Chemical Forums
Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: MightyMan on October 03, 2009, 05:26:43 PM
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Hey guys,
I was hoping you could help me out with a problem.
The question reads,
"What is the solubility of silver sulphide in pure water? HINT: This is actually an acid-base equilibrium situation involving a diprotic acid and combined equilibria. Ksp (Ag2S) = 6 x 10-50 M3"
When I first looked at it, I thought it was a simple solubility question.
Ag2S dissociates into 2Ag+ and S2-
and using an ICE table and the Ksp, you figure out the solubility.
however my prof keeps talking about how you need to combine equilibria, solve 6 unknowns using 6 equations, etc.
so i get the feeling there is more to this question (plus its out of 15 marks...)
i know that i can find the Ka for the dissociation of diprotic H2S, etc.
but my intuition still tells me that solubility is related to Ksp and since i'm given Ksp, i can just find solubility using an ICE table, etc...
can anyone point me in the right direction?
thanks in advanced.
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Once Ag2S dissolve, S2- - which is a strong base - reacts with water, creating HS-. That removes S2- from the solution and shifts solubility right. HS- can hydrolize as well, so the situation in the solution is rather compilcated, as it combines several equilibria.
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Thanks Borek,
Does the silver do anything in the solution at all? Or does it remain as an ion?
I can't think of any reactions that could use the silver ions..
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To some extent it reacts with OH- creating AgOH and Ag(OH)2-, but you better pretend you have no idea abut these reactions.
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Hahaha,
alrighty
i'll give it a try
thanks again for the help
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So I'm really stumped at this point..
I've got all the chemical equations of the reactions that I think take place:
Ag2S -----> 2Ag++S2- (Ksp=6x10-50)
S2-+H2O ------> OH-+HS- (Kb=1)
HS-+H2O ------> S2-+H3O+ (Ka=1x10-14)
HS-+H2O ------> OH-+H2S (Kb=1.05x10-7)
H2S+H2O ------> HS-+H3O (Ka=9.5x10-8)
2H2O -----> H3O++OH- (Kw=1x10-14)
Now this is where I am stumped: im unsure of what assumptions to make.
Is it safe to say that only the first 2 equations play a vital role in determining the value of the solubility, as the rest have significantly lower K values compared to the S2- reacting with water?
Then from there, the only source of S2- is from the dissociation of silver sulfide,
how do i figure out how much S2- is needed to get that system to equilibrium?
thanks in advanced
Edit: tags corrected
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S2-+H2O ------> OH-+HS- (Kb=1)
HS-+H2O ------> S2-+H3O+ (Ka=1x10-14)
You don't need both reactions - one of these will do to describe what is happening. You are throwing in water ion product and it 'takes care' of combining both dissociation and hydrolysis.
HS-+H2O ------> OH-+H2S (Kb=1.05x10-7)
H2S+H2O ------> HS-+H3O (Ka=9.5x10-8)
Same problem.
2H2O -----> H3O++OH- (Kw=1x10-14)
Don't write it this way - stay with H+ and OH-, K for the reaction you wrote is not 10-14.
Now this is where I am stumped: im unsure of what assumptions to make.
Do you have to make any?
Were you told to solve the question analytically, or are you just asked to calculate concentrations? Analytical solution will require assumptions, but iterative approach (or more generally numerical one) will give the answer without any assumptions, although it is not a thing you want to do by hand...
Note: assumptions will have to be done not to these equations, but to mass balances that you have not listed.
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I believe I have to solve this problem analytically, although the question just asks for solubility (concentration).
Alright, I've changed my list of equations per your advice.
Ag2S -----> 2Ag++S2- (Ksp=6x10-50)
S2-+H2O ------> OH-+HS- (Kb=1)
HS-+H2O ------> OH-+H2S (Kb=1.05x10-7)
H2O -----> H3O++OH- (Kw=1x10-14)
I've also created a mass balance formula..
F(S2-) = M(S2-)+M(HS-)+M(H2S)
I can simplify it to:
F(S2-) =M(S2-)+M(H2S)
if i combine the two protonation reactions
now i think that M(S2-) is what im supposed to calculate (x).
However, when i create an ICE (equilibrium table) for the dissociation of silver sulfide,
i only have one variable, I'm unsure how to take into account the fact that S2- will be disappearing....
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I would not call these mass balance. What are F() & M()?
Solubility is total amount of salt dissolved - how is it related to concentration of silver? How does amount of sulfides in the solution depend on the amount of silver (note: this is nothing else but mass balance.
You will need charge balance and equation that will bind concentration of Ag+ with all forms of H2S present in the solution.
I don't think ICE will work here, at least not in an obvious way.
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I had set F() as the formal concentration and M() as the concentrations after equilibrium.
I know stoichiometrically, there will be 2 moles of silver for every mole of sulfide.
I know you mentioned ICE would not work here,
but i combined the three formulas to get:
Ag2S + 3H2) -----> 2Ag++2OH-+H2S (K=6.3x10-57)
I believe this equation includes everything I need to take into account.
I set up an ICE table and solved for the concentration of Ag+
Is there any error in my method?..
Thanks again for all your help.
