Kd vs Ksp

[DoctorRealtor]

I am getting these two terms confused.

Kd is for larger complexes/proteins/antibodies and the equation is written as

Kd=( [A] ) / AB

Is Ksp the same thing except for ionic compounds?

Thanks
DR

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[Thoroughbred_Med]

If I am not mistaken isn't Kd just the dissociation constant between an antibody and and an antigen? While Ksp the equilibrium constant in solution chemistry?

 

[aldol16]

Kd is a dissociation constant. So if A + B ---> AB, then the Kd would be [A]/[AB] at equilibrium. This is just the inverse of the association constant. These are just plain equilibrium constants.

Ksp is different. Ksp is the solubility product. The key difference is that first, it refers only to a saturated solution. Now, the second difference is that say you're dissolving a solid AB into solution. The equilibrium constant would be [A]/[AB] (note the similarity in form here to the dissociation constant). Now, what's [AB]? Well, it's not the concentration of AB in solution because we assume AB dissolves in solution. It's also not the amount of solid AB because solids don't affect solution equilibria much. So what it is is actually the concentration of AB in solid AB. Weird, I know. But all you should know is that that's a constant. So you multiple both sides of the equilibrium expression by [AB] and you get [A] = K x [AB] and we call the right side of this expression the solubility product constant. So it's a very specific application of equilibrium constants.

 

[FutureD0C17]

Kd is a dissociation constant. So if A + B ---> AB, then the Kd would be [A]/[AB] at equilibrium. This is just the inverse of the association constant. These are just plain equilibrium constants.

Good explanation and prob just a typo, but shouldn't Kd be [A] /[AB]?

Weird! The forum won't even allow A * B in the numerator. It automatically defaults to [A]/[AB]

 

[aldol16]

Good explanation and prob just a typo, but shouldn't Kd be [A] /[AB]?

Weird! The forum won't even allow A * B in the numerator. It automatically defaults to [A]/[AB]

Lol I did type that but it did that weird thing on me.

 

[theonlytycrane]

[A]/[AB]

ah! I had to try it. I was also wondering about that term for Kd

 

[betterfuture]

I have a similar question. I became a little confused on the concept of Keq. So far my understanding of Keq is as follows:

∙If Keq>1 then ∆G<0 (spontaneous)

∙If Keq<1 then ∆G>0 (nonspontaneous)

So my questions are:
1) When Keq>1, this describes the reaction as having MORE products than reactants, correct, since Keq is products/reactants, (larger numerator)?
But what I don't understand is when they define a reaction as "spontaneous", does it mean that a reaction is regarded as "spontaneous" when a reaction goes proceeds from reactant ---> products and NOT FROM products---->reactants?

2)And wouldn't a spontaneous reaction favor FORMING PRODUCTS, so why do they term a reaction WHICH HAS ALREADY MADE MORE PRODUCTS THAN REACTANTS as spontaneous when it has already completed forming reactants --->products?

I am really trying to put 2 and 2 together but I don't seem to quite understand this concept at the level which I want to. It would be much appreciated if anyone would take the time to explain this confusing topic. Thank you.

 

[theonlytycrane]

For a reaction A -> B, A is the reactant and B is the product. At equilibrium, K = B (conc) / A (conc). If K > 1, then the reaction is favored in the forward direction (products). If K < 1, the reaction is favored in the reverse direction (reactants).

Say K > 1, then the reaction is favored in the forward direction which must mean that deltaG in the forward direction < 0 (spontaneous).

You can also look at deltaG = -RTln(K) at equilibrium. As K gets bigger, deltaG gets more negative (more spontaneous).

 

[wheatthinners]

.

 

[aldol16]

Edit: Wow, ok. The formatting made me look stupid. I'm assuming the equilibrium constant is A times B over AB, correct? But the other questions I'm still confused about.

Yes, something is clearly wrong with the formatting.

Another question, why do you include solid AB in the expression? I learned that you should ignore the solids when writing these. Is this a specific case or a different case?

This will be helpful to you: http://www.chemteam.info/Equilibrium/Writing-Ksp-expression.html. The reason you are taught to "ignore" solids and liquids is because their activities are arbitrarily defined to be unity.