The weak acid conjugate base - something i can't understand.

In summary: Basically, in the first reaction, the concentration of acid is so high that almost all of the conjugate base dissociates, leading to a very low concentration of HA. In the second reaction, the concentration of base is much lower, so more of the conjugate base dissociates, and the concentration of HA is higher.
  • #1
christian0710
409
9
I seem to have a problem understanding this: If 1% of a weak acid dissociates in pure water. I would assume that 99% of it’s conjugate base would dissociate to form HA in pure water, but this is not the case: I tried to set up a situation below: Please help me understand what I’m misunderstanding


Let’s assume we have a concentration of 1,0M acid with a Ka=10^-8 then we can discard waters Kw (due to the high concentration of acid=) and calculate the concentration of HA, A(-) and OH (-)

HA + H20 = A(-) + H(+)

Ka = x^2/(1,0-x)

So at equilibrium we get
x= 0,0001 =[A-] =[OH-]
[HA] = 1,0 – 0,0001 = 0,9998

Now in situation 2, we add 1.0M of the conjugate base of the above acid to pure water and we get.

The Kb= kw/Ka=10-6

A(-) + H2O = HA + OH(-)

Kb=x^2/(1,0-x)
x= 0.001= [HA]=[OH


My question is this: How come the concentration of HA at equilibrium in the first acid + water reaction is different than the concentration of HA in the second base + water reaction? If the acid strength is inversely proportional to the conjugate base strength, should the amount of acid dissociating from from situation 1 not equal the amount of HA formed from situation 2? If 1,3% of an acid dissociates into a conjugate base, should 98,7% of it’s conjugate base not dissociate in water? What is the reason this is not the case?
 
Chemistry news on Phys.org
  • #2
I can see now that Kb > Ka so more base would dissociate more than the acid but with such a small kb and ka the amount of conjugate acid formed by a 1M base base reaction with water let's say 2% could never equal the amount of conjugate base formed by the same concentration of of that acid (so 98% of the base could impossibly dissociate with that Ka) So i guess there exists no such relationship.
 
  • #3
As far as I can tell (I am not entirely sure I understand your posts), there is no such relationship.

Only for acid with Ka=10-7 Ka=Kb and your expect some symmetry.

However, if you compare ammonia and acetic acid, their Ka and Kb (respectively) are almost identical, so any calculations done for acetic acid (that is, pH found as a function of the acid concentration) is mirrored by the calculations for ammonia (that is, pOH found as a function of the base concentration).
 
  • #4
Yea that makes good sense - Thank's for clearing that up for me :)
 
  • #5
Hey Borek! I understand why this symmetry can't possibly exist :D

In the reaction R1
HCN + H20 = CN(-) + H(+) the Ka = 6.2*10^-10 this a very weak acid, and CN(-) quiet a strong base

But in the reverse reaction R2
CN(-) + H20 = HCN + OH Kb is 1.6*10^-5 This suggest that CN(-) in R2 is quite a weak base with respect to the first reaction: if we had the same amount of moles in both reactions, the amount of Base (CN-) formed from R1 is less than the amount of base which remains on base-form (CN(-)) in R2.

Why is this the case?
In R1 the Base CN(-) is competing againt H2O For H(+), and CN(-) has a stronger affnity for H(+) than H2O. So CN(-) pushes the equilibrium far to the left.

In R2: The base CN(-) is competing against OH(-) for H(+) and here OH(-) has a stronger affinity for H(+) than does CN(-), so CN(-) is a weak base relative to OH(-) but a strong base relative to H2O. So in R2 OH(-) pushes the quilibrium far to the left.

Is that correct?
 
Last edited:

1. What is a weak acid?

A weak acid is a substance that only partially dissociates in water, meaning it does not completely break apart into ions. This results in a lower concentration of hydrogen ions in the solution compared to a strong acid.

2. What is a conjugate base?

A conjugate base is the species that remains after a weak acid loses a hydrogen ion. It can also be thought of as the anion formed from the acid's dissociation. For example, the conjugate base of acetic acid (CH3COOH) is acetate (CH3COO-).

3. How are weak acids and conjugate bases related?

Weak acids and their conjugate bases are part of a conjugate acid-base pair. This means that they differ by only one hydrogen ion. As a weak acid loses a hydrogen ion, its conjugate base gains one. This equilibrium between the weak acid and its conjugate base is what defines its strength.

4. What is the significance of the equilibrium between a weak acid and its conjugate base?

The equilibrium between a weak acid and its conjugate base determines its acidity. A stronger acid will have a higher equilibrium constant, meaning it will dissociate more and have a higher concentration of hydrogen ions. A weaker acid will have a lower equilibrium constant and a lower concentration of hydrogen ions.

5. How does the strength of a weak acid and its conjugate base affect its pH?

The strength of a weak acid and its conjugate base ultimately determines the pH of a solution. A stronger acid will have a lower pH (more acidic) compared to a weaker acid with a higher pH (less acidic). Similarly, the conjugate base of a strong acid will have a higher pH compared to the conjugate base of a weak acid.

Similar threads

  • Chemistry
Replies
1
Views
2K
Replies
6
Views
2K
Replies
14
Views
2K
  • Chemistry
Replies
3
Views
2K
Replies
2
Views
1K
Replies
6
Views
6K
Replies
5
Views
2K
  • Chemistry
Replies
6
Views
2K
Replies
4
Views
6K
Back
Top