One path is to investigate just add KI (or NaI) and see if any bright yellow PbI2 is formed (see https://en.wikipedia.org/wiki/Lead(II)_iodide ).
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Also, more reagent friendly, but needs to be tested for minor detection ability is just add NaOCl to a soluble Pb salt. For example:
Pb(HCO3)2 + 2 NaOCl -> Pb(OCl)2 (s) + 2 NaHCO3
A white precipitate of Lead hypochlorite, Pb(ClO)2 forms which on mild heating is converted into red PbO2. Source, please see http://bcs.whfreeman.com/WebPub/Chemistry/ichem5e/Videos/Hyp... , to quote:
"Initially, lead(II) ion reacts with hypochlorite ion to give a white precipitate of lead(II) hypochlorite:
Pb2++(aq) + 2 ClO-(aq) --> Pb(ClO)2(s)
The lead(II) hypochlorite is then oxidized to lead(IV) oxide. Reaction is slow at room temperature but much faster when the tube is placed in boiling
water.
Pb2+(aq) + 2 H2O(l) --> PbO2(s) + 4 H+(aq) + 2 e-
ClO-(aq) + 2 H+(aq) + 2 e- --> Cl-(aq) + H2O(l) "
Unfortunately, common chlorine based bleach has added NaOH for stability, and as a result expect also:
Pb(HCO3)2 + 2 NaOH -> Pb(OH)2 (s) + 2 NaHCO3
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I suspect that a Marsh like test (see https://en.wikipedia.org/wiki/Marsh_test) employed here for Pb in place of As likely does not work.
If it does, the chemistry would be based on the creation of the hydrogen radical (.H) from say the action of NaOH or HCl on Zn (or, may I suggest
pehaps Al in place of Zn).
The underlying chemistry could then proceed as follows:
.H <=> H+ + e-
Pb(HCO3)2 + 2 .H -> Pb (s) + 2 H2O + 2 CO2 (g)
Note: problematic formation of Plumbane, PbH4, is not likely, unlike AsH4. See https://en.wikipedia.org/wiki/Plumbane .
So, the question still remains, how does one make the minor amount of Pb metal (or its ions) more visible with common inexpensive chemicals other than
an I- or OCl-? |