Why do we neglect the effect of conjugate base when measuring the pH of weak acid?

  • Thread starter gladius999
  • Start date
  • #1
60
0

Main Question or Discussion Point

for example, ethanoic acid in water,

"="means equilibrium arrows

CH3COOH(aq) + H2O(l) = CH3COO-(aq) + H3O(aq)

we take the pH as the -log of the concentration of H3O ions.

BUT why do we not take in account the effect of the ethanoate ion with water which produces OH-???

CH3COO-(aq) + H2O(l) = CH3COOH(aq) + OH-(aq)

surely this should impact the pH of the solution right? If it does, how would I calculate the
concentration of each of the species if the both affect each other in equilibrium? Do I multiply the equilibrium constants together or something?

Thanks
 

Answers and Replies

  • #2
274
0


1) CH3COOH(aq) + H2O(l) = CH3COO-(aq) + H3O+(aq)

2) CH3COO-(aq) + H2O(l) = CH3COOH(aq) + OH-(aq)
3) H3O+(aq) + OH-(aq) = 2H2O(l)
-------------------------------------------------
net of 2 and 3: CH3COO-(aq) + H3O+(aq) = CH3COOH(aq) + H2O(l)
which is the reverse of 1.

So your second equation is already embodied in equation 1.
 
  • #3
60
0


i see you have gotten the resultant equation 1 from adding 2 and 3 but don't you need to include the equation 1 aswell? I know you have gotten equation 1 as a resultant but don't you need to include it aswell or else we are assuming it doesn't happen?

btw is equation 2 the autoprotolysis of water? or is the H3O from the ethanoic acid?

Thanks for your reply
 
  • #4
Borek
Mentor
28,326
2,713


i see you have gotten the resultant equation 1 from adding 2 and 3 but don't you need to include the equation 1 aswell? I know you have gotten equation 1 as a resultant but don't you need to include it aswell or else we are assuming it doesn't happen?
There are three equations, but only two of them matter - and in fact which two you select doesn't matter. That is, there are only two independent equations and equilbrium constants. If you will calculate equilibrium using two of these euqations, you will find that calculated concentrations fit third equilibrium.

Imagine question:

George has $5, John has $3 more, Jimmy has $2 more than John. How many dollars Jimmy has?

Obviously $10. Now I modify the question to

George has $5, John has $3 more, Jimmy has $2 more than John and Jimy has $5 more that George. How many dollars Jimmy has?

Have I changed anything? No. I have added information that was already there, just not given directly. That's the same problem as with your solution, just in the case of of acetate solution it is less obvious.

btw is equation 2 the autoprotolysis of water? or is the H3O from the ethanoic acid?
It doesn't matter where the H3O+ is from - it is indistingushable and only total concentration of H3O+ is used in equilibrium calculation.

And the equation you refer to is called water ionization or dissociation, or autodissociation.

Note: you are probably referring to pH calculation, not measurement.

--
methods
 
  • #5
60
0


There are three equations, but only two of them matter - and in fact which two you select doesn't matter. That is, there are only two independent equations and equilbrium constants. If you will calculate equilibrium using two of these euqations, you will find that calculated concentrations fit third equilibrium.

Imagine question:

George has $5, John has $3 more, Jimmy has $2 more than John. How many dollars Jimmy has?

Obviously $10. Now I modify the question to

George has $5, John has $3 more, Jimmy has $2 more than John and Jimy has $5 more that George. How many dollars Jimmy has?

Have I changed anything? No. I have added information that was already there, just not given directly. That's the same problem as with your solution, just in the case of of acetate solution it is less obvious.



It doesn't matter where the H3O+ is from - it is indistingushable and only total concentration of H3O+ is used in equilibrium calculation.

And the equation you refer to is called water ionization or dissociation, or autodissociation.

Note: you are probably referring to pH calculation, not measurement.

--
chemical calculators - buffer calculator, concentration calculator
www.titrations.info - all about titration methods
I apologize for my level of understanding. I am currently taking a Chemistry course in the first year of University. The textbook just gives the simple dissociation of the acid and the other reactions resulting from the dissociation confuses me.

What I still don't understand is that when ethanoic acid dissociates into ethanoate ions and H+ ions, the equilibrium reaction of the ethanoate ions and water begins generating ethanoic acid and OH- ions. There would be a decrease in ethanoate ions and therefore shouldn't it affect the dissociation of ethanoic acid? On top of this, wouldn't the H+ ions from acid react with the OH- ions from the equilibrium reaction between ethanoate ion and water which further disrupts the equilibrium concentration of the other 2 reactions?

Maybe it would be easier to explain in an equilibrium involving 2 reactions.
Say haber process,
N2+3H2=2NH3
say there was some water present in the mixture which further reacts with the NH3
NH3+H2O=NH4+ + OH-
As this reaction proceeds, the NH3 would decrease, thus disturbing the concentration of
NH3 in the first reaction. Therefore shoudn't there be a change in equilibrium concentration of NH3?

Thank you very much for your time
 
  • #6
Borek
Mentor
28,326
2,713


What I still don't understand is that when ethanoic acid dissociates into ethanoate ions and H+ ions, the equilibrium reaction of the ethanoate ions and water begins generating ethanoic acid and OH- ions. There would be a decrease in ethanoate ions and therefore shouldn't it affect the dissociation of ethanoic acid? On top of this, wouldn't the H+ ions from acid react with the OH- ions from the equilibrium reaction between ethanoate ion and water which further disrupts the equilibrium concentration of the other 2 reactions?
It all happens, but as I already explained, there are only two independent equilibria present. Each of the reactions you have listed has its own equilibrium constant, but knowing any two of them you can calculate the rest. Taking them into account you don't add any new information to the system. See again dollars example.

N2+3H2=2NH3
say there was some water present in the mixture which further reacts with the NH3
NH3+H2O=NH4+ + OH-
As this reaction proceeds, the NH3 would decrease, thus disturbing the concentration of
NH3 in the first reaction. Therefore shoudn't there be a change in equilibrium concentration of NH3?
Initial reaction was described by one equilibrium. You have added second reaction, there are two equilbria present. So far they are independent. This is not a good example, as it doesn't create additional dependent reactions.

--
 
  • #7
834
1


Therefore shoudn't there be a change in equilibrium concentration of NH3?
Let's say we have some equilibrium reaction A + B → C with an equilibrium constant K1. Given initial concentrations you can now evaluate the equilibrium concentration of A, B or C. In this case, [tex][A]=\frac{[C]}{K_1}[/tex] where all concentrations are at equilibrium.

Now we add another equilibrium reaction, say A + D → 2C with an equilibrium constant K2. Now we have two equilibrium reactions, and these have a mutual component A. To see how these work together, we just add them together: 2A + B + D → 3C. Convince yourself that the equilibrium constant for this reaction is K3=K1K2. If we now were to calculate the equilibrium concentration of A, we would find [tex][A]=\sqrt{\frac{[C]^3}{[D]K_1K_2}}[/tex] which is generally different from the one we found above.

The only conclusion is that adding the second equilibrium reaction to the mix changes the equilibrium concentration of A.
 
  • #8
60
0


It all happens, but as I already explained, there are only two independent equilibria present. Each of the reactions you have listed has its own equilibrium constant, but knowing any two of them you can calculate the rest. Taking them into account you don't add any new information to the system. See again dollars example.



Initial reaction was described by one equilibrium. You have added second reaction, there are two equilbria present. So far they are independent. This is not a good example, as it doesn't create additional dependent reactions.

--
buffer calculator, concentration calculator
pH calculator, stoichiometry calculator
Let's say we have some equilibrium reaction A + B → C with an equilibrium constant K1. Given initial concentrations you can now evaluate the equilibrium concentration of A, B or C. In this case, [tex][A]=\frac{[C]}{K_1}[/tex] where all concentrations are at equilibrium.

Now we add another equilibrium reaction, say A + D → 2C with an equilibrium constant K2. Now we have two equilibrium reactions, and these have a mutual component A. To see how these work together, we just add them together: 2A + B + D → 3C. Convince yourself that the equilibrium constant for this reaction is K3=K1K2. If we now were to calculate the equilibrium concentration of A, we would find [tex][A]=\sqrt{\frac{[C]^3}{[D]K_1K_2}}[/tex] which is generally different from the one we found above.

The only conclusion is that adding the second equilibrium reaction to the mix changes the equilibrium concentration of A.


I think I am almost understanding it now! The dissociation of ethanoic acid is special in the case that the third reaction generated when added to the equilibria cancelled itself and the second equation therefore 1 equation takes in account the other two.

In the case of the 2 reactions of NH3, there is no third reaction which cancels out another reaction, therefore we need to add the equations up and multiply there respective equilibria constants to form a resultant equilibrium reaction to calculate equilibria?

N2+3H2+NH3+H2O=2NH3+NH4++OH-

and the equilibria constant being K1xK2?

Thanks
 
  • #9
Borek
Mentor
28,326
2,713


Very convoluted what you wrote, but from what I was able to guesstimate, you are getting it.

--
methods
 

Related Threads for: Why do we neglect the effect of conjugate base when measuring the pH of weak acid?

  • Last Post
Replies
3
Views
16K
Replies
4
Views
2K
  • Last Post
Replies
2
Views
16K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
13
Views
7K
  • Poll
  • Last Post
Replies
7
Views
651
Replies
6
Views
2K
Top