Sciencemadness Discussion Board

Physical properties of HgCO3, mercuric carbonate

prole - 26-2-2008 at 10:38

I've been searching for the physical properties of HgCO3 for an upcoming project, and can't find anything. It's not listed in Merck (12th ed), and google had nothing. Can someone direct me to the information? Thanks so much.

Xenoid - 26-2-2008 at 10:59

I think you are out of luck!

From my Mellor's:

"Mercuric carbonate has not been prepared, and only basic mercuric carbonates are known. The addition of potassium carbonate to a solution of mercuric nitrate, gives a brown precipitate of HgCO3.2HgO; whilst that of potassium bicarbonate gives a brown precipitate of HgCO3.3HgO."

prole - 28-2-2008 at 05:54

Here's the balanced equation I'm working from:

HgSO4 + 2NaHCO3 --> HgCO3 + Na2SO4 + CO2 + H20

So the carbonate produced should read more like HgCO3.3HgO? I've never seen a compound written this way. How do you pronounce this? How does this affect equation balance?

I would like to understand all I can about this reaction before I proceed with it. My efforts will lead ultimately to HgCl2, so the carbonate/oxide revelation is totally acceptable here.

not_important - 28-2-2008 at 08:53

Usually known as "basic carbonates". Magnesium and zinc behave similarly, addition of Na2CO3 to a solution of a Mg or Zn salt gives one or another basic carbonate. However with those metals using NaHCO3 (bicarbonate), possibly saturated with CO2, will give the simple MCO3 carbonate, this does not work with mercury or Cu(II).

Think of it as forming the normal carbonate, which then immediately loses some CO2 to form the basic carbonate - mixed carbonate-oxide.

So, while not the actual reaction,

Hg(NO3)2 + Na2CO3 => HgCO3 + 2 NaNO3

3 HgCO3 => HgCO3.2HgO + 2 CO2

for an overall

3 Hg(NO3)2 + 3 Na2CO3 => HgCO3.2HgO + 6 NaNO3 + 2 CO2

Note that it will fizz from the CO2, try it with copper sulfate and see.

PHILOU Zrealone - 28-2-2008 at 09:09

Originally posted by prole
My efforts will lead ultimately to HgCl2, so the carbonate/oxide revelation is totally acceptable here.

-Maybe start from Hg, Hg(OH)2, HgS or HgO with HCl instead of messing with HgCO3?
-Also with a fairly unsoluble HgSO4, the reaction you stated will only proceed under such a high heat that any HgCO3 formed will decompose into HgO and finally into Hg and O2....

Xenoid - 28-2-2008 at 11:27

Some further notes, again from Mellor's:

Mercuric chloride, (corrosive sublimate), HgCl2 was formerly made on a commercial scale, by heating a mixture of mercuric sulphate and sodium chloride, a little manganese dioxide usually being added to prevent the formation of mercurous chloride.

2NaCl + HgSO4 --> Na2SO4 + HgCl2

Mercuric chloride sublimes as a white translucent mass.

Mercury reacts rapidly with chlorine, especially when heated, and mercuric chloride is now made this way commercially.

vulture - 28-2-2008 at 15:17

Your equation is wrong. Mercuric sulfate is not acidic, so it won't react with (hydrogen)carbonate in the way you propose.

prole - 18-3-2008 at 08:45

OK, Vulture, I see that mercuric sulfate isn't acidic. Perhaps the Hg is replaced with sodium because of their positions in the activity series? I was working from Rhodium's mercuric choride prep. and came up with the following three reactions along the route to the title compound:

1) Hg + H2SO4 --> HgSO4 (0)SO2 + 2H

2) HgSO4 + 2NaHCO3 --> HgCO3 + Na2SO4 + CO2 + H2O

3) HgCO3 + 2HCl --> HgCl2 + CO2 + H2O

I checked these equations at the online balancer at They will tell you if an equation is impossible until you enter the right formula. So, are they wrong? What am I not getting? And why is there a zero in front of SO2 if copious amounts are evolved? I do want to understand what's going on here, even though I am considering combining chlorine and mercury directly to get the chloride.

UnintentionalChaos - 18-3-2008 at 10:39

Do you know much about redox chemistry prole? I will briefly explain the method below. Also, a reactivity/electromotive series would be useful if you don't have one for reference. Here's one for you to use: I apologize If I underestimate what you know, but Im going to assume you know nothing to make sure you can't get lost.

Metals and elements in a pure state have an oxidation number of 0. If you form compounds, metals will tend to have a positive oxidation number and nonmetals a negative oxidation number. In sodium chloride for example, the sodium ion has a charge of +1. Sodium is in the +1 oxidation state. In order to neutralize this +1 and make the salt charge-neutral, the chlorine atom exists as chloride with a -1 oxidation state, and -1 charge.
In terms of practical chemical knowledge, all alkali metals only can be in 0 or +1 oxidation states. All alkaline earth metals are either 0 or +2. The transition metals, like iron, copper, mercury are a bit uglier since they can possibly be in a few oxidation states. The right side of the periodic table becomes more predictable, but most can exist in quite a few oxidation states. An important one to remember is oxygen which is almost always 0 or -2.The only other really important oxidation state it can be in is +1, which it is in peroxides.

To oxidize a compound means that its state is getting more positive. Sodium metal (0) gets oxidized to a sodium cation (+1) Iodide (-1) can get oxidized to elemental iodine (0).

Lets look at your very first equation:
Hg + H2SO4 --> HgSO4 (0)SO2 + 2H

What ozidation state is mercury on the left side in? We're talking about elemental mercury metal, so it is in the 0 oxidation state. Now look at the sulfuric acid. Hydrogen almost always forms +1 ions (except in metal hydrides where it is a -1). If we take the two hydrogen off of sulfuric acid, we get a sulfate anion, which has a charge of -2. This anion has 4 oxygen and 1 sulfur. What is the oxidation state of each kind of atom?

Well, the oxygen aren't peroxides, so we assume they have a charge of -2.
4(-2) + z = -2 (Where z is the charge on sulfur)
z= +6, so sulfur in sulfate is in the +6 oxidation state.

Look at the electromotive series. This is a fairly simple tool to use, but it doesn't always work as easily in practice as in theory for some reactions. All of these should be ordered by their electrode potential to get the bigger picture. Things at the top are powerful reducing agents (They like to be oxidized and in the process reduce other things) Metals below hydrogen are called noble metals and are not attacked by simple acids. Nonmetals at the bottom (like the halogens) are powerful oxidizing agents.

Why dont metals below hydrogen react with simple acids? I'll show you in a second after I explain how to write a simple redox equation.

Lets say you want to make sodium chloride from the pure elements.
2Na (0) + Cl2 (0) ---> 2NaCl

Do you see how both sides are charge neutral? We can break up that equation into what are called half reactions showing the oxidation and reduction of the reactants.

Na (0) ---> Na (+1)

This is the half reaction for sodium. You can see that its oxidation state went up, so it has been oxidized. In order to keep the equation charge neutral, we add an electron (denoted by e-), which has a charge of -1 to the right side.

Na (0) ---> Na (+1) + e-

Now for the chlorine

Cl2 (0) ---> Cl (-1)

We need to conserve the number of atoms of chlorine so put a 2 in front of the chloride.

Cl2 (0) ---> 2Cl (-1)

Now we balance the charge by subtracting 2 electrons the right side or adding two to the left side, since we have two chloride ions with a total charge of -2.

Cl2 (0) ---> 2Cl (-1) -2e-

Now look at the two finished half reactions. Chlorine has gone down in oxidation state and therefore was reduced (It behaves as an oxidizing agent) Also note that If we simply recombine them, the charges wont be balanced. To fix this, take 2x the first equation and add it to the second equation. The electrons cancel out and you get:

2Na (0) + Cl2 (0) ---> 2NaCl

Now we can move on to slightly more complicated equations like the one you are struggling with.

Hg + H2SO4 --> HgSO4 + 2H

The equation is not this simple. Do you see how the sulfate anion is on both sides unchanged? We call this a spectator ion and it can be removed from the equation. Also, hydrogen exists as diatomic H2, so write the 2H as H2 instead. This gives.

Hg (0) + 2H (+1) ---> Hg (+2) + H2 (0)

Do you see how the charge is still balanced, but it isn't zero? This is perfectly acceptable for equations. Now, it is time to look at the electromotive series. Do you see how mercury is located below hydrogen? This means that hydrogen would assume the role of reducing agent and mercury of oxidizing agent if you mixed the two. if you look at the equation above, you can see that mercury needs to be oxidized (Its state needs to go from 0 to +2) and thus it needs to be a reducing agent. This is in conflict with the electromotive series. The noble metals can't reduce hydrogen ions to hydrogen gas and get oxidized. When a simple acid reacts with a metal, this is what is happeneing. The anion from the acid does nothing.

Now, if the hydrogen ions must not be doing the oxidizing, something else has to. The only other thing present is the sulfate anion. Now, the oxygens on it are pretty happy being -2, but the sulfur in the +6 oxidation state can change. If SO2 is produced in the reaction, then what is the oxidation state of sulfur in SO2? We use the same method as before.
2(-2) + z = 0
z = +4

Aha! So, the sulfur is being reduced as the mercury is oxidized instead of the hydrogen. We need to write our unbalanced half reactions first.

Hg (0) ---> Hg (+2)
SO4 (-2) ---> SO2

Take the first one. This is pretty simple. There is one atom of mercury on each side so all the components are balanced, but the charge is not, so we need to add 2 electrons to the right side and balance the charge.

Hg (0) ---> Hg (+2) + 2e-

Now for the next half. The rules for these more complicated ones are as follows.
1: balance the central atom (In this case sulfur) if it isn't balanced.
2: Balance the oxygens by adding H2O on either side if they aren't balanced
3: Balance the hydrogens by adding H (+1) ions if they aren't balanced.
4: Count up all the charges on everything and balance the charge by adding electrons.

So, for SO4 (-2) ---> SO2

1: The sulfur is balanced already. One on each side.
2: The oxygen is not balanced. There are 4 on the left and two on the right. So we add two water to the right side.

SO4 (-2) ---> SO2 + 2H2O

3: The hydrogen are not balanced. We have 4 on the right and 0 on the left, so we add 4 H (+1) to the left side.

4H (+1) + SO4 (-2) ---> SO2 + 2H2O

4: The charges are not balanced. The left side adds up to +2 while the right side adds up to 0, so we need to add 2 electrons the the left side.

2e- + 4H (+1) + SO4 (-2) ---> SO2 + 2H2O

Now, we look at our two half reactions.

2e- + 4H (+1) + SO4 (-2) ---> SO2 + 2H2O

Hg (0) ---> Hg (+2) + 2e-

If we just add them to each other, the electrons will cancel out, so we don't need to multiply either one first. We get:

4H (+1) + SO4 (-2) + Hg (0) ---> SO2 + 2H2O + Hg (+2)

For seeing a reaction simply, we can do a bit better, but a professor would want to see the above equation only. Since of course, we could also mix sodium sulfate to give a sulfate with hydrochloric acid to supply the hydrogen ions and it would work just the same. To explain the reaction you're doing though, we can combine 2 of the hydrogen ions with the sulfate anion on the left to tidy things up a bit giving:

2H (+1) H2SO4 + Hg (0) ---> SO2 + 2H2O + Hg (+2)

Of course the only thing that can be supplying the hydrogen ions in this case is sulfuric acid, so we can add the sulfate spectator ion back on both sides, giving:

2H2SO4 + Hg ---> SO2 + 2H2O + HgSO4

I hope that explains things a bit better. If you're confused, I suggest you go find a textbook and read up on this since it is an important skill to have.

However, read this MSDS for mercuric sulfate:

It decomposes on contact with water giving an insoluble basic sulfate and sulfuric acid. You won't be able to get the bicarbonate to react with the sulfate since it'll form an insoluble compound in aqueous solution.

[Edited on 3-18-08 by UnintentionalChaos]

prole - 19-3-2008 at 07:33

Thank you, UnintentionalChaos, for your reply. I am self-employed as a Consulting Arborist and am an expert in the field. Chemistry is not my first language, so to speak; trees are. I am making mercuric chloride because 1) it looks like fun 2) for use as an intesifier in my darkroom. 3) shipping is costly. In the process, I want to understand what it is I am doing to learn and to maximize safety. I will review my textbooks and your reply for redox chemistry and get my answers from them.