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I had set F() as the formal concentration and M() as the concentrations after equilibrium.
Formal - so it is not F(S2-), just F... That will be the solubility.
I know stoichiometrically, there will be 2 moles of silver for every mole of sulfide.
Correct.
I know you mentioned ICE would not work here,
but i combined the three formulas to get:
Ag2S + 3H2) -----> 2Ag++2OH-+H2S (K=6.3x10-57)
I believe this equation includes everything I need to take into account.
No idea how you have combined the formulas, but I doubt it is OK. I mean - could be K value is OK, but I doubt it is enough to correctly solve the question.
Show how you got there.
What was the result you got?
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I simple added:
Ag2S ---> 2Ag+ +S2-
S2-+H2O ---> OH-+HS-
HS-+H2O ---> OH-+H2S
the resulting equation:
Ag2S + 3H2O -----> 2Ag++2OH-+H2S
the equation is charged balanced, so I figured it was OK.
the final K, i just multiplied the 3 K values together (combined equilibria).
i figured the combined equilibria would take into account all forms of H2S
My final answer for the concentration of Ag was 6.61x10-12
I'm not too sure how to use mass balances, we haven't covered this in our lectures yet.
I think we are supposed to solve this problem with ICE (although I'm not certain) using assumptions that greatly simplify the problem.
My prof keeps explaining how easy it is, make assumptions, and you'll slap yourself in the face a couple of times when you figure it out..
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As I wrote earlier - this is equation that is technically correct, but it doesn't describe whole situation in solution.
Assuming this equation is correct, [Ag+] = [OH-], right? So pOH = 11.2, pH = 2.8 - but obviously when dissolving Ag2S solution can't become acidic, a little bit alkalic perhaps.
Mass balance will be:
2[Ag+] = [S2-] + [HS-] + [H2S]
It can be also expressed as two separate equations for F - but they can be easily combined and one unknown, F is automatically eliminated.
Write charge balance equation for the solution (that is - solution must be electrically neutral).
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The charge balance is Ag++H+=HS-+2S2-+OH-
now i also figured the concentration of [OH-]=[H+]=1x10-7
as so little sulfide is produced from the solubility
using the Kb values and the known [OH-] values,
i was able to write everything in terms of S2-
[HS-]=10000000[S2-]
[H2S]=10499999.48[S2-]
now i can substitute these equations into my charge balance and mass balance equation and solve for either [Ag+] or [S2-]:
2[Ag+]=[S2-]+10000000[S2-]+10499999.48[S2-]
[Ag+]=10000000[S2-]+2[S2-]
However, i noticed when i substitute the equations to solve, my variable gets cancelled out...
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[OH-]=[H+]=1x10-7
as so little sulfide is produced from the solubility
Very doubtfull assumption.
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I know Borek..
However I spoke to my professor and that is the assumption that he wants us to make..
I still cant figure out why my variables are cancelling!
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My guess is that if they cancel out you are 'oversubstituting' - that is a good way of proving 1=1. It works almost always ;)
Take a look at equation 9.13:
http://www.chembuddy.com/?left=pH-calculation&right=pH-polyprotic-acid-base
it may come handy. Note, that assumption you are asked to make means pH=7.00.
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I'm not too sure about what you mean by over-substituting,
well, i do know what you mean, but are my equations over-substituted?
i mean, there are two variables and i have two equations
I know that those equations can be used to solve the question,
I've been told that they will work...
I took a look at the equation you suggested,
it tells me that [S2-]=2[Ag+],
which is equal to the solubility of silver sulfide
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I took a look at the equation you suggested,
it tells me that [S2-]=2[Ag+],
which is equal to the solubility of silver sulfide
You have missed the main point - it gives you [S2-] as a function of TOTAL concentration of all forms of dissolved hydrogen sulfide. If pH is 7 everything else is known. But I won't tell anything more, as at this moment it is 30 sec work.
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Do you mean that [S2-]=[H2S]+[HS-]+[S2-]?
I'm still uncertain on how to calculate the concentration of the different forms of hydrogen sulfide.
Does pH=7 allow me to assume that [HS-]=[H2S]=1x10-7?
I can then find the concentration of the reactant:
[S2-]=1x10-14
[HS-]=9.524x10-8
[H2S]=1x10-7
Then the sum of these will equal the total S2- in the solution,
and therefore the solubility of silver sulfide=1.952x10-7?
I apologize if it seems as if I have no background knowledge on this subject,
it is an extra-work question and it seems to be a tad advanced, but i'd still like to understand how to solve it..
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Nevermind Borek,
I figured out the answer using my original method.
I understood what you meant by over-substituting....
Thanks again for all your help,
I really appreciate it.
Just out of curiousity,
should the mass balance be 0.5[Ag+]?
as there are 2 mols of silver per mole of sulfide
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Hello!
I am trying to solve this problem as well.
I'm still very lost.
How did you go about solving this?
I got till here and I'm stuck..
Ag2S --> 2Ag+ + S2- Ksp 6x10^-50, where unknown x = [S2-]
S2- + H2O --> OH- + HS- Kb = 1
I assumed all other Ka and Kb don't play a big role since their values are less than 10^-4.