# Using Dissociation Constant Conceptually?

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I’m completely stumped on how to use the Kd to gauge relative concentration of bound vs free protein. I was told something along these lines (I used "conc." instead of brackets because they weren't showing up properly):

Kd = ( conc. A x conc. B  ) / conc. AB

"If concentration A is fixed and relatively low, and concentration of B  is lower than Kd, then there will be relatively low AB and mostly free A."

How does this make sense conceptually? I've tried to model it out, but I can't seem to get anywhere. Can anyone please help?

Edited by linda43
re-wording
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I'm rushing here as I won't be back tonight, but here is a start.

Let the initial concentration of AB be q and the concentration of A be p then

Equilibrium equation is

$AB \leftrightarrow A + B$

Dissociation equation is

${K_d} = \frac{{\left[ A \right] + \left[ B \right]}}{{\left[ {AB} \right]}}$

Also

$\left[ A \right] = \left[ B \right] = p$

and

${\left[ {AB} \right]_{initial}} = q$

So

${K_d} = \frac{{2p}}{{\left( {q - p} \right)}}$

Now I think this is right but I'll check again later unless Babcock or CharonY call in with a better idea first.

Edited by studiot
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2 hours ago, studiot said:

I'm rushing here as I won't be back tonight, but here is a start.

Let the initial concentration of AB be q and the concentration of A be p then

Equilibrium equation is

ABA+B

Dissociation equation is

Kd=[A]+[B][AB]

Also

[A]=[B]=p

and

[AB]initial=q

So

Kd=2p(qp)

Now I think this is right but I'll check again later unless Babcock or CharonY call in with a better idea first.

Erm the numerator should be the product of the concentrations of the dissociated ions: [ A ]x[ B ]/[AB] = Kd

And I think the questioner is asking about the situation in which [A]=/= [ B ] , e.g. an acetic acid/acetate buffer, or something like that.

I suppose what the statement in question is driving at is that is if [ B ] << Kd, then the ratio [A]/[AB] will have to be >> Kd, so that when multiplied by [ B ] it gives Kd.

But since, for weak acids and bases, K <<1 , it is not necessary for [A] >> [AB] to achieve that.

.......I think......

For example, with acetic acid, Kd = 1.8 x 10⁻⁵ = [H+] x [acetate]/[acetic acid].

So if, for the sake of argument,  [H+] is also 1.8 x 10⁻⁵, then the [acetate]/acetic acid]  ratio would be 1/1.

So I too am a bit baffled by the statement our questioner is querying.

Edited by exchemist
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5 hours ago, exchemist said:

Erm the numerator should be the product of the concentrations of the dissociated ions: [ A ]x[ B ]/[AB] = Kd

Thanks +1

Told you I was in a rush, don't know now where that + idea came from.

So the top is p2, not 2p.

Moral never rush.

😳

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On 9/11/2021 at 11:35 PM, studiot said:

Thanks +1

Told you I was in a rush, don't know now where that + idea came from.

So the top is p2, not 2p.

Moral never rush.

😳

It's not p² if A and B are not present at the same concentration, which they are not in a buffer solution, for instance. As I say, I suspect the questioner's problem related to something like that. But since they have not been back, we can't know for sure.

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2 hours ago, exchemist said:

It's not p² if A and B are not present at the same concentration, which they are not in a buffer solution, for instance. As I say, I suspect the questioner's problem related to something like that. But since they have not been back, we can't know for sure.

Yes Linda is not recorded as having returned since before I posted my original erroneous reply.

And yes you are correct in assessing a difficulty with the original statement in the OP about the concentrations of A and B.

That is why I started with the chemical equation AB = A + B

This is what is implied in the OP

It is also implied that the equation could not be for instance A2B = 2A + B.

So unless there is an unknown source of A and/or B, or some further reaction involving A or B,  the concentrations of A and B must be equal as the only known source is AB.

Of course the OP is at variance with this so has not been properly formulated.

So it may well be as you suspect since this is a biochemical reaction.

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