Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: zephyrblows 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 (K_{1}*K_{2}*...)?
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?
Thanks for your *delete me*!

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 (http://www.chembuddy.com/?left=pHcalculation&right=generalpHcalculation) lecture and read first paragraph :)

Sometimes you can combine them. Assume the following:
H_{3}PO_{4} + H_{2}O <> H_{2}P_{O4}^{} + H_{3}O^{+ } > K_{1}
H_{2}PO_{4}^{} + H_{2}O <> HPO_{4}^{2} + H_{3}O^{+} > K_{2}
Express the reaction quotients:
K_{1} = [H_{3}O^{+}][H_{2}PO_{4}^{}]/[H_{3}PO_{4}]
K_{2} = [H_{3}O^{+}][HPO_{4}^{2}]/[H_{2}PO_{4}^{}]
Now you say from the expression of K_{1} that [H_{2}PO_{4}^{}] = K_{1} * [H_{3}PO_{4}]/[H_{3}O^{+}] and substitue that in the other equilibrium:
K_{2} = [H_{3}O^{+}][HPO_{4}^{2}]/( K_{1} * [H_{3}PO_{4}] / [H_{3}O^{+}] )
Which yields:
K_{1} * K_{2} = ( [H_{3}O^{+}]^{2}[HPO_{4}^{2}] ) / [H_{3}PO_{4}]

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.
K_{1} * K_{2} = ( [H_{3}O^{+}]^{2}[HPO_{4}^{2}] ) / [H_{3}PO_{4}]
I wonder how you are going to use this equation to calculate H_{2}PO_{4}^{} 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 K_{1}*K_{2} together with equation for K_{1}, 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.

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 NH_{3} solution is?
All I can find are:
1.7*10^{10} = [Ag^{+}] [Cl^{}]
1.7*10^{7} = [Ag(NH_{3})_{2}^{+}] / [Ag^{+}] [NH_{3}]^{2}
[Cl^{}] = [Ag(NH_{3})_{2}^{+}] + [Ag^{+}]
[NH_{3}]_{0} = [NH_{3}] + 2[Ag(NH_{3})_{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!

Well, I've tried to solve it by myself...
I assumed [Cl^{}] = [Ag(NH_{3})_{2}^{+}]
because Ag^{+} is almost used up...
So I got the solubility = [Cl^{}] = 0.049M

I assumed [Cl^{}] = [Ag(NH_{3})_{2}^{+}]
Correct idea.
So I got the solubility = [Cl^{}] = 0.049M
Show your calculations.

All I can find are:
1.7*10^{10} = [Ag^{+}] [Cl^{}]
1.7*10^{7} = [Ag(NH_{3})_{2}^{+}] / [Ag^{+}] [NH_{3}]^{2}
[Cl^{}] = [Ag(NH_{3})_{2}^{+}] + [Ag^{+}]
[NH_{3}]_{0} = [NH_{3}] + 2[Ag(NH_{3})_{2}^{+}]
Are you sure about the Kvalues? 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.

Sorry, I typed the wrong K_{sp} 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*10^{7} [NH_{3}]^{2} = [Cl^{}]^{2} / 1.7*10^{10}
[NH_{3}] = 18.6 [Cl^{}]
Substitution:
1 = 2[Cl^{}] + 18.6 [Cl^{}] = 20.6 [Cl^{}]
[Cl^{}] = 0.049M

