Hi UltimaOnline,
For this question, I am more interested in the mechanics of the
experiment rather than the answer.
If I had understood this procedure correctly, it is the part where
âsaturating the solution with hydrogen sulphideâ that is providing
the S2- ions which is used to precipitate ZnS, followed by
MnS.
A: Now, what is the purpose of passing HCl into the solution prior
to the saturation of H2S? My thinking is that it suppresses the
dissociation of H2S (by LCP), thereby ensuring that only a very
minute of S2- is present in the solution when it is added to
prevent both precipitates from forming at the same time, given
their very small Ksp values?
B: Next, how is the pH raised during the experiment to allow for
the maximum effective precipitation (just right before MnS starts
to precipitate)? My understanding for such reactions is that OH-
ions will be supplied to decrease [H+] but this is not stated in
the question. Is my understanding correct?
C: As I add more H2S into the solution, does the pH increase? I
think it does because, the dissociation of H+ ions from H2S is too
insignificant to counteract the decrease of [H+] from the dilution,
hence resulting to a lower [H+] and thereby a higher pH.
Yo Gohby!
A : Correct.
B : Correct.
C : Depends on the molarity of the H2S(aq) added. If H2S(g) is
added, then the pH will still decrease (unless the solution is
already saturated with H2S, then any additional H2S added to the
solution will be forced to leave as H2S(g), so the pH won't
change).
Happy New Year, UltimaOnline! :)
Why does the pH decrease with the addition of H2S?
Wouldnât the mixing of a strong acid (HCl) with a weak acid (H2S)
result to a lower [H+], thereby increasing the pH of the
solution?
Now if the pH does decrease with the addition of H2S,
and given that we need to increase the pH of the solution to point
when MnS starts to precipitate by adding OH-, procedurally how does
the experiment work â I assume that I add the H2S and OH-
simultaneously? The question doesnât offer a very clear picture
tbh.
(P.S: I am deviating so much from the original question
and I feel that I am being too scholastic⌠but I think a sound
understanding of how this experiment works will aid one in arriving
at the solution..)
You too, Gohby! :)
For such experiments, you can saturate the soln first with H2S,
then add OH- from eg. NaOH(aq) to control the pH to control
precipitation. For this qn however, instead of adding OH-, the qn
says you simply add H2S until sufficient S2- ions are present to
precipitate out the less soluble metal sulfide. In other words, no
OH- is added. You simply slowly add H2S until you get a ppt.
H2S is only acidic, thus even if it's a weak acidic, and a strong
acid is already present, there's no way the pH can increase, unless
the molarity is sufficiently low such that (as you previously
mentioned) the dilution exceeds the proton dissociation.
Only when adding a basic (eg. S2-) or amphiprotic species (eg. HS-)
to a strong acid, would the pH increase (as H+ is removed from soln
by the amphiprotic species). But adding H2S (which is only acidic,
not amphiprotic or basic) only serves to add more H+ into soln,
thus decreasing pH further, but as stated previously, depending on
(g) vs (aq), and depending on molarity of (aq) added.
