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Offline MightyMan

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Help understanding Ksp
« on: December 01, 2007, 03:48:43 PM »
Hi guys, my first post so i hope i dont mess up too much  :-\

I recently did a lab involving mixing two solutions and getting a precipitate, yada, yada

After finishing the experiment i realized as you increase the concentration of the two solutions, the Ksp value increases.  At first I said "okay", but then i think i began to confuse myself...
If Ksp is the threshold level (minimum concentration is need to make a precipitate), wouldn't the Ksp DECREASE as you increase the concentrations of the reactants?  Since there is more reactants to react and from a precipitate, wouldn't the "threshold level" be lower?... \

but my observations show that as you DECREASE the concentrations, the Ksp DECREASES as well...

argg, my head hurts...

Offline Borek

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Re: Help understanding Ksp
« Reply #1 on: December 01, 2007, 04:18:50 PM »
Ksp - like every equilibrium constant - doesn't decrease nor increase. Reaction quotient does.
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Offline vcalder

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Re: Help understanding Ksp
« Reply #2 on: December 01, 2007, 04:20:45 PM »
The "solubility product constant" is a CONSTANT at a given temperature assuming that the solute being dissolved forms an "ideal" solution. It derives from thermodynamics of the solution process: MmXx(solid) <----> m M(+x) + x X(-m) where the +x and -m when the product "m*(+x)" and "x*(-m)" are equal. That is the solution remains electrically neutral. Without going through the derivation, which you can find in many General Chemistry texts, and all Physical Chemistry texts is" Ksp = [M]^m * [X]^x where the terms in square brackets are molar concentations of M and X and m and x are the coefficients of the balanced chemical equation:
MmXx(solid) <----> m M(+x) + x X(-m)
If you change the concentration of [M] and/or [X] by independently changing the concentration of one by adding a salt that contributes one of the ions but not the other, the concentration of the other counter ion decreases according to the equation:  
Ksp = [M]^m * [X]^x. The value of Ksp remains unchanged.

Offline MightyMan

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Re: Help understanding Ksp
« Reply #3 on: December 01, 2007, 04:29:09 PM »
RIGHT
its the EXPERIMENTAL Ksp (Q) (same as reaction quotient?..) that is changing...

so, correct me if I'm wrong, but the only reason the Ksp is "changing" is because I'm increasing the concentrations of BOTH reactants, thus "increasing" the Q.

Offline Padfoot

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Re: Help understanding Ksp
« Reply #4 on: December 01, 2007, 10:39:02 PM »
RIGHT
its the EXPERIMENTAL Ksp (Q) (same as reaction quotient?..) that is changing...

so, correct me if I'm wrong, but the only reason the Ksp is "changing" is because I'm increasing the concentrations of BOTH reactants, thus "increasing" the Q.
What do you mean by EXPERIMENTAL Ksp?  Ksp is constant (at given temp) as mentioned previously.  The ionic product (you should probably call it this instead of reaction quotient), at equilibruim is equal to Ksp.

Offline Borek

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Re: Help understanding Ksp
« Reply #5 on: December 02, 2007, 04:30:13 AM »
The ionic product (you should probably call it this instead of reaction quotient)

You can, although you should remember that it is just an approximation that assumes activity of the solid phase equals 1.

AgCl(s) = Ag+ + Cl-

K = aAg+ * aCl- / aAgCl(s)

When talking about Kso we usually omit denominator, as it is close to 1 - but the general formula is still quotient. Additionally we so often use just concentrations instead of activities that many people don't remember about the difference and possible implications.
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Offline MightyMan

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Re: Help understanding Ksp
« Reply #6 on: December 02, 2007, 01:00:44 PM »
Yeah, i understand it now

by experimental Ksp, i was referring to the reaction quotient (Q)

my teacher calls in experimental Ksp but my textbook calls it the reaction quotient so i kinda use both..

thanks for all the help

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