Equation for reaction between EDTA and Lead Nitrate (NOT NITRIC ACID)
Mr Semiwestern
Nitrate (NOT NITRIC ACID as I said before)
Could someone please tell me the equation for the reaction
between Lead Nitrate and EDTA?
Thanks
Dr. Henry
--
Dr. Henry Boyter, Jr. Ph.D. Chemist
The opinions of Dr. Boyter are provided for informational
purposes only and should not be used as advice. No
warranty or expression of professionalism is implied.
**********************************************************
Mr Semiwestern wrote in message <35d1dce6...@news.easynet.co.uk>...
Dr. Henry
Eric Lucas
EDTA. Pb(+4) is a fairly strong Lewis acid, and presumably acidifies the
bound EDTA, preferring to bind Pb(+4) as EDTA(-4) or H2EDTA(-2), releasing
free H+ to be absorbed by the amines of a free EDTA. On the other hand,
the soft Pb(+2) is much less acidic, and is likely much happier bound to
H4EDTA or H2EDTA(-2). The free H+ ions then have an effect on the
titration of the excess EDTA. Just a guess, however.
Eric Lucas
Don Wilkins wrote:
> On Wed, 12 Aug 1998 15:37:18 -0400, "Dr. Henry" <hbo...@cstone.net>
> wrote:
>
> >What valence of lead?
>
> One would assume that the valence was +2 but pray tell what difference
> does it make? It is not a redox reaction.
>
> One EDTA per one lead ion in solution.
Don Wilkins
Don Wilkins
Semiwestern) wrote:
>I am trying to calculate the excess EDTA in a titration with Lead
>Nitrate (NOT NITRIC ACID as I said before)
>Could someone please tell me the equation for the reaction
>between Lead Nitrate and EDTA?
There will be one mole of EDTA per mole of lead.
Don Wilkins
wrote:
>It does, however, probably affect the state of protonation of the bound
>EDTA. Pb(+4) is a fairly strong Lewis acid, and presumably acidifies the
>bound EDTA, preferring to bind Pb(+4) as EDTA(-4) or H2EDTA(-2), releasing
>free H+ to be absorbed by the amines of a free EDTA. On the other hand,
>the soft Pb(+2) is much less acidic, and is likely much happier bound to
>H4EDTA or H2EDTA(-2). The free H+ ions then have an effect on the
>titration of the excess EDTA. Just a guess, however.
I don't know of any quantitative titration where a metal ion is bound
to EDTA without displacing some (most of the) hydrogen ions and at the
acidity where you can "absorb" free H+ on the nitrogens of EDTA there
are damn few metals (if any) which will form a complex which is of
any significance in an analytical titration. When you get to this
stage of protonation you have in effect a highly hindered
ethylenediamine molecule.
Almost every EDTA titration that I know is run in a buffered solution
so free H+ concentration is determined by the buffer not the ion being
titrated. Very few are run in even moderately acidic solutions or in
unbuffered solutions for the reason you mention. In a solution where
H4EDTA becomes a significant factor there are very few metal ions
which can be titrated. In addition the indicators available for these
titrations do not consider an acidic solution to be a very friendly
environment to grab metal ions.
If you do the calculations using the dissociation constants for the
H4EDTA and the formation constant of the metal complex it becomes
apparent why most of the titrations are done in solutions well
buffered somewhere between pH 4.5 and pH 10 depending on the ion being
determined, the indicator, other metals present,etc. Most of this work
was published in the late 50s and early 60s.
Since the original poster was using a lead nitrate solution as the
titrant in a back titration of excess EDTA (it is a common practice to
use Pb, Zn or Cu solutions) the probability of he/she using a Pb+4
nitrate solution as a standard solution is rather slim (to none) and
that is why I felt the valence of the lead was not relevant aside from
the fact that the ratio will be 1:1 no matter which was used.
In addition a Pb+4 nitrate aqueous solution would need to have
something to stabilize the Pb+4. This would mean a competing
complexing agent of some type which of course would screw up any
advantage of Pb+4 forming more stable complexes with EDTA.
>
> Eric Lucas
>
>Don Wilkins wrote:
>