Ksp...beats me! Help!!!

[yorkiepoo]

Which of the following best expresses the algebraic
relationship between molar solubility, X, of Ca3(PO4)2
and its solubility product constant Ksp?
A. Ksp = 108X^5
B. Ksp = 72X^5
C. Ksp = 32X^5
D. Ksp = 81X^5
E. None of the above


ans:

The expression asked for can be derived by setting up the equilibrium expression for the dissolution of calcium
phosphate, which will produce 3x moles of calcium ions and 2x moles of phosphate for each x moles of calcium phosphate
dissolved, and algebraically rearranging to solve for Ksp in terms of molar solubility:
Ca3(PO4)2 3 Ca^2+ + 2 PO4^3-
Ksp = [Ca2+]^3[PO43-]^2 = (3x)^3(2x)62 = (27x^3)(4x^2) = 108x^5.


But why isn't it just x^5????


I can't seem to find any info about Ksp besides that.

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[yorkiepoo]

basically what i mean is how do they go

from: [Ca2+]^3[PO43-]^2 TO: (3x)^3(2x)^2 = (27x^3)(4x^2)

?

why is the exponant broughten down like that?

 

[thefifthbeatle]

basically what i mean is how do they go

from: [Ca2+]^3[PO43-]^2 TO: (3x)^3(2x)^2 = (27x^3)(4x^2)

?

why is the exponant broughten down like that?

yeah these give me trouble to..from my understanding:

when you see the subscript, such as Ca3, x is not only raised to the three, it is multiplied...it is like that for every number. AgNO3 would be (1x)^1 * (1x)^1 = x^2......therefore, for this problem, (3x)^3 * (2x)^2....I don't fully understand Ksp, I just memorize it.

 

[DMD to Be]

There is an easy way to do this. Just memorize.

If there are two ions that disassociate like LiCl, It will always be x^2

If there are three ions like Ca2Cl, it will always be 4x^3

If there are four ions, it will always be 27x^4

If there are five ions, it will always be 108x^5.

The most I have seen is five.

 

[lifeisgood]

Just remember for Ksp problems of this nature to take the subscript and multiply it by x then take the product of the two to the power of the subscript.

It's the same idea as memorizing the above post, but you only have to remember one idea then do a little math.

 

[yorkiepoo]

thanks.. yeah that's kinda what I thought.... cram time!