Common Ion Effect

[scotties123]

im confused...and i blame EK.
They say (pg 76 if u have it) that common ions do NOT effect Ksp, but then they also say that adding a common ion disturbs the equilibrium.....help? thanks.

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[deleted21595]

Ksp is a 'static' (not really) measure of the ratio of products to reactants at equilbrium (think Keq). The only difference is that since there are only solid reactants, and you don't look at immisceable solids in your equilibria ratios, the only components you care about are the products (numerator of Keq). So Ksp is what you're SUPPOSED to have at equilbrium in terms of ratio (think ratio of 1 man to 1 woman in a happy marriage).

That's where the name Ksp comes from. This differs from what you actually have going on. In any particular reaction mixture, you can have too many of the product ions. This means that your Ion Product (IP) are combined greater than what they should be (Ksp). Think of this as having the extra girlfriend on the side. you're only supposed to have 1 man and 1 woman, for a total product (Ksp) of 1x1=1. Instead you have 1 man and 2 women for a total Ion product (IP) of 2. In this case you've got too many females, and one will need to revert back to reactant (single) status.

As much as I'd hate to say this (kidding) EK is right.

 

[scotties123]

Ksp is a 'static' (not really) measure of the ratio of products to reactants at equilbrium (think Keq). The only difference is that since there are only solid reactants, and you don't look at immisceable solids in your equilibria ratios, the only components you care about are the products (numerator of Keq). So Ksp is what you're SUPPOSED to have at equilbrium in terms of ratio (think ratio of 1 man to 1 woman in a happy marriage).

That's where the name Ksp comes from. This differs from what you actually have going on. In any particular reaction mixture, you can have too many of the product ions. This means that your Ion Product (IP) are combined greater than what they should be (Ksp). Think of this as having the extra girlfriend on the side. you're only supposed to have 1 man and 1 woman, for a total product (Ksp) of 1x1=1. Instead you have 1 man and 2 women for a total Ion product (IP) of 2. In this case you've got too many females, and one will need to revert back to reactant (single) status.

As much as I'd hate to say this (kidding) EK is right.

thank u for helping, and u have an excellent analogy, but im still not 100% on this. so it DOES effect ksp? or does Ksp remain constant and it just prevents further dissolution?

 

[midn]

Ksp is an inherent property and is a function of temperature and probably pressure too. It does NOT change with the addition of a common ion or really any substance for that matter.

Ksp is just an equilibrium constant. Think of it this way, when you're at equilibrium and you add a certain substance, does that change the K value? No, the K value defines what the solution would "like" to be at at a given temperature and pressure. Therefore, the reaction will reverse or go forward to re-establish the concentrations divided by the K value.

This is basically the same thing the last guy/gal said, but I hope this will clear things up.

 

[vicinihil]

Like he said....Ksp Ka Kb Keq etc etc are only depending upon temperature. The Keq of H2O changes depending on temperature.

Also like what midn said, when you add stuff to a solution, the solution plus the new stuff will strive for equilibrium. In equilibrium, at 25 degrees C, the Keq value doesn't change no matter how much you have. Think about the equation.

Keq = [prod]/[reactant]...if you add reactant (demoninator increases) you automatically make more product due to le chatlier principle (numerator increases)....now if both the bottom and top increase, 2nd grade math tells us that the answer doesn't change

HOWEVER, when you change the quantity of either the product or reactant, Q does change because Q is [prod]/[reactant] NOT AT equilibrium. So immediately after you add reactant, the denominator increases, causing Q to decrease (due to increasing denominator and constant numerator) the reaction hasn't had time to equilibriate yet.

Any questions? Good. Btw I teach gen chem for TPR so I **** you not.

 

[deleted21595]

thank u for helping, and u have an excellent analogy, but im still not 100% on this. so it DOES effect ksp? or does Ksp remain constant and it just prevents further dissolution?

You're welcome, but since my analogy didn't help, lets try another. Imagine you have a slinky (spring) and you the equilibrium position. This middle resting equilibrium position is the Keq, no matter how much you pull or compress the spring, the resting position is still in the middle and that's where the spring will all eventually end up.

Now that's not necessarily where the spring is at any given point in time. That current position is your Ion Product (also called Q). If your spring is too compressed, then it'll try to uncompress back to the middle (keq). If its too stretched, it'll try to compress back to the middle (Keq). either way, all roads lead to Keq.

Ksp is the name for Keq for solutions.

 

[AnEyeLikeMars]

thank u for helping, and u have an excellent analogy, but im still not 100% on this. so it DOES effect ksp? or does Ksp remain constant and it just prevents further dissolution?

the bolded one

 

[bjb305]

common ion does NOT affect Ksp.. it affects the molar solubility (pretty much how much stuff you can have in solution before you get a solid)

 

[scotties123]

cool. thanks alot guys i get it now. little concepts seem to confuse me but i have no problem with a 20 step equation lol.