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Topic: Solubility Question.. or is it?  (Read 16319 times)

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Offline MightyMan

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Solubility Question.. or is it?
« on: October 03, 2009, 05:26:43 PM »
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.

Offline Borek

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Re: Solubility Question.. or is it?
« Reply #1 on: October 03, 2009, 05:40:52 PM »
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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Offline MightyMan

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Re: Solubility Question.. or is it?
« Reply #2 on: October 04, 2009, 08:32:03 PM »
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..

Offline Borek

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Re: Solubility Question.. or is it?
« Reply #3 on: October 05, 2009, 02:51:59 AM »
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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Offline MightyMan

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Re: Solubility Question.. or is it?
« Reply #4 on: October 05, 2009, 11:49:46 AM »
Hahaha,
alrighty
i'll give it a try
thanks again for the help

Offline MightyMan

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Re: Solubility Question.. or is it?
« Reply #5 on: October 06, 2009, 07:36:57 PM »
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
« Last Edit: October 07, 2009, 03:33:36 AM by Borek »

Offline Borek

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Re: Solubility Question.. or is it?
« Reply #6 on: October 07, 2009, 03:50:44 AM »
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.

Quote
HS-+H2O ------> OH-+H2S (Kb=1.05x10-7)
H2S+H2O ------> HS-+H3O (Ka=9.5x10-8)

Same problem.

Quote
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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Offline MightyMan

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Re: Solubility Question.. or is it?
« Reply #7 on: October 07, 2009, 07:51:18 AM »
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....

Offline Borek

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Re: Solubility Question.. or is it?
« Reply #8 on: October 07, 2009, 08:18:19 AM »
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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Offline MightyMan

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Re: Solubility Question.. or is it?
« Reply #9 on: October 07, 2009, 08:37:40 AM »
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.

Offline Borek

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Re: Solubility Question.. or is it?
« Reply #10 on: October 07, 2009, 08:45:19 AM »
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.

Quote
I know stoichiometrically, there will be 2 moles of silver for every mole of sulfide.

Correct.

Quote
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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Offline MightyMan

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Re: Solubility Question.. or is it?
« Reply #11 on: October 07, 2009, 08:58:15 AM »
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..

Offline Borek

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Re: Solubility Question.. or is it?
« Reply #12 on: October 07, 2009, 09:23:40 AM »
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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Offline MightyMan

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Re: Solubility Question.. or is it?
« Reply #13 on: October 07, 2009, 03:05:11 PM »
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...

Offline Borek

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Re: Solubility Question.. or is it?
« Reply #14 on: October 07, 2009, 03:33:14 PM »
[OH-]=[H+]=1x10-7
as so little sulfide is produced from the solubility

Very doubtfull assumption.
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