Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: ritwik06 on June 12, 2009, 01:58:42 PM
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What are the conditions for two equilibria to be added?
Actually I got stuck with the example of H2S. I was asked the following question:
Calculate the H+ ion concentration of 0.1 M H2S solution. Ka1 and Ka2 are respectively 10 -7 and 1.2*10 -13.
Method 1:
I consider two separate equilibria:
H2S ::equil:: HS- + H+
HS- ::equil:: H+ + S-2
Since the Ka1 >> Ka2 , I assume all H+ to be coming from the first ionisation:
0.1*a^2=10 -7
from 1st equlibrium where a is the degree of ionisation. and 1-a is approximately =1, since a is small.
a=10 -3
H+=10 -4
from second equilibrium
[S-2]=Ka2*[HS-]/[H+]
[S-2]=1.2*10 -13
Method 2:
I add the two equilibria mentioned above:
H2S ::equil:: 2 H+ + S-2
0.1(1-b) 0.2*b 0.1
K=Ka1*Ka2
K={(0.2b)2*0.1*b}/{0.1(1-b)}
Since b is small 1-b ~ 1
K=0.04 b3
[H+]=1.34* 10 -7
[S-2]=6.69*10 -8
The values from the two methods mismatch. Why is the method in which I added the two equilibria didnt work?
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The first method is better. You can omit the second ionisation. But I don't understand your calculation.
0.1*a^2=10 -7
From where did you take that equation? Write the equation for Ka1, then make ICE tabel for process.
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What is the fallacy in method 2 then?
Method 1:
I consider two separate equilibria:
H2S ::equil:: HS- + H+
HS- ::equil:: H+ + S-2
Since the Ka1 >> Ka2 , I assume all H+ to be coming from the first ionisation:
0.1*a^2=10 -7
This is the equation I got from the first ionisation of H2S.
Ka1=[HS-][H+]/[H2S]
from 1st equlibrium where a is the degree of ionisation. and 1-a is approximately =1, since a is small.
a=10 -3
H+=10 -4
from second equilibrium
[S-2]=Ka2*[HS-]/[H+]
[S-2]=1.2*10 -13
This is the equation I got from the first ionisation of H2S.
Ka2=[S-2][H+]/[HS-]
Method 2:
I add the two equilibria mentioned above:
H2S ::equil:: 2 H+ + S-2
0.1(1-b) 0.2*b 0.1
K=Ka1*Ka2
K={(0.2b)2*0.1*b}/{0.1(1-b)}
Since b is small 1-b ~ 1
K=0.04 b3
[H+]=1.34* 10 -7
[S-2]=6.69*10 -8
I dont know about ICE tables though. But I cannot understand why method 2 does not work? Will you please specify the condition for adding 2 equilibria? Thanks a lot for the ~heLp~!
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I don't have the figures to hand, but what is the magnitude of the second disassociation constant?
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Magnitude of second dissociationconstant=1.2*10 -9
As per my calculations.
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Define b.
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Define b.
In method 2, I added the two equilibria to get a single equilibrium. I got K (new equilibrium constant) by multiplying the equilibrium constants of the individual equilibria. And then I assumed the dissociation constant for the overall reaction
H2S ::equil:: 2H+ + S-2
to be b. Then I framed the equation
K=[H+]2*[S-2]/[H2S] .(i)
If the dissociation constant be b, then [H+]=2*[H2S]0*b
and [S-2]=[H2S]0*b
and the final H2S concentration=[H2S]0(1-b)
I put these in equation (i) and calculated the value of b. Then I found out the concentration of [H+]=2*[H2S]0*b
and [S-2]=[H2S]0*b
by substituting b. And surprisingly I got different and wrong results as from method 1.
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I added the two equilibria to get a single equilibrium. I got K (new equilibrium constant) by multiplying the equilibrium constants of the individual equilibria.
This is called overall dissociation constants.
And then I assumed the dissociation constant for the overall reaction
H2S ::equil:: 2H+ + S-2
to be b.
This is an overall dissociation equation.
If so, b = K.
[H+]=2*[H2S]0*b
that makes
b = [H+]/(2*[H2S]0)
This is not an overall dissociation constant.
I still have no idea what b is.
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I added the two equilibria to get a single equilibrium. I got K (new equilibrium constant) by multiplying the equilibrium constants of the individual equilibria.
This is called overall dissociation constants.
And then I assumed the dissociation constant for the overall reaction
H2S ::equil:: 2H+ + S-2
to be b.
This is an overall dissociation equation.
If so, b = K.
[H+]=2*[H2S]0*b
that makes
b = [H+]/(2*[H2S]0)
This is not an overall dissociation constant.
I still have no idea what b is.
I am so sorry, I just got confused with the names.
Let me define b. 'b' is the number of moles that undergo the reaction mentioned out of every 1 mole of the reactant present(which in my case is H2S). I think, now I have made it clear. right?
I just want to know why the method 2 does not work. Is there any condition to be applied before adding equlibria?
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The method 2 assumes that [H+]=2[S2-] It isn't true becasue in the first step of dissociation come into being the protons.
H2S + H2O ::equil:: H3O+ + HS-
Based on Le Chatelier-Braun principle we know that the second step of dissociation must be withdrawn (the protons from the first step move reaction to left). So [S2-]<[HS-]. The concentration of protons is the sum: [H+]=[HS-]+[S2-]. As soon as [S2-]<[HS-] is the truth, it won't be possible [H+]=2[S2-]
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The concentration of protons is the sum: [H+]=[HS-]+[S2-].
Close, but not exact:
[H+]=[HS-]+2[S2-]
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[H+]=[HS-]+2[S2-]
Yes, of course. Stupid mistake. Now it is even more visible that the second method is incorrect.