With respect to my good friend Rubidium, there will be NO reaction for these reactants. Neither is soluble in water, and the possible products are MORE insoluble than the reactants. Therefore, there will be NO reaction.
You need to know something about the Mercury (I) ion. It exists as Hg2^2+, not as single Hg+ ions.
The second issue is that both of these reactants are quite insoluble in water. Therefore whatever reaction might occur, would occur with very few ions in solution. You might write the following "reaction", but keep in mind that any actual "reaction" will depend on the relative solubilities of these reactants and hypothetical products, the Ksp's.
Pb(OH)2(aq) + Hg2S(aq) --> PbS(s) + Hg2(OH)2(s)
So back to first point. A compound of mercury (I) hydroxide must be written as Hg2(OH)2.
When we look at the actual values of Ksp for the reactants and the hypothetical products, we get:
Pb(OH)2 1.2 x 10-15
Hg2S 1.0 x 10-47
PbS 3 x 10-29
Hg2(OH)2 ???
I haven't found the Ksp for mercury (I) hydroxide assuming this compound may actually exist.
Notice that the most insoluble compound of the three that we have Ksp's for is Hg2S. Therefore, with these ions Hg2+2 and S2- or OH-, then it's going to form Hg2S and not Hg2(OH)2. Therefore there is not going to be a reaction. Especially when we consider the reactants are very insoluble in the first place.
The bottom line is that the entire premise of reaction between lead (II) hydroxide and mercury (I) sulfide is bogus.
========== Follow up ===========
Mercury (I) hydroxide is Hg2(OH)2 and lead (II) sulfide is PbS.
This signals the formation of a precipitate. You should know your solubility rules so you know which stuff precipitate out of solution and which don't.
The solubility rules tell you which are soluble in water and which aren't. Since some substances aren't soluble in water, they precipitate out of solution. You can find a copy of the solubility rules in any chemistry textbook.
Anyway, what you have here is a double-displacement reaction. Just interchange the places of the two cations (Pb2+ and Hg2+) involved.
PbS (lead (II) sulfide, which will precipitate out of solution) + HgOH (mercury (I) hydroxide, which has a low solubility and will probably form some precipitate)
right here is the balanced equation: Na2CO3 + 2HCl --------> 2NaCl + H2CO3 You sort sodium chloride and carbonic acid. Carbonic acid isn't good at room temperature so it could greater beneficial dissociate as Cheryl N reported. desire this enables.
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Verified answer
With respect to my good friend Rubidium, there will be NO reaction for these reactants. Neither is soluble in water, and the possible products are MORE insoluble than the reactants. Therefore, there will be NO reaction.
You need to know something about the Mercury (I) ion. It exists as Hg2^2+, not as single Hg+ ions.
The second issue is that both of these reactants are quite insoluble in water. Therefore whatever reaction might occur, would occur with very few ions in solution. You might write the following "reaction", but keep in mind that any actual "reaction" will depend on the relative solubilities of these reactants and hypothetical products, the Ksp's.
Pb(OH)2(aq) + Hg2S(aq) --> PbS(s) + Hg2(OH)2(s)
So back to first point. A compound of mercury (I) hydroxide must be written as Hg2(OH)2.
When we look at the actual values of Ksp for the reactants and the hypothetical products, we get:
Pb(OH)2 1.2 x 10-15
Hg2S 1.0 x 10-47
PbS 3 x 10-29
Hg2(OH)2 ???
I haven't found the Ksp for mercury (I) hydroxide assuming this compound may actually exist.
Notice that the most insoluble compound of the three that we have Ksp's for is Hg2S. Therefore, with these ions Hg2+2 and S2- or OH-, then it's going to form Hg2S and not Hg2(OH)2. Therefore there is not going to be a reaction. Especially when we consider the reactants are very insoluble in the first place.
The bottom line is that the entire premise of reaction between lead (II) hydroxide and mercury (I) sulfide is bogus.
========== Follow up ===========
Mercury (I) hydroxide is Hg2(OH)2 and lead (II) sulfide is PbS.
This signals the formation of a precipitate. You should know your solubility rules so you know which stuff precipitate out of solution and which don't.
The solubility rules tell you which are soluble in water and which aren't. Since some substances aren't soluble in water, they precipitate out of solution. You can find a copy of the solubility rules in any chemistry textbook.
Anyway, what you have here is a double-displacement reaction. Just interchange the places of the two cations (Pb2+ and Hg2+) involved.
2Pb(OH)2 + Hg2S --> Pb2S + 2Hg(OH)2
Pb2S precipitates out of solution.
:)
PbS (lead (II) sulfide, which will precipitate out of solution) + HgOH (mercury (I) hydroxide, which has a low solubility and will probably form some precipitate)
right here is the balanced equation: Na2CO3 + 2HCl --------> 2NaCl + H2CO3 You sort sodium chloride and carbonic acid. Carbonic acid isn't good at room temperature so it could greater beneficial dissociate as Cheryl N reported. desire this enables.