Subject: Titrations

The following comments relating to the pH at the equivalence point in
a titration apply to monoprotic acids and "monoprotic" bases (bases
that react with only 1 H+, such as NH3, also referred to as monobasic).

Let's say we are titrating HF with NaOH.  At the equivalence point all
HF will be converted over to F-,

        HF  +  NaOH  ->  Na+   +   F-   +  H2O

At the eq. pt. you now have the salt NaF.  The Na+ ions are neutral
(generally, cations cause the solution to be acidic but group 1A and
2A cations are an exception and are neutral).  The F- ions are the
conjugate base of HF.  Remember anions in general cause the solution
to be basic (except the conj. bases of the 7 strong acids, 6 which
are neutral and one, H2SO4, is acidic).  F- will act as a base and
the equilibrium at the eq. pt. that determines the pH is:


        F-   +  H2O   <->   HF   +  OH-

This means since we have a base, F-, at the eq. pt. the solution
will be basic.

To calculate the conc. of the F- at the eq. pt. you need to remember
that all the HF has been converted to F-.  Thus, the number of moles
of F- will be equal to the number of moles of HF you started with
(Ma*Va).  The molarity of the F- will be the moles of F- divided by
the total volume of solution at the eq. pt., Vt, given by the following,

	Vt = Va + Vb

where Vb is the volume of base added during the titration,

Therefore, at the eq. pt. you can easily calculate the conc. of F-,

	[F-] = (Ma*Va)/Vt,

where Ma and Va are the conc. and volume of acid, HF, you started with
and Vt = Va + Vb  (where Vb is the volume of base you added).

The volume of base added during the titration, Vb, can be found by
using,

	Ma * Va = Mb * Vb

where the Ma and Va are the conc. and volume of acid, HF, you started
with and Mb is the conc. of OH- (not just the conc. of the base - think
about the difference using NaOH and Ca(OH)2).

When using the above equations you can use mL for the volumes, you don't
have to convert them to liters.

Remember the following about the pH at the eq. pt.:

titrate a strong monoprotic acid with strong base (and vice versa):  pH = 7

titrate a weak acid with strong base:  pH > 7 (solution is basic
	since in this case you have the conj. base of the weak acid
	(an anion) you titrated and it will make the solution basic.)

titrate a weak base with strong acid:  pH < 7 (solution is acidic
        since in this case you have the conj. acid of the weak base
        (a cation) you titrated and it will make the solution acidic.)

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At the beggining of a titration, before you start titrating, it's
just a strong or weak acid/base problem.

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During a titration, and before you reach the eq. pt., it's a common-ion
problem.

You use a stoichiometry table (in moles) when dealing with strong acid-
strong base, weak acid/strong base, weak base/strong acid reactions.

Things like:

        HCl + NaOH -> NaCl + H2O        (strong acid/strong base)

        HF + NaOH -> NaF + H2O          (weak acid/strong base)

        NH3 + HCl -> NH4Cl              (weak base/strong acid)

These all go to completion (no reactants left).

You use molarities in an equilibrium (ICE) table to do the resulting
common-ion problems.

At the midpoint of the titration (1/2-way to the eq. pt.) you have
pH = pKa.  Near this point you have a buffer system and you can
use the Henderson-Hasselbalch eqn.

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Use molarities in an equilibrium (ICE) table (weak acids or bases in
solution, common ion problems, buffers, pH calculations in titrations
between the start of the titration and the equivalence pt.).