used for most of Part A, M2*V2 = M1*V1.
That's pretty much it and then of course write the K and substitute your
conc.

Hopefully this helps.


Dr. Zellmer