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### Topic: many equilibriums in one system  (Read 18488 times)

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#### zephyrblows

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##### many equilibriums in one system
« on: August 14, 2006, 09:04:29 PM »
Hello,

If there is a material that takes part in more than one reactions occuring in a system, should I solve K equations individually or combine them into one equation (K1*K2*...)?

What makes me confused is, if two reactions have very different K values, when I combine them into one, will the stoichiometric relations in the net equation become invalid?

#### Borek

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##### Re: many equilibriums in one system
« Reply #1 on: August 15, 2006, 03:47:28 AM »
If there is a material that takes part in more than one reactions occuring in a system, should I solve K equations individually

Yes. Take a look at pH calculation - general case lecture and read first paragraph
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#### sdekivit

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##### Re: many equilibriums in one system
« Reply #2 on: August 15, 2006, 05:04:13 AM »
Sometimes you can combine them. Assume the following:

H3PO4 + H2O <--> H2PO4- + H3O+   --> K1

H2PO4- + H2O <--> HPO42- + H3O+ --> K2

Express the reaction quotients:

K1 = [H3O+][H2PO4-]/[H3PO4]

K2 = [H3O+][HPO42-]/[H2PO4-]

Now you say from the expression of K1 that [H2PO4-] = K1 * [H3PO4]/[H3O+] and substitue that in the other equilibrium:

K2 = [H3O+][HPO42-]/( K1 * [H3PO4] / [H3O+] )

Which yields:

K1 * K2 = ( [H3O+]2[HPO42-] ) / [H3PO4]
« Last Edit: August 15, 2006, 05:10:37 AM by sdekivit »

#### Borek

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##### Re: many equilibriums in one system
« Reply #3 on: August 15, 2006, 05:34:48 AM »
Your post can be confusing for the OP.

Sometimes you can combine them.

At the cost of loosing part of information and loosing chances of solving the system.

Quote
K1 * K2 = ( [H3O+]2[HPO42-] ) / [H3PO4]

I wonder how you are going to use this equation to calculate H2PO4- concentration that was described by the original set of equations, but is not present in your equation now.

Sure, you can use your new equation for K1*K2 together with equation for K1, but apart from changing from the stepwise dissociation constants to overall dissociation constants - which is just different formalism - you have not changed anything. You still need the same number of equations as before so you have hardly combined them into one, as OP asked.
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#### zephyrblows

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##### Re: many equilibriums in one system
« Reply #4 on: August 15, 2006, 09:12:16 AM »
Borek,thank you!^^

Could you show how to solve this following problem in a simplified way for example?

The solubility of AgCl in 1 liter of 1M NH3 solution is?

All I can find are:
1.7*1010 = [Ag+] [Cl-]
1.7*107 = [Ag(NH3)2+] / [Ag+] [NH3]2
[Cl-] = [Ag(NH3)2+] + [Ag+]
[NH3]0 = [NH3] + 2[Ag(NH3)2+]

But I don't know how to simplify it so I can solve it by hand.
And, I'm not quite sure about how to find mass and charge relations...FAST...

Thanks a lot!

#### zephyrblows

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##### Re: many equilibriums in one system
« Reply #5 on: August 15, 2006, 10:07:32 AM »
Well, I've tried to solve it by myself...

I assumed [Cl-] = [Ag(NH3)2+]
because Ag+ is almost used up...

So I got the solubility = [Cl-] = 0.049M
« Last Edit: August 17, 2006, 09:17:12 AM by zephyrblows »

#### Borek

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##### Re: many equilibriums in one system
« Reply #6 on: August 15, 2006, 10:46:36 AM »
I assumed [Cl-] = [Ag(NH3)2+]

Correct idea.

Quote
So I got the solubility = [Cl-] = 0.049M

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#### sdekivit

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##### Re: many equilibriums in one system
« Reply #7 on: August 15, 2006, 12:24:56 PM »
All I can find are:
1.7*1010 = [Ag+] [Cl-]
1.7*107 = [Ag(NH3)2+] / [Ag+] [NH3]2
[Cl-] = [Ag(NH3)2+] + [Ag+]
[NH3]0 = [NH3] + 2[Ag(NH3)2+]

Are you sure about the K-values? According to my tables the solubility product of AgCl is 1,8 x 10^-10.

The answer is 5,0 x 10^-2 M though.

And you need the method in my post, which would be confusing, to solve this problem The keyword is: substitution. And yes, this is a complicated problem.
« Last Edit: August 15, 2006, 01:10:47 PM by sdekivit »

#### zephyrblows

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##### Re: many equilibriums in one system
« Reply #8 on: August 15, 2006, 09:19:51 PM »
Sorry, I typed the wrong Ksp of AgCl... My table says it's 1.7*10-10...

My calculations are:

What I want to find is [Cl-]:

So,
[Ag+] = 1.7*10-10 / [Cl-]

1.7*107 [NH3]2 = [Cl-]2 / 1.7*10-10

[NH3] = 18.6 [Cl-]

Substitution:
1 = 2[Cl-] + 18.6 [Cl-] = 20.6 [Cl-]
[Cl-] = 0.049M
« Last Edit: August 17, 2006, 09:17:40 AM by zephyrblows »

#### sdekivit

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##### Re: many equilibriums in one system
« Reply #9 on: August 16, 2006, 03:05:04 AM »

This is what school teachers teach us (without explanation):
The net equation is, Ag+ + 2NH3 <-> Ag(NH3)2+ + Cl-      K = KspKf = 2.89*10-3
According to the net equation, make x = [Cl-] = [Ag(NH3)2+]
Solve: 2.89*10-3 = x2 / (1-2x)2
And get x = 0.049M

this is correct. I'll try to explain this method to you. When dissolving AgCl in a solution containing 1 M NH3 we get the following 2 equilibria:

AgCl(s) <--> Ag+ + Cl- : Ks = 1,7 x 10-10
Ag+ + 2 NH3 --> Ag(NH3)2+ : Ks = 1,7 x 107

So the total reaction occuring here is:

AgCl + 2 NH3 <--> Ag(NH3)2+ + Cl- : Ks = ?

Now we must find this Ks-value for the toal reaction and to get this one, we need to combine these equilibria by using substitution:

Ks = [Ag+][Cl-] = 1,7 x 10-10
Ks = [Ag(NH3)2+] / ( [Ag+][NH3]2 ) = 1,7 x 107

The combined Ks:

Ks = [Ag(NH3)2+][Cl-] /  ([Ag+][NH3]2 )

Because 1,7 x 107 >> 1,7 x 10-10 we can say, according the LeChatelier principle, All the Ag+ formed during solvation is complexed. So we don't know what [Ag+] is ( we know [Cl-] = [Ag(NH3)2+], as you correctly stated)

But from the The fact that from dissolving the complex Ks = [Ag(NH3)2+] / ( [Ag+][NH3]2 ), we can also say:

Ks * [Ag+] * [NH3]2 [Ag(NH3)2+] --> [Ag+] = [Ag(NH3)2+] / (1,7 x 107 * [NH3]2)

Let's put that in the other Ks, [Ag+][Cl-]:

1,7 x 10-10 = [Ag(NH3)2+] * [Cl-] / (1,7 x 107 * [NH3]2)

Thus:

1,7 x 10-10 * 1,7 x 107 = [Ag(NH3)2+] * [Cl-] / [NH3]2) = 2,89 x 10-3

You're with me so far ? We only used the solubility product of the complex ion and modified it so that we could put in the solubility product of AgCl.

Now let's assume we dissolve x mol AgCl in 1 L 1 M NH3. Then, as already said 1,7 x 107 >> 1,7 x 10-10 and LeChatelier principle, [Ag+] = 0 and [Cl-] = [Ag(NH3)2+] = x.

Then we only need to see that x mol Ag+ reacts with 2x mol NH3. We started with 1 mol / L NH3, so at equilibrium we have (1-2x) M NH3.

Let's plug that into our derived equation:

2,89 x 10-3 = x2 / (1-2x)2

Taking the square root on both sides:

5,38 x 10-2 = x / (1-2x)

--> 5,38 x 10-2 - 1,08 x 10-1x = x

--> 5,38 x 10-2 = 1,11x

--> x = [Cl-] = [Ag(NH3)2+] = 0,049 M

I hope this makes sense  to you. I must say: this is very tough and rigorous maths for highschool chemistry. But i think you can just assume that when adding 2 equilibria, the equilibrium constant of the summed reaction is the product of the equilibrium constants of the separate reactions.

#### sdekivit

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##### Re: many equilibriums in one system
« Reply #10 on: August 16, 2006, 03:51:56 AM »
what have you tried to calculate the solubility of PbS in 1 M HCl ? In any case, the solubility should improve much

#### Borek

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##### Re: many equilibriums in one system
« Reply #11 on: August 16, 2006, 12:46:05 PM »
OK, finally I had some time to put my hands on the first question.

(1) Kso=[Ag+][Cl-]
(2) K=[Ag(NH3)2+]/([Ag+][NH3]2)
(3) 2[Ag(NH3)2+] + [NH3] = 1M
(4) [Ag(NH3)2+] + [Ag+] = [Cl-]

As you have already correctly spotted, 4th equation can be simplified as

(4') [Ag(NH3)2+] = [Cl-]

And that's all. From now on you can forget about chemistry, all you have to do is some math. We need just [Ag(NH3)2+] so lets try to eliminate everything else.

Solve (1) for [Ag+].
Solve (3) for [NH3].

Insert [Ag+] and [NH3] into (2) (note - I have moved Kso to the left side):

K*Kso = [Ag(NH3)2+][Cl-]/(1M - 2[Ag(NH3)2+])2

Finally, use information from (4'):

K*Kso = [Ag(NH3)2+]2/(1M - 2[Ag(NH3)2+])2

This is exactly the same equation sdekivit got to (take square root of both sides).

Which approach is better? I don't know, they are just different. My approach is general, thus gives universal tool that should work always, not only in specific cases. But then sometimes it is easier to use some spefic method that can give faster answer - if you see such approach, do not hesitate to use it. It is result that is important, not the way it was calculated (as long as the way was right and result is correct).
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#### Borek

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##### Re: many equilibriums in one system
« Reply #12 on: August 16, 2006, 01:00:08 PM »
My assumptions is, if:
1.a material takes part in two reaction
2.it is produced in the first reaction and consumed in the second reaction
3.value of K of the second reaction is much larger than the K of the first reaction
then, the concentration of this material can be ignored in the mass-balance and charge-balance equations we get.

You are probably right, although such generalizations are difficult to prove and to use - you can't remember them all, especially when you need them I would rather think in terms of hydrogen sulfide being so weak an acid that it is completely protonated in pH = 0 solution.

Quote
Using this way to calculate the solubility of PbS in 1 M HCl, [HS-] and [S2-] can be neglected:

In mass balance. OK.

Quote
1 = [H+] + [H2S]
[Pb2+] + [H+] = 1

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#### sdekivit

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##### Re: many equilibriums in one system
« Reply #13 on: August 16, 2006, 01:29:49 PM »
the reaction occuring when PbS is dissolved in 1 M HCl-solution is:

PbS + 2H+ --> Pb2+ + H2

thus:

(1) PbS --> Pb2+ + S2-<--> K = 3,2 x 10-28

(2) S2- + H+ <--> HS- K = 9,1 x 1011 (note: 1/Ka of HS-)

(3) HS- + H+ <--> H2S K = 1,1 x 107 (= 1/Ka H2S)

Now you know all the equilibrium quotients and can substitute for [S2-], because summation of reaction 2 and 3 has aK equal to 9,1 x 1011 * 1,1 x 107 = 1,0 x 1019

#### zephyrblows

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##### Re: many equilibriums in one system
« Reply #14 on: August 16, 2006, 10:49:49 PM »
I think the universal approach is what I should learn, and sdekivit's method is a decent reference ,too.
I guess even my method is usable, this is still a rude method...so, I'll still choose Borek's method if I have enough time.

« Last Edit: August 17, 2006, 09:29:11 AM by zephyrblows »