The Nobel Prize and Aqua Regia - ChemPRIME

The Nobel Prize and Aqua Regia

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Work in Progress. Rhetue 16:40, 22 February 2011 (UTC)

The sweep of Nazi Germany across Europe began with the invasion of Poland in 1939 and was soon followed by Operation Weserübung, or the assault against Denmark and Norway. As the bombing of Copenhagen began, scientist George de Hevesy at the University of Copenhagen feared for the safety of the Nobel Prizes of fellow Danes Max von Laue and James Franck, who had both opposed the Nazi regime. In order to prevent the Germans from confiscating the medals, de Hevesy dissolved them in concentrated solutions of aqua regia (an acidic mixture) and fled Copenhagen. When the Germans swept through the university, all they found in de Hevesy's lab were hundreds of solutions in jars and beakers. The German soldiers, who had neglected to take chemistry classes, failed to identify the aqua regia solution and left Copenhagen medal-less. When de Hevesy returned after the war, he precipitated the gold out of the solution and presented it back to the Nobel Foundation, who recast the medals and gave them to their rightful owners.

Contents

The Aqueous Chemistry of Aqua Regia and Noble Metals

Why did de Hevesy not resort to another method to chemically hide the gold medals from the Germans? Obviously, the particular shape of the medals before dissolving was lost upon precipitation, so why not use "gentler" chemistry? The problem is that gold is one of the noble metals, a classification of transition metals that are exceptionally resistant to corrosion and solvation. An exceptionally strong acid is required to dissolve these metals in solution, but not just any acid will suffice.

Example

Name several strong acids.


ANSWER

Possible answers include hydronium ions, nitric acid, sulfuric acid, chloric acid, perchloric acid, hydrochloric acid, hydrobromic acid, and hydroiodic acid.


Gold lives up to its title as a noble metal as it actually does not dissolve in any of these strong acids. If de Hevesy even had access to the then little-known superacids in 1940, it is likely that none of these acids would have worked either. Gold, as an element, is not a base; it certainly does not accept hydrogen atoms as a Brøonsted-Lowry base, nor does it readily accept electron pairs as a Lewis base. Because gold is so inert as a metal, it must be converted into an ion first, though this process is almost as difficult.

Aqua regia's dissolving power is due to its properties as a mixture; it is comprised of both nitric acid and hydrochloric acid in a certain ratio. Each acid carries out a different purpose when dissolving gold. Nitric acid ionizes neutral gold atoms to gold ions in a redox reaction, then hydrochloric acid provides chloride ions to form a stable aqueous species in solution. Let's look at these steps more closely.

Redox

The conversion of a neutral metal into a positively charged cation is called oxidation and occurs due to the loss of the metal's electrons. More generally, a redox reaction includes any reaction where elements change oxidation states by the transfer of electrons. The oxidation of Au to Au3+, for example, requires an Au atom to lose three electrons, as given in the following reaction:

Au (s) → Au3+ (aq) + 3e-

A reaction in which only oxidation occurs is called a half-reaction, because it must be paired with a reduction reaction in which another compound or element gains the electrons and is reduced. The reduction half-reaction paired with the oxidation of atomic gold is the reduction of the nitrate ion to nitrogen dioxide.

3e- + 3 NO3- (aq) + 6 H+ (aq) → 3 NO2 (g) + 3 H2O (l)


These half-reactions combine to give the full redox reaction of gold in aqua regia:


Au (s) + 3 NO3- (aq) + 6 H+ (aq) → Au3+ (aq) + 3 NO2 (g) + 3 H2O (l)

How can we keep track of the flow of electrons in redox reactions, and identify which species are reduced or oxidized? A useful method is to identify the oxidation numbers of each element in every compound of the redox reaction. Oxidation numbers describe the number of valence electrons an element contains relative to its normal valence electron state. For example, the oxidation number of Au3+ is +3 because it has lost 3 valence electrons.

For example, in NO3 the nitrogen is assigned an oxidation number of +5 and each oxygen an oxidation number of –2. This arbitrary assignment corresponds to the nitrogen having lost its original five valence electrons to the electronegative oxygens. Note that the overall charge of the molecule agrees with the individual oxidation numbers of each element. The three oxygens have a combined "charge" of -6, combined with the +5 of nitrogen, contribute to the -1 charge of the entire molecule. In NO2, on the other hand, the neutrality of the compound implies that nitrogen has an oxidation number of + 4 and may be thought of as having one valence electron for itself, that is, one more electron than it had in NO3.

The oxidation number of an element can vary between compounds, as shown in the case of nitrogen. Usually, it is simpler to obtain the oxidation number of elements by using the charge of the compound as a whole and the oxidation number of a few elements that always have the same oxidation number:

  • Oxygen always has an oxidation number of -2.
  • Alkali metals always have an oxidation number of +1.
  • Alkaline earth metals always have an oxidation number of +2.
  • Elements in atomic form (not ionized and not bonded to other elements) always have an oxidation number of 0.

Several other elements have very few exceptions to their general oxidation numbers:

  • Hydrogen nearly always has an oxidation number of +1.
  • Halogens nearly always have an oxidation number of -1.

Example

Identify the oxidation numbers of every element in the aqua regia oxidation of gold:

Au (s) + 3 NO3- (aq) + 6 H+ (aq) → Au3+ (aq) + 3 NO2 (g) + 3 H2O (l)

ANSWER

Compound Au NO3- H+ Au3+ NO2 H2O
Element Au N O H Au N O H O
Oxidation Number 0 5 -2 1 3 4 -2 1 -2

Identifying oxidation numbers allows the identification of elements that are reduced and oxidized. Nitrogen goes from a +5 to a +4 state and is reduced; gold goes from a 0 to +3 state and is oxidized.


With gold oxidized, Cl- ions in solution from hydrochloric acid combine with Au3+ to form AuCl4-, the chloroaurate ion. The solution may then be evaporated, and because aqueous ions cannot exist otuside of water, the chloroaurate ion combines with H+ from hydrochloric acid to form the solid HAuCl4. This compound may be redissolved in water and precipitated to give elemental gold.

Na2SO3 + 2 H+ → 2 Na+ + H2O + SO2

2 AuCl4- + 3 SO2 + 6 H2O → 2 Au (s) + 3 SO42- + 8 Cl- + 12 H+

Precipitation

The precipitation of solid gold from aqua regia is rather different from the type of precipitation reactions we are going to discuss. de Hevesy probably used a powerful solid reducing agent, such as sodium borohydride, to reduce HAuCl4 to elemental Au. We will study the precipitation of a solid by the addition of two aqueous solutions, which involve spectator ions and solubility rules.

Initially, each solution consists of ions that are completely soluble in water. When these solutions are mixed, some of these ions may form an insoluble compound and precipitate out of solution, while some remain soluble.

Instead of gold, let's look at the precipitation reaction that produces solid AgCl from solutions of silver nitrate and barium chloride. Each solution contains free aqueous ions, but when combined, silver and chloride ions form the insoluble AgNO3:

2AgNO3(aq) + CaCl2(aq) → 2AgCl(s) + Ca(NO3)2(aq)      

Note the designation of phases in this equation. Initially, each compound is aqueous, but one of the products is in the solid state. The other produce, calcium nitrate, is still soluble, and thus really consists of free Ca2+ and NO3- ions.

Since aqueous compounds form dissociative ions, we can write each ionic species out separately in an ionic equation:

2Ag+(aq) + 2NO3(aq) + Ca2+(aq) + 2Cl(aq) → 2AgCl(s) + Ca2+(aq) + 2NO3(aq)      

The ionic equation clearly shows that some ionic species appear on both sides of the equation and do not actually participate in the reaction. In this case, NO3 and Ca2+ remain as aqueous species throughout the reaction and are not part of the actual precipitation reaction. We can remove these spectator ions from the equation to obtain the net ionic equation.

Ag+(aq) + Cl(aq) → AgCl(s)      

Solids, gases, and liquids (not aqueous ions) are undissolved species that take ions out of solution and thus always appear in net ionic equations. It is usually easy to identify gases, such as CO2 and liquids, such as H2O, but identifying whether a precipitate will form from available aqueous ions requires a degree of memorization. Because of the general utility of precipitates in chemistry, it is worth having at least a rough idea of which common classes of compounds can be precipitated from solution and which cannot. The following table gives a list of rules which enable us to predict the solubility of the most commonly encountered substances. Use of this table is illustrated in the following example.

Soluble in Water Important Exceptions (insoluble)
All Group IA and NH4+ salts
All nitrates, chlorates, perchlorates and acetates
All sulfates CaSO4, BaSO4, SrSO4, PbSO4
All chlorides, bromides, and iodides AgX, Hg2X2, PbX2 (X= Cl, Br, or I)
Sparingly Soluble in Water Important Exceptions (soluble)
All carbonates and phosphates Group IA and NH4+ salts
All hydroxides Group IA and NH4+ salts; Ba2+, Sr2+, Ca2+ sparingly soluble
All sulfides Group IA, IIA and NH4+ salts; MgS, CaS, BaS sparingly soluble
All oxalates Group IA and NH4+ salts
The following electrolytes are of only moderate solubility in water:
CH3COOAg, Ag2SO4, KClO4
They will precipitate only if rather concentrated solutions are used



Example

Write the balanced net ionic equations for any precipitation reactions that occur when the following solutions are mixed. Designate the phases of each species in your equations.

a) 0.1 M SrI2 (aq) + 0.1 M Ba(OH)2 (aq)

b) 0.1 M FeSO4 (aq) + 0.1 M Ba(OH)2 (aq)

c) 0.2 M NH4NO3 (aq) + 0.1 M Mg(ClO4)2 (aq)

d) 0.2 M AgCl (aq) + Cu(CH3COO)2 (aq)


ANSWERS

a) There is no net ionic equation, as no precipitate forms. The two possible candidates, BaI2 and Sr(OH)2, are both soluble.

b) Fe2+ (aq) + SO42- (aq) + Ba2+ + 2OH- → BaSO4 (s) + Fe(OH)2 (s)

c) There is no net ionic equation, as no precipitate forms. The two possible candidates, NH4ClO4 and Mg(NO3)2, are both soluble.

d) Ag+ (aq) + CH3COO- → AgCH3COO (s)

The importance of the molarity of solution is simply to ensure that the balance of ions is stoichiometric. For example, in d), CH3COO- reacts in a 1:1 ratio with Ag+ to give the product precipitate. However, for every one mole of Cu(CH3COO)2, there are two moles of CH3COO-. Thus, twice the amount of AgCl is needed to obtain an equal amount of Ag+ and CH3COO-.

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