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

1. Feb 20, 2014

### christian0710

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?

2. Feb 20, 2014

### christian0710

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. Feb 20, 2014

### Staff: Mentor

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. Feb 21, 2014

### christian0710

Yea that makes good sense - Thank's for clearing that up for me :)

5. Feb 22, 2014

### christian0710

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: Feb 22, 2014