[ritwik06]

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  HS^- + H⁺

HS^-  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  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?

[plankk]

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.

[ritwik06]

What is the fallacy in method 2 then?


Method 1:
I consider two separate equilibria:
H₂S  HS^- + H⁺

HS^-  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  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~!

[sjb]

I don't have the figures to hand, but what is the magnitude of the second disassociation constant?

[ritwik06]

Magnitude of second dissociationconstant=1.2*10 ^-9
As per my calculations.

[Borek]

Define b.

[ritwik06]

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   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.

[Borek]

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   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.

[ritwik06]

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   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?

[plankk]

The method 2 assumes that [H⁺]=2[S^2-] It isn't true becasue in the first step of dissociation come into being the protons.
H₂S + H₂O 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 [S^2-]<[HS^-]. The concentration of protons is the sum: [H⁺]=[HS^-]+[S^2-]. As soon as [S2-]<[HS-] is the truth, it won't be possible [H⁺]=2[S^2-]

[Borek]

The concentration of protons is the sum: [H⁺]=[HS^-]+[S^2-].

Close, but not exact:

[H⁺]=[HS^-]+2[S^2-]

[plankk]

[H⁺]=[HS^-]+2[S^2-]

Yes, of course. Stupid mistake. Now it is even more visible that the second method is incorrect.