This is what school teachers teach us (without explanation):
The net equation is, Ag^{+} + 2NH_{3} <> Ag(NH_{3})_{2}^{+} + Cl^{} K = K_{sp}K_{f} = 2.89*10^{3}
According to the net equation, make x = [Cl^{}] = [Ag(NH_{3})_{2}^{+}]
Solve: 2.89*10^{3} = x^{2} / (12x)^{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^{} : K^{s} = 1,7 x 10^{10}
Ag^{+} + 2 NH_{3} > Ag(NH3)_{2}^{+} : Ks = 1,7 x 10^{7}
So the total reaction occuring here is:
AgCl + 2 NH_{3} <> Ag(NH_{3})_{2}^{+} + Cl^{} : K_{s } = ?
Now we must find this Ksvalue for the toal reaction and to get this one, we need to combine these equilibria by using substitution:
K_{s} = [Ag^{+}][Cl^{}] = 1,7 x 10^{10}
K_{s} = [Ag(NH_{3})_{2}^{+}] / ( [Ag^{+}][NH_{3}]^{2} ) = 1,7 x 107
The combined K_{s}:
K_{s} = [Ag(NH_{3})_{2}^{+}][Cl^{}] / ([Ag^{+}][NH_{3}]^{2} )
Because 1,7 x 10^{7} >> 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(NH_{3})_{2}^{+}], as you correctly stated)
But from the The fact that from dissolving the complex K_{s} = [Ag(NH_{3})_{2}^{+}] / ( [Ag^{+}][NH_{3}]^{2} ), we can also say:
Ks * [Ag^{+}] * [NH_{3}]^{2} [Ag(NH_{3})_{2}^{+}] > [Ag^{+}] = [Ag(NH_{3})_{2}^{+}] / (1,7 x 10^{7} * [NH_{3}]^{2})
Let's put that in the other Ks, [Ag^{+}][Cl^{}]:
1,7 x 1010 = [Ag(NH_{3})_{2}^{+}] * [Cl^{}] / (1,7 x 10^{7} * [NH_{3}]^{2})
Thus:
1,7 x 10^{10 } * 1,7 x 10^{7} = [Ag(NH_{3})_{2}^{+}] * [Cl^{}] / [NH_{3}]^{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 10^{7} >> 1,7 x 10^{10} and LeChatelier principle, [Ag^{+}] = 0 and [Cl] = [Ag(NH_{3})_{2}^{+}] = x.
Then we only need to see that x mol Ag^{+} reacts with 2x mol NH_{3}. We started with 1 mol / L NH_{3}, so at equilibrium we have (12x) M NH_{3}.
Let's plug that into our derived equation:
2,89 x 10^{3} = x_{2} / (12x)^{2}
Taking the square root on both sides:
5,38 x 10^{2} = x / (12x)
> 5,38 x 10^{2}  1,08 x 10^{1}x = x
> 5,38 x 10^{2} = 1,11x
> x = [Cl^{}] = [Ag(NH_{3})_{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.

what have you tried to calculate the solubility of PbS in 1 M HCl ? In any case, the solubility should improve much :)

OK, finally I had some time to put my hands on the first question.
You start with set of four equations, that describe solution:
(1) K_{so}=[Ag^{+}][Cl^{}]
(2) K=[Ag(NH_{3})_{2}^{+}]/([Ag^{+}][NH_{3}]^{2})
(3) 2[Ag(NH_{3})_{2}^{+}] + [NH_{3}] = 1M
(4) [Ag(NH_{3})_{2}^{+}] + [Ag^{+}] = [Cl^{}]
As you have already correctly spotted, 4^{th} equation can be simplified as
(4') [Ag(NH_{3})_{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(NH_{3})_{2}^{+}] so lets try to eliminate everything else.
Solve (1) for [Ag^{+}].
Solve (3) for [NH_{3}].
Insert [Ag^{+}] and [NH_{3}] into (2) (note  I have moved K_{so} to the left side):
K*K_{so} = [Ag(NH_{3})_{2}^{+}][Cl^{}]/(1M  2[Ag(NH_{3})_{2}^{+}])^{2}
Finally, use information from (4'):
K*K_{so} = [Ag(NH_{3})_{2}^{+}]^{2}/(1M  2[Ag(NH_{3})_{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).

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 massbalance and chargebalance 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.
Using this way to calculate the solubility of PbS in 1 M HCl, [HS^{}] and [S^{2}] can be neglected:
In mass balance. OK.
1 = [H^{+}] + [H_{2}S]
[Pb^{2+}] + [H^{+}] = 1
You have already lost me.

the reaction occuring when PbS is dissolved in 1 M HClsolution is:
PbS + 2H^{+} > Pb^{2+} + H_{2}S
thus:
(1) PbS > Pb^{2+} + S^{2}<> K = 3,2 x 10^{28}
(2) S^{2} + H^{+ } <> HS^{} K = 9,1 x 10^{11} (note: 1/K_{a} of HS^{})
(3) HS^{ } + H^{+} <> H_{2}S K = 1,1 x 10^{7} (= 1/K_{a} H2S)
Now you know all the equilibrium quotients and can substitute for [S^{2}], because summation of reaction 2 and 3 has aK equal to 9,1 x 10^{11} * 1,1 x 10^{7} = 1,0 x 10^{19}

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.

charge balance:
[Cl^{}] + [S^{2}] + [HS^{}] = [H^{+}] + [Pb^{2+}]
I am not checking anything else  but think it over... Especially S^{2} and Pb^{2+} part.
Also note that in general you have omitted one ion (and one equilibrium). As this ion is in minute concentrations at pH 1 and is not taking part in any of the equilbria important for your calculations, that's OK. This ion (plus one other) has been omitted in the previous question for the similar reasons.

charge balance:
[Cl^{}] + [S^{2}] + [HS^{}] = [H^{+}] + [Pb^{2+}]
I am not checking anything else  but think it over... Especially S^{2} and Pb^{2+} part.
Ow...then I must have had some wrong conceptions...
If I put all the cations in one side of chargebalance equation and anions in the other, will something go wrong?

well the method of Borek has the same maths as the method i use, only i substitue other things to become the same equation. Both ways are correct as stated by borek and it's up to you which method you prefer. In any case, the solubility of PbS in 1 M HCl is indeed 8,9 x 10^{4} M. I'll write my calculation here so you can see how it can be calculated:
8 x 10^{28} = [Pb^{2+}] [S^{2}]
Then for the other equilibria where S^{2 } is involved:
10^{14} = [HS^{}] / ( [S^{2}][H^{+}] )
10^{7} = [H_{2}S] / ( [HS^{}] [H^{+}] )
Because in the first equation we want to express [S^{2}] in terms of [H_{2}S], we use the above 2 equations:
[S^{2}] = [HS^{}] / ( [H^{+}]*10^{14} )
And also:
[HS^{}] = [H_{2}S] / ( [H^{+}] * 10^{7} )
Now substitue [HS^{}] in the expression of [S^{2}] to become:
[S^{2}] = [H_{2}S] / ( [H^{+}]^{2} * 1 x 10^{21} )
This can be plugged in into the solubility product of PbS in water to become:
8 x 10^{28} = [Pb^{2+}] * [H_{2}S] / ( [H^{+}]^{2} * 1 x 10^{21} )
Now we have our equation and an equilibrium:
PbS + 2 H^{+} <> Pb^{2+} + H_{2}S
When dissolving x mol in 1 L 1 M HCl, we gain x mol/L Pb^{2+} and H_{2}S. The concentration H^{+} will be 1  2x M
Plug these into our equation we get exactly the same as with the other question:
8 x 10^{7} = x^{2} / (12x)^{2}
Solve for x to find the solubility of 8,9 x 10^{4} M

If I put all the cations in one side of chargebalance equation and anions in the other, will something go wrong?
That's OK, but you have to account for the fact that some ions have different charges. Think in terms of CaCl_{2} solution. Is charge balance
[Ca^{2+}] = [Cl^{}]
or
2[Ca^{2+}] = [Cl^{}] ?

Aw, right! I've totally missed that!!
But I don't understand what you said:
Also note that in general you have omitted one ion (and one equilibrium). As this ion is in minute concentrations at pH 1 and is not taking part in any of the equilbria important for your calculations, that's OK. This ion (plus one other) has been omitted in the previous question for the similar reasons.
You mean, I don't need to put [Cl] in the equations? Without [Cl^{}], how should I derive the charge equation?

You mean, I don't need to put [Cl] in the equations?
No. OH^{} and K_{w}.

Borek, sorry but I can't understand it intuitively :X...
If the charge of Cl^{} is neglected, this solution no more holds a neutral charge...
So, I shouldn't use "putting all the cations in one side of chargebalance equation and anions in the other"...
And I don't know what to do now...
All I can write down is:
PbS <> Pb^{2+} + S^{2} (8*10^{28})
H^{+} + S^{2} <> HS^{}
H^{+} + HS^{} <> H_{2}S
H2O <>H^{+} + OH^{}
mass: [OH^{}] + [H^{+}]_{0} = [H^{+}] + [HS^{}] + [H_{2}S]
charge:(???) [H^{+}] + 2[Pb^{2+}] = [OH^{}] + [HS^{}] + 2[H_{2}S]
I know there must be something wrong...Is there another method to build charge equation without considering the net charge of the system?

Wipe everything about Cl^{}, I have not said a word about it.
I was referring JUST to the fact, that every water solution contains both H^{+} and OH^{} and their concentrations are connected through K_{w}=[H^{+}][OH^{}]. No idea where did you get this Cl^{} thing  it must be some your misconception that I am not able to understand.
mass: [OH^{}] + [H^{+}]_{0} = [H^{+}] + [HS^{}] + [H_{2}S]
These should be separate mass balances for different substances. Like:
[Pb^{2+}] = [S^{2}] + [HS^{}] + [H_{2}S]
which reflects the fact that there is exactly the same amount of sulfide as lead in the solution.
charge:(???) [H^{+}] + 2[Pb^{2+}] = [OH^{}] + [HS^{}] + 2[H_{2}S]
In the charge balance you should use ONLY charged particles. Or was 2[H_{2}S] just a typo? And Cl^{} must be included:
[H^{+}] + 2[Pb^{2+}] = [OH^{}] + [HS^{}] + 2[S^{2}] + [Cl^{}]
Now, what you can safely omit here is OH^{}. Its concentration is much lower that that of the other things and it doesn't play any role in the whole system that we are interested in.

No wonder...finally I get what you mean.
Sorry, my English is not good enough, so I misunderstood some of your words.
It's not so difficult as I thought.
Thanks a lot !! ;D