so i was experimenting with H2S gas
and lead some into MgSO4 and a precipitate formed Mg(OH) i think
where did the sulfate go did it form H2SO4? normally ammonium hydroxide precipitates Mg(OH) forming ammonium sulfate?
[Edited on 4-10-2011 by symboom]blogfast25 - 4-10-2011 at 06:16
We can imagine the reactions to be as follows:
Dissociation (deprotonation) of hydrogen sulphide:
H2S(aq) + 2 H2O < === > S2-(aq) + 2 H3O+(aq)
Formation of magnesium sulphide:
Mg2+(aq) + S2-(aq) === > MgS(s)
Hydrolysis of magnesium sulphide:
MgS(s) + 2 H2O(l) === > Mg(OH)2(s) + H2S(aq)
The sulphate ions present play no part. They’re also known as spectator ions. Mixell - 4-10-2011 at 08:25
Blogfast, I have read somewhere that the discrete S2- anion exists only in extremely basic aqueous solutions, shouldn't the dissociation proceed like
this:
H2S(aq) +H2O <===> HS-(aq) + H3O+(aq) ?
Although there is probably an equilibrium that shifts strongly to the left: HS-(aq) +H2O <===> S2-(aq) + H3O+(aq).Nicodem - 4-10-2011 at 10:19
Furthermore, Mg(OH)<sub>2</sub> redissolves in acidic medium, so the products of any such reaction can not be
Mg(OH)<sub>2</sub> + 2H3O<sup>+</sup> like blogfast suggests above.
Symboom, Mg(OH) does not exist as a compound! blogfast25 - 4-10-2011 at 12:31
Mixell and Nicodem:
I was trying to explain the formation of this precipitate, which thinking about it some more shouldn’t really even form!
Sulphide ion concentration is very low except at very high pHs, but is nonetheless high enough to precipitate several metal sulphides (CuS e.g.) from
neutral solutions.
I suspect the observed precipitate may be MgCO3 (or a basic variant of it), caused perhaps by CO2 contamination of the H2S? Or some contamination of
the MgSO4?
Pure MgSO4 solution and pure H2S should not cause any precipitate to form. rstar - 11-10-2011 at 03:55
It might be some insoluble hydrate of MgSO3, but that has a low chance of formation.
but I think your reactants are kinda' contaminated, may be with some Ca(2+) blogfast25 - 11-10-2011 at 04:33
