dilution and eq. pt. eqns

[Zellmer, Robert]

I've received some questions about two very similar equations
you've seen in class, M2*V2=M1*V2 and Ma*Va=Mb*Vb.  They look
similar but are different and that difference is important.

In Chem 1210 (or the equivalent) you learned about dilution and
using the eqn. for dilution,  M2*V2 = M1* V1.

Last week we discussed acid-base titrations and the fact at the
equivalence pt. of a titration the moles of acid equals the moles
of base.  Technically, this is only true for monoprotic acids and
monobasic bases (contribute 1 OH-).  More generally, it's the
moles of H+ equals the moles of OH- at the eq. pt. (which will be
the same as moles of acid = moles of base for a monoprotic
acid reacting with a monobasic base or when titrating a
polyprotic acid to the first eq. pt.).  We saw that based on
this we have an eqn that looks similar to the one above,

   mol H+ = mol OH-

mol acid = mole base (for monoprotic acids and monobasic bases)

    Ma*Va = Mb*Vb.

While the two eqns look the same they are NOT.  There are
different meanings for the M and V.

For dilution, M2*V2 = M1* V1,

    M1 = initial molarity of solute in solution
    V1 = initial volume of the solution
    M2 = final molarity of solute after dilution
    V2 = final TOTAL volume of the new solution

Remember, it's not dilution only when water is added.  If one
solution is added to another (like the BAR exp) the molarity of
everything changes (the final molarities of everything are decreased).

At the eq. pt. in an acid-base titration,

    moles acid = moles base  (technically, mol H+ = mol OH-)

      Ma*Va    =    Mb*Vb,

    Ma = molarity of the acid (technically the H+)
    Va = volume of the acid solution
    Mb = molarity of the base (technically the OH-)
    Vb = volume of the base solution

Note, neither volume is total volume of the solution resulting
from mixing the acid and base solutions during the titration.
This eqn can be used to determine the volume of base needed
to reach the eq. pt. when an acid (strong or weak) is titrated
by a base (the titrant).  In this case Ma & Va are for the acid
being titrated and the Mb & Vb are for the base titrant.  It can
also be used to determine the volume of acid needed to reach
the eq. pt. when a base (strong or weak) is titrated by an acid
(the titrant).  In this case Mb & Vb are for the base being
titrated and the Ma & Va are for the acid titrant.

Understanding the difference is important. If you do, you know
when to use which (i.e. is it a dilution problem or acid-base
neutralization problem).  A lot of people don't like to use the
Ma*Va = Mb*Vb equation for neutralization problems and titrations.
Some people don't like using the M2*V2 = M1*V1 eqn either.
If you understand each, how to use them and when to use them you
should be fine and it makes it more convenient.

Also, for a polyprotic acid you have to be careful with this eqn.  It
can be used to determine the volume of titrant required to reach each
eq. pt.  You use it using the conc. of the acid and base to find the
volume to the first eq. pt.   The volume of titrant required to reach
the second eq. pt. will  be the same as the volume to reach the first
eq. pt.  This is true for  each eq. pt.  If you want to determine the
volume of base to completely  neutralize a polyprotic acid (react with
all the acidic protons) you can  think of what I just said and use that.

Alternatively, you could keep in mind for complete neutralization of an
acid the Ma is the molarity of the H+.  The Mb is the molarity of the
OH-.  This would allow one to use the Ma*Va=Mb*Vb eqn for complete
neutralization.  For example, to completely neutralize H2SO4 with NaOH
it will take 2 moles of NaOH for every one mole of H2SO4 because there
are 2 protons in the H2SO4 which have to react.  The neutralization
reaction is,

    H2SO4   +   2 NaOH  --->   2 Na+   +   SO4^2-   + 2 H2O

You can use this eqn to determine the volume of NaOH it would take to
get to the 2nd eq. pt. by doing a stoichiometry problem or you could use
the MaVa = MbVb eqn with making sure the Ma is the conc. of H+.

For example, lets say you are titrating 30 mL of 0.10 M H2SO4 with 0.15 M
NaOH to the 2nd eq. pt. (complete neutralization).  You need to find the
volume of NaOH it will take to get to the 2nd eq. pt. Lets do it both ways
as described above.

Using the neutralization eqn and stoichiometry:

                            0.10 mol H2SO4      2 mol NaOH      1 L NaOH soln
? L NaOH = 0.030 L H2SO4 x ----------------- x ------------- x ---------------
                            1 L H2SO4 soln      1 mol H2SO4     0.15 mol NaOH

                 =  0.040 L NaOH soln   (40 mL)

        (it takes 20 mL to get to the first eq. pt. and another 20 mL to
          get to the 2nd eq. pt.)


Using the Ma*Va=Mb*Vb eqn for Ma you need to account for the fact
there are 2 H+ ions in H2SO4 which will ultimately react,

        Ma = 2(0.10 M) = 0.20 M in H+ ions

even though in solution they don't both come off completely, in the
titration they both react completely with the strong base to get to
the 2nd eq. pt.

        Ma * Va = Mb * Vb

    Vb = (Ma * Va)/Mb = {(0.20 M H+)(30 mL)}/(0.15 M OH-) = 40 mL

You get the same answer.

If only going to the first eq. pt. you would use Ma = 0.10 M
because only the first proton is reacting and the Vb would be
20 mL (and it takes another 20 mL to get to the 2nd eq. pt., i.e.
react with the 2nd proton).

When using this Ma * Va = Mb * Vb eqn for a strong or weak acid
being titrated by a strong base the Ma and Va are for the acid being
titrated and the Mb and Vb are for the strong base used as the titrant.
Sometimes people think the Mb and Vb are for the conjugate base of
the acid (particularly when it's a weak acid) that's being titrated.

For a strong or weak base being titrated by a strong acid the Mb and
Vb would be for the base being titrated and the Ma and Va are for the
strong acid used as the titrant.

For titrations we are generally using this eqn to determine the
volume of titrant needed to reach the eq. pt.  When you know where
the eq. pt. is (in terms of the volume of titrant required to reach
it) determining other things (pH at various points) becomes much
easier.  For instance if you know it takes 30 mL to reach the
eq. pt. in a WA-SB titration then if you want to know the pH at
15 mL you know this is halfway to the eq. pt. and thus pH = pKa.
You also know at the eq. pt. what reaction is taking place (hydrolysis,
a salt soln problem). In the case of a WA-SB titration it would be the
hydrolysis rxn for the conj. base of the WA and thus the soln should
be basic and you need the total volume of solution in order to
calculate the concentration of this weak base (i.e. you need the
volume of SB required to reach the eq. pt. so you can add it to the
volume of WA you started with to get the total volume).  In the case
of a WB-SA titration it would be the hydrolysis rxn for the conj. acid
of the WB and thus the soln should be acidic and you need the total
volume of solution in order to calculate the concentration of this
weak acid at the eq. pt.

Hopefully this makes sense, particularly if you've been in lecture.
I've gone over titrations more than once in lecture at this point so
with a little work things shouldn't be too bad.


Dr. Zellmer